COMPENDIUM AIME

1983 – 2020

Gerard Romo Garrido

Toomates Coolección

Los documentos de Toomates son materiales digitales y gratuitos. Son digitales porque están pensados para ser consultados

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hecho. Lo que no es aceptable, por inmoral y mezquino, es el modelo de las llamadas "licencias digitales" con las que las editoriales

pretenden cobrar a los estudiantes, una y otra vez, por acceder a los mismos contenidos (unos contenidos que, además, son de una bajísima calidad). Este modelo de negocio es miserable, pues impide el compartir un mismo libro, incluso entre dos hermanos,

pretende convertir a los estudiantes en un mercado cautivo, exige a los estudiantes y a las escuelas costosísimas líneas de Internet,

pretende pervertir el conocimiento, que es algo social, público, convirtiéndolo en un producto de propiedad privada, accesible solo a aquellos que se lo puedan permitir, y solo de una manera encapsulada, fragmentada, impidiendo el derecho del alumno de poseer

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se ofrecen en dos versiones: En formato “pdf” para una cómoda lectura y en el formato “doc” de MSWord para permitir y facilitar su edición y generar versiones parcial o totalmente modificadas. Se agradecerá cualquier observación, comentario o colaboración a

toomates@gmail.com

La biblioteca Toomates Coolección consta de los siguientes libros:

Problem-solving:

Geometría Axiomática GA pdf 1 2 ... 23 portada

Problemas de Geometría PG pdf 1 2 3 4 5 6 7 8 9

Introducción a la Geometría PI pdf doc

Teoría de números AR pdf 1 2 3

Trigonometría PT pdf doc

Desigualdades DE pdf doc

Números complejos PZ pdf doc

Álgebra PA pdf doc

Combinatoria PC pdf doc

Probabilidad PR pdf doc

Guía del estudiante de Olimpiadas Matemáticas OM pdf

Libros de texto (en catalán):

Àlgebra AG pdf 1 2 3 4

Funcions FU pdf doc

Combinatòria i Probabilitat CP pdf doc

Geometria analítica GN pdf 1 2

Trigonometria TR

pdf doc

Nombres complexos CO pdf doc

Àlgebra Lineal 2n batxillerat AL pdf doc

Geometria Lineal 2n batxillerat GL pdf doc

Càlcul Infinitesimal 2n batxillerat CI pdf 1 2

Programació Lineal 2n batxillerat PL pdf doc

Recopilaciones de pruebas PAU España:

Catalunya TEC ST , Catalunya CCSS SC , Galicia SG

Recopilaciones de pruebas PAU Europa:

Portugal A SP, Portugal B SQ

Recopilaciones de problemas olímpicos y preolímpicos:

IMO SI, OME SE, OMI SD, AIME SA , Cangur SR , Canguro SG , Kangourou SK ,

AMC12 (2008-2020) SM , SMT SF

Versión de este documento: 26/01/2021

Todos estos documentos se actualizan constantemente. ¡No utilices una versión anticuada! Descarga gratis la última versión de los documentos en los enlaces superiores.

www.toomates.net

Índice.

AIME I AIME II Enunciados Enunciados Soluciones Enunciados Enunciados Soluciones

en inglés en español en español en inglés en español en español

1983 4

1984 8

1985 12 1986 16

1987 19 1988 23

1989 27

1990 31 1991 35

1992 39

1993 42

1994 46

1995 50

1996 54 1997 57

1998 61

1999 64 2000 68 72

2001 75 78

2002 82 86 2003 90 93

2004 96 100

2005 104 107 2006 110 114

2007 118 123

2008 127 131 2009 135 138

2010 142 145

2011 148 152 2012 156 160

2013 164 168

2014 171 175 2015 179 183

2016 187 191

2017 195 199 2018 203 295 298 207 315 318

2019 211 258 262 215 276 280

2020 218 224 227 221 239 242

Respuestas a todos los problemas en la página 339

Este documento es una compilación de los archivos que se encuentran en la página web

https://artofproblemsolving.com/wiki/index.php/AIME_Problems_and_Solutions

Agrupados en un único archivo “pdf” mediante la aplicación https://pdfjoiner.com/

Se puede acceder a las soluciones de la web AoPS clicando en la palabra “Solution” que aparece al final de cada enunciado.

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1983 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1983)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See Also

Let , and all exceed and let be a positive number such that , and .Find .

Solution

Let , where . Determine the minimum value taken by for in the interval .

Solution

What is the product of the real roots of the equation ?

Solution

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is cm, the length of is cm and that of is cm. The angle is a right angle. Find the square of the distance (in centimeters) from to the

center of the circle.

1983 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

Suppose that the sum of the squares of two complex numbers and is and the sum of the cubes is . What is the largestreal value that can have?

Solution

Let . Determine the remainder on dividing by .

Solution

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equallylikely - and are sent off to slay a troublesome dragon. Let be the probability that at least two of the three had been sitting next toeach other. If is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution

What is the largest -digit prime factor of the integer ?

Solution

Find the minimum value of for .

Solution

The numbers , and have something in common: each is a -digit number beginning with that has exactlytwo identical digits. How many such numbers are there?

Solution

The solid shown has a square base of side length . The upper edge is parallel to the base and has length . All other edges have

length . Given that , what is the volume of the solid?

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Solution

Diameter of a circle has length a -digit integer (base ten). Reversing the digits gives the length of the perpendicular chord . The distance from their intersection point to the center is a positive rational number. Determine the length of .

Solution

For and each of its nonempty subsets a unique alternating sum is defined as follows. Arrange the numbers inthe subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example,the alternating sum for is and for it is simply . Find the sum of all suchalternating sums for .

Solution

In the adjoining figure, two circles with radii and are drawn with their centers units apart. At , one of the points ofintersection, a line is drawn in such a way that the chords and have equal length. Find the square of the length of .

Solution

Problem 12

Problem 13

Problem 14

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The adjoining figure shows two intersecting chords in a circle, with on minor arc . Suppose that the radius of the circle is ,that , and that is bisected by . Suppose further that is the only chord starting at which is bisected by

. It follows that the sine of the central angle of minor arc is a rational number. If this number is expressed as a fraction

in lowest terms, what is the product ?

Solution

1983 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1983))

Preceded byFirst AIME

Followed by1984 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems&oldid=102995"

Problem 15

See Also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems 1/4

1984 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1984)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find the value of if , , is an arithmetic progression with common difference1, and .

Solution

The integer is the smallest positive multiple of such that every digit of is either or . Compute .

Solution

A point is chosen in the interior of such that when lines are drawn through parallel to the sides of , theresulting smaller triangles , , and in the figure, have areas , , and , respectively. Find the area of .

1984 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

Let be a list of positive integers - not necessarily distinct - in which the number appears. The arithmetic mean of thenumbers in is . However, if is removed, the arithmetic mean of the numbers is . What's the largest number that canappear in ?

Solution

Determine the value of if and .

Solution

Three circles, each of radius , are drawn with centers at , , and . A line passing through is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the threecircles to the other side of it. What is the absolute value of the slope of this line?

Solution

The function f is defined on the set of integers and satisfies

Find .

Solution

The equation has complex roots with argument between and in the complex plane. Determinethe degree measure of .

Solution

In tetrahedron , edge has length 3 cm. The area of face is and the area of face is . These two faces meet each other at a angle. Find the volume of the tetrahedron in .

Solution

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over . From this, Johnwas able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over , Johncould not have determined this. What was Mary's score? (Recall that the AHSME consists of multiple choice problems and that

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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one's score, , is computed by the formula , where is the number of correct answers and is the numberof wrong answers. Students are not penalized for problems left unanswered.)

Solution

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement

being equally likely. Let in lowest terms be the probability that no two birch trees are next to one another. Find .

Solution

A function is defined for all real numbers and satisfies and for all .If is a root for , what is the least number of roots must have in the interval

?

Solution

Find the value of

Solution

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution

Determine if

Solution

1984 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1984))

Preceded by1983 AIME Problems

Followed by1985 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems 4/4

Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems&oldid=97712"

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems 1/4

1985 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1985)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Let , and for let . Calculate the product .

Solution

When a right triangle is rotated about one leg, the volume of the cone produced is . When the triangle is rotated aboutthe other leg, the volume of the cone produced is . What is the length (in cm) of the hypotenuse of the triangle?

Solution

Find if , , and are positive integers which satisfy , where .

Solution

1985 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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A small square is constructed inside a square of area by dividing each side of the unit square into equal parts, and thenconnecting the vertices to the division points closest to the opposite vertices, as shown in the figure. Find the value of if the the

area of the small square is exactly .

Solution

A sequence of integers is chosen so that for each . What is the sum of the first terms of this sequence if the sum of the first terms is , and the sum of the first terms is ?

Solution

As shown in the figure, is divided into six smaller triangles by lines drawn from the vertices through a common interiorpoint. The areas of four of these triangles are as indicated. Find the area of .

Solution

Assume that , , and are positive integers such that , and . Determine .

Solution

The sum of the following seven numbers is exactly 19: , , , , , , . It is desired to replace each by an integer approximation , , so that the sum of the

's is also and so that , the maximum of the "errors" , is as small as possible. For this minimum , what is ?

Solution

In a circle, parallel chords of lengths , , and determine central angles of , and radians, respectively, where . If , which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its

numerator and denominator?

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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How many of the first positive integers can be expressed in the form

,

where is a real number, and denotes the greatest integer less than or equal to ?

Solution

An ellipse has foci at and in the -plane and is tangent to the -axis. What is the length of its major axis?

Solution

Let , , and be the vertices of a regular tetrahedron, each of whose edges measures meter. A bug, starting from vertex , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally

likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let be the probability that the bug is at

vertex when it has crawled exactly meters. Find the value of .

Solution

The numbers in the sequence , , , , are of the form , where . Foreach , let be the greatest common divisor of and . Find the maximum value of as ranges through the positiveintegers.

Solution

In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1

point, the loser got 0 points, and each of the two players earned point if the game was a tie. After the completion of the

tournament, it was found that exactly half of the points earned by each player were earned in games against the ten players withthe least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nineof the ten). What was the total number of players in the tournament?

Solution

Three cm cm squares are each cut into two pieces and , as shown in the first figure below, by joining the midpointsof two adjacent sides. These six pieces are then attached to a regular hexagon, as shown in the second figure, so as to fold into apolyhedron. What is the volume (in ) of this polyhedron?

Solution

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems 4/4

Copyright © 2020 Art of Problem Solving

1985 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1985))

Preceded by1984 AIME Problems

Followed by1986 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

1985 AIMEAmerican Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems&oldid=97713"

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems 1/3

1986 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1986)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

What is the sum of the solutions to the equation ?

Solution

Evaluate the product .

Solution

If and , what is ?

Solution

Determine if , , , , and satisfy the system of equations below.

1986 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

What is that largest positive integer for which is divisible by ?

Solution

The pages of a book are numbered through . When the page numbers of the book were added, one of the page numbers wasmistakenly added twice, resulting in an incorrect sum of . What was the number of the page that was added twice?

Solution

The increasing sequence consists of all those positive integers which are powers of 3 or sums of

distinct powers of 3. Find the term of this sequence.

Solution

Let be the sum of the base logarithms of all the proper divisors of . What is the integer nearest to ?

Solution

In , , , and . An interior point is then drawn, and segments are drawnthrough parallel to the sides of the triangle. If these three segments are of an equal length , find .

Solution

In a parlor game, the magician asks one of the participants to think of a three digit number where , , and representdigits in base in the order indicated. The magician then asks this person to form the numbers , , , ,and , to add these five numbers, and to reveal their sum, . If told the value of , the magician can identify the originalnumber, . Play the role of the magician and determine the if .

Solution

The polynomial may be written in the form , where and the 's are constants. Find the value of .

Solution

Let the sum of a set of numbers be the sum of its elements. Let be a set of positive integers, none greater than 15. Suppose notwo disjoint subsets of have the same sum. What is the largest sum a set with these properties can have?

Solution

In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head isimmediately followed by a head, and etc. We denote these by TH, HH, and etc. For example, in the sequence TTTHHTHTTTHHTTHof 15 coin tosses we observe that there are two HH, three HT, four TH, and five TT subsequences. How many different sequences

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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Copyright © 2020 Art of Problem Solving

of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT subsequences?

Solution

The shortest distances between an interior diagonal of a rectangular parallelepiped, , and the edges it does not meet are ,

, and . Determine the volume of .

Solution

Let triangle be a right triangle in the -plane with a right angle at . Given that the length of the hypotenuse is ,and that the medians through and lie along the lines and respectively, find the area of triangle

.

Solution

1986 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1986))

Preceded by1985 AIME Problems

Followed by1987 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems&oldid=97714"

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems 1/4

1987 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1987)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

An ordered pair of non-negative integers is called "simple" if the addition in base requires no carrying. Findthe number of simple ordered pairs of non-negative integers that sum to .

Solution

What is the largest possible distance between two points, one on the sphere of radius 19 with center and theother on the sphere of radius 87 with center ?

Solution

By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural numbergreater than 1 will be called "nice" if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nicenumbers?

Solution

Find the area of the region enclosed by the graph of

1987 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

Find if and are integers such that .

Solution

Rectangle is divided into four parts of equal area by five segments as shown in the figure, where , and is parallel to . Find the length of

(in cm) if cm and cm.

Solution

Let denote the least common multiple of positive integers and . Find the number of ordered triples of positiveintegers for which , , and .

Solution

What is the largest positive integer for which there is a unique integer such that ?

Solution

Triangle has right angle at , and contains a point for which , , and . Find .

Solution

Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of theescalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many stepsare visible on the escalator at a given time? (Assume that this value is constant.)

Solution

Find the largest possible value of for which is expressible as the sum of consecutive positive integers.

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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Let be the smallest integer whose cube root is of the form , where is a positive integer and is a positive real numberless than . Find .

Solution

A given sequence of distinct real numbers can be put in ascending order by means of one or more "bubblepasses". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging themif and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if thethird term is smaller, and so on in order, through comparing the last term, , with its current predecessor and exchanging them ifand only if the last term is smaller.

The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numberscompared at each step are underlined.

Suppose that , and that the terms of the initial sequence are distinct from one another and are inrandom order. Let , in lowest terms, be the probability that the number that begins as will end up, after one bubble pass, in

the place. Find .

Solution

Compute

.

Solution

Squares and are inscribed in right triangle , as shown in the figures below. Find if area and area .

Solution

1987 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1987))

Preceded by1986 AIME Problems

Followed by1988 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

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Problem 13

Problem 14

Problem 15

See also

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The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=1988)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

One commercially available ten-button lock may be opened by depressing -- in any order -- the correct five buttons. The sampleshown below has as its combination. Suppose that these locks are redesigned so that sets of as many as ninebuttons or as few as one button could serve as combinations. How many additional combinations would this allow?

Solution

For any positive integer , let denote the square of the sum of the digits of . For , let . Find .

Solution

1988 AIME Problems

Contents

Problem 1

Problem 2

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Find if .

Solution

Suppose that for . Suppose further that

.

What is the smallest possible value of ?

Solution

Let , in lowest terms, be the probability that a randomly chosen positive divisor of is an integer multiple of . Find

.

Solution

It is possible to place positive integers into the vacant twenty-one squares of the 5 times 5 square shown below so that thenumbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by theasterisk (*).

Solution

In triangle , , and the altitude from divides into segments of length and . What isthe area of triangle ?

Solution

The function , defined on the set of ordered pairs of positive integers, satisfies the following properties:

Calculate .

Solution

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Find the smallest positive integer whose cube ends in .

Solution

A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedronone square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of thepolyhedron rather than along an edge or a face?

Solution

Let be complex numbers. A line in the complex plane is called a mean line for the points if contains points (complex numbers) such that

For the numbers , , , , and , there is a unique mean line with y-intercept . Find the slope of this mean line.

Solution

Let be an interior point of triangle and extend lines from the vertices through to the opposite sides. Let , , , and denote the lengths of the segments indicated in the figure. Find the product if and .

Solution

Find if and are integers such that is a factor of .

Solution

Let be the graph of , and denote by the reflection of in the line . Let the equation of be written inthe form

Find the product .

Solution

In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of thepile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters tobe typed during the day, and the boss delivers them in the order 1, 2, 3, 4, 5, 6, 7, 8, 9.

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about themorning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will betyped. Based upon the above information, how many such after-lunch typing orders are possible? (That there are no letters left tobe typed is one of the possibilities.)

Solution

1988 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1988))

Preceded by1987 AIME Problems

Followed by1989 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems&oldid=97716"

See also

5/3/2020 Art of Problem Solving

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1989 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1989)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Compute .

Solution

Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) ofthe ten points as vertices?

Solution

Suppose is a positive integer and is a single digit in base 10. Find if

Solution

1989 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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If are consecutive positive integers such that is a perfect square and is a perfect cube, what is the smallest possible value of ?

Solution

When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to and is the same as

that of getting heads exactly twice. Let , in lowest terms, be the probability that the coin comes up heads in exactly out of

flips. Find .

Solution

Two skaters, Allie and Billie, are at points and , respectively, on a flat, frozen lake. The distance between and is meters. Allie leaves and skates at a speed of meters per second on a straight line that makes a angle with . At thesame time Allie leaves , Billie leaves at a speed of meters per second and follows the straight path that produces theearliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

Solution

If the integer is added to each of the numbers , , and , one obtains the squares of three consecutive terms of anarithmetic series. Find .

Solution

Assume that are real numbers such that

Find the value of .

Solution

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positiveinteger such that . Find the value of .

Solution

Let , , be the three sides of a triangle, and let , , , be the angles opposite them. If , find

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode(most frequent value). Let be the difference between the mode and the arithmetic mean of the sample. What is the largestpossible value of ? (For real , is the greatest integer less than or equal to .)

Solution

Let be a tetrahedron with , , , , , and , asshown in the figure. Let be the distance between the midpoints of edges and . Find .

Solution

Let be a subset of such that no two members of differ by or . What is the largest number ofelements can have?

Solution

Given a positive integer , it can be shown that every complex number of the form , where and are integers, can beuniquely expressed in the base using the integers as digits. That is, the equation

is true for a unique choice of non-negative integer and digits chosen from the set ,with . We write

to denote the base expansion of . There are only finitely many integers that have four-digit expansions

Find the sum of all such .

Solution

Point is inside . Line segments , , and are drawn with on , on , and on (see the figure at right). Given that , , , , and , find the area of

.

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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Solution

1989 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1989))

Preceded by1988 AIME Problems

Followed by1990 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems&oldid=97717"

See also

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1990 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1990)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

The increasing sequence consists of all positive integers that are neither the square nor the cube ofa positive integer. Find the 500th term of this sequence.

Solution

Find the value of .

Solution

Let be a regular and be a regular such that each interior angle of is as large as

each interior angle of . What's the largest possible value of ?

Solution

Find the positive solution to

1990 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

Let be the smallest positive integer that is a multiple of and has exactly positive integral divisors, including and itself.

Find .

Solution

A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, andreleases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate thenumber of fish in the lake on May 1, she assumes that 25% of these fish are no longer in the lake on September 1 (because ofdeath and emigrations), that 40% of the fish were not in the lake May 1 (because of births and immigrations), and that the numberof untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologistcalculate for the number of fish in the lake on May 1?

Solution

A triangle has vertices , , and . The equation of the bisector of canbe written in the form . Find .

Solution

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets.A marksman is to break all the targets according to the following rules:

1) The marksman first chooses a column from which a target is to be broken.

2) The marksman must then break the lowest remaining target in the chosen column.

If the rules are followed, in how many different orders can the eight targets be broken?

Solution

A fair coin is to be tossed times. Let , in lowest terms, be the probability that heads never occur on consecutive tosses.Find .

Solution

The sets and are both sets of complex roots of unity. The set is also a set of complex roots of unity. How many distinct elements are in ?

Solution

Someone observed that . Find the largest positive integer for which can be expressed as the product of consecutive positive integers.

Solution

A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be writtenin the form

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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where , , , and are positive integers. Find .

Solution

Let . Given that has 3817 digits and that its first (leftmost) digit is9, how many elements of have 9 as their leftmost digit?

Solution

The rectangle below has dimensions and . Diagonals and intersect at . If triangle is cut out and removed, edges and are joined, and the figure is then creased along segments

and , we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.

Solution

Find if the real numbers , , , and satisfy the equations

Solution

1990 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1990))

Preceded by1989 AIME Problems

Followed by1991 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 13

Problem 14

Problem 15

See also

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https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems 4/4

Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems&oldid=97718"

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1991 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1991)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find if and are positive integers such that

Solution

Rectangle has sides of length 4 and of length 3. Divide into 168 congruent segments with points , and divide into 168 congruent segments with points

. For , draw the segments . Repeat this construction on the sides and , and then draw the diagonal . Find the sum of the lengths of the 335 parallel segments drawn.

Solution

Expanding by the binomial theorem and doing no further manipulation gives

1991 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

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where for . For which is the largest?

Solution

How many real numbers satisfy the equation ?

Solution

Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator anddenominator. For how many rational numbers between 0 and 1 will be the resulting product?

Solution

Suppose is a real number for which

Find . (For real , is the greatest integer less than or equal to .)

Solution

Find , where is the sum of the absolute values of all roots of the following equation:

Solution

For how many real numbers does the quadratic equation have only integer roots for ?

Solution

Suppose that and that where is in lowest terms. Find

Solution

Two three-letter strings, and , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment,each of the six letters has a 1/3 chance of being received incorrectly, as an when it should have been a , or as a when itshould be an . However, whether a given letter is received correctly or incorrectly is independent of the reception of any otherletter. Let be the three-letter string received when is transmitted and let be the three-letter string received when istransmitted. Let be the probability that comes before in alphabetical order. When is written as a fraction in lowestterms, what is its numerator?

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Twelve congruent disks are placed on a circle of radius 1 in such a way that the twelve disks cover , no two of the disksoverlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in thefigure below. The sum of the areas of the twelve disks can be written in the form , where are positiveintegers and is not divisible by the square of any prime. Find .

Solution

Rhombus is inscribed in rectangle so that vertices , , , and are interior points on sides , , , and , respectively. It is given that , , , and . Let , in lowest

terms, denote the perimeter of . Find .

Solution

A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected

randomly without replacement, there is a probability of exactly that both are red or both are blue. What is the largest possible

number of red socks in the drawer that is consistent with this data?

Solution

A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by , has length 31. Find the sum ofthe lengths of the three diagonals that can be drawn from .

Solution

For positive integer , define to be the minimum value of the sum

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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where are positive real numbers whose sum is 17. There is a unique positive integer for which is also aninteger. Find this .

Solution

1991 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1991))

Preceded by1990 AIME Problems

Followed by1992 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems&oldid=112622"

See also

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1992 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1992)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms.

Solution

A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than anydigit to its right. How many ascending positive integers are there?

Solution

A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she hasplayed. At the start of a weekend, her win ratio is exactly . During the weekend, she plays four matches, winning three andlosing one. At the end of the weekend, her win ratio is greater than . What's the largest number of matches she could've wonbefore the weekend began?

Solution

In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutiveentries occur that are in the ratio ?

1992 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

Let be the set of all rational numbers , , that have a repeating decimal expansion in the form , where the digits , , and are not necessarily distinct. To write the elements of as fractions

in lowest terms, how many different numerators are required?

Solution

For how many pairs of consecutive integers in is no carrying required when the twointegers are added?

Solution

Faces and of tetrahedron meet at an angle of . The area of face is , the area of face is , and . Find the volume of the tetrahedron.

Solution

For any sequence of real numbers , define to be the sequence , whose term is . Suppose that all of the terms of the sequence

are , and that . Find .

Solution

Trapezoid has sides , , , and , with parallel to . A circle

with center on is drawn tangent to and . Given that , where and are relatively prime positive

integers, find .

Solution

Consider the region in the complex plane that consists of all points such that both and have real and imaginary parts

between and , inclusive. What is the integer that is nearest the area of ?

Solution

Lines and both pass through the origin and make first-quadrant angles of and radians, respectively, with the positive

x-axis. For any line , the transformation produces another line as follows: is reflected in , and the resulting line is

reflected in . Let and . Given that is the line , find the

smallest positive integer for which .

Solution

In a game of Chomp, two players alternately take bites from a 5-by-7 grid of unit squares. To take a bite, a player chooses one ofthe remaining squares, then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the loweredge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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remove the shaded square and the four squares marked by (The squares with two or more dotted edges have been removedfrom the original board in previous moves.)

The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board andempty board in your count.

Solution

Triangle has and . What's the largest area that this triangle can have?

Solution

In triangle , , , and are on the sides , , and , respectively. Given that , , and are

concurrent at the point , and that , find .

Solution

Define a positive integer to be a factorial tail if there is some positive integer such that the decimal representation of ends with exactly zeroes. How many positive integers less than are not factorial tails?

Solution

1992 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1992))

Preceded by1991 AIME Problems

Followed by1993 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems&oldid=97720"

Problem 13

Problem 14

Problem 15

See also

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1993 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1993)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

How many even integers between 4000 and 7000 have four different digits?

Solution

During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of thetour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the

candidate went miles on the day of this tour, how many miles was he from his starting point at the end of the day?

Solution

The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestantscaught fish for various values of .

In the newspaper story covering the event, it was reported that

(a) the winner caught fish;

1993 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

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(b) those who caught or more fish averaged fish each;(c) those who caught or fewer fish averaged fish each.

What was the total number of fish caught during the festival?

Solution

How many ordered four-tuples of integers with satisfy and ?

Solution

Let . For integers , define . What is thecoefficient of in ?

Solution

What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutiveintegers, and the sum of eleven consecutive integers?

Solution

Three numbers, , , , are drawn randomly and without replacement from the set . Three othernumbers, , , , are then drawn randomly and without replacement from the remaining set of 997 numbers. Let be theprobability that, after a suitable rotation, a brick of dimensions can be enclosed in a box of dimensions

, with the sides of the brick parallel to the sides of the box. If is written as a fraction in lowest terms, what is thesum of the numerator and denominator?

Solution

Let be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of so that theunion of the two subsets is ? The order of selection does not matter; for example, the pair of subsets , represents the same selection as the pair , .

Solution

Two thousand points are given on a circle. Label one of the points 1. From this point, count 2 points in the clockwise direction andlabel this point 2. From the point labeled 2, count 3 points in the clockwise direction and label this point 3. (See figure.) Continuethis process until the labels are all used. Some of the points on the circle will have more than one label andsome points will not have a label. What is the smallest integer that labels the same point as 1993?

Solution

Euler's formula states that for a convex polyhedron with vertices, edges, and faces, . A particularconvex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its vertices, triangular faces and

pentagonal faces meet. What is the value of ?

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain ahead. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game.Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is , where and arerelatively prime positive integers. What are the last three digits of ?

Solution

The vertices of are , , and . The six faces of a die are labeled with two 's, two 's, and two 's. Point is chosen in the interior of , and points , , are

generated by rolling the die repeatedly and applying the rule: If the die shows label , where , and is the

most recently obtained point, then is the midpoint of . Given that , what is ?

Solution

Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel pathsthat are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when thebuilding first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let be the amount of time, in seconds,before Jenny and Kenny can see each other again. If is written as a fraction in lowest terms, what is the sum of the numeratorand denominator?

Solution

A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (howeverslightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck

in a 6 by 8 rectangle, the smallest perimeter has the form , for a positive integer . Find .

Solution

Let be an altitude of . Let and be the points where the circles inscribed in the triangles and are tangent to . If , , and , then can be expressed as , where and are relatively prime integers. Find .

Solution

1993 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1993))

Preceded by1992 AIME Problems

Followed by1994 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems&oldid=97721"

Problem 12

Problem 13

Problem 14

Problem 15

See also

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Copyright © 2020 Art of Problem Solving

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1994 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1994)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

The increasing sequence consists of those positive multiples of 3 that are one less than a perfect square.What is the remainder when the 1994th term of the sequence is divided by 1000?

Solution

A circle with diameter of length 10 is internally tangent at to a circle of radius 20. Square is constructed with and on the larger circle, tangent at to the smaller circle, and the smaller circle outside . The length of

can be written in the form , where and are integers. Find .

Solution

The function has the property that, for each real number

.

If what is the remainder when is divided by 1000?

Solution

1994 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

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Find the positive integer for which

.

(For real , is the greatest integer )

Solution

Given a positive integer , let be the product of the non-zero digits of . (If has only one digit, then is equal to thatdigit.) Let

.

What is the largest prime factor of ?

Solution

The graphs of the equations

are drawn in the coordinate plane for These 63 lines cut part of the plane into equilateral

triangles of side . How many such triangles are formed?

Solution

For certain ordered pairs of real numbers, the system of equations

has at least one solution, and each solution is an ordered pair of integers. How many such ordered pairs are there?

Solution

The points , , and are the vertices of an equilateral triangle. Find the value of .

Solution

A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one ata time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. Thegame ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty. Theprobability that the bag will be emptied is where and are relatively prime positive integers. Find

Solution

In triangle angle is a right angle and the altitude from meets at The lengths of the sides of areintegers, and , where and are relatively prime positive integers. Find

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Ninety-four bricks, each measuring are to be stacked one on top of another to form a tower 94 bricks tall.Each brick can be oriented so it contributes or or to the total height of the tower. How many different tower heightscan be achieved using all 94 of the bricks?

Solution

A fenced, rectangular field measures 24 meters by 52 meters. An agricultural researcher has 1994 meters of fence that can beused for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides ofthe squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can bepartitioned using all or some of the 1994 meters of fence?

Solution

The equation

has 10 complex roots where the bar denotes complex conjugation. Find the value of

Solution

A beam of light strikes at point with angle of incidence and reflects with an equal angle of reflection asshown. The light beam continues its path, reflecting off line segments and according to the rule: angle of incidenceequals angle of reflection. Given that and determine the number of times the lightbeam will bounce off the two line segments. Include the first reflection at in your count.

Solution

Given a point on a triangular piece of paper consider the creases that are formed in the paper when and arefolded onto Let us call a fold point of if these creases, which number three unless is one of the vertices, donot intersect. Suppose that and Then the area of the set of all fold points of

can be written in the form where and are positive integers and is not divisible by the square ofany prime. What is ?

Solution

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems 4/4

Copyright © 2020 Art of Problem Solving

1994 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1994))

Preceded by1993 AIME Problems

Followed by1995 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems&oldid=97722"

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1995 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1995)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Square is For the lengths of the sides of square are half the lengths of the sides of square twoadjacent sides of square are perpendicular bisectors of two adjacent sides of square and the other two sides of square

are the perpendicular bisectors of two adjacent sides of square The total area enclosed by at least one of can be written in the form where and are relatively prime positive integers. Find

Solution

Find the last three digits of the product of the positive roots of

1995 AIME Problems

Contents

Problem 1

Problem 2

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Solution

Starting at an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up,or down, all four equally likely. Let be the probability that the object reaches in six or fewer steps. Given that can bewritten in the form where and are relatively prime positive integers, find

Solution

Circles of radius and are externally tangent to each other and are internally tangent to a circle of radius . The circle of radius has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.

Solution

For certain real values of and the equation has four non-real roots. Theproduct of two of these roots is and the sum of the other two roots is where Find

Solution

Let How many positive integer divisors of are less than but do not divide ?

Solution

Given that and

where and are positive integers with and relatively prime, find

Solution

For how many ordered pairs of positive integers with are both and integers?

Solution

Triangle is isosceles, with and altitude Suppose that there is a point on with

and Then the perimeter of may be written in the form where and are integers. Find

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Solution

What is the largest positive integer that is not the sum of a positive integral multiple of 42 and a positive composite integer?

Solution

A right rectangular prism (i.e., a rectangular parallelepiped) has sides of integral length with A planeparallel to one of the faces of cuts into two prisms, one of which is similar to and both of which have nonzero volume.Given that for how many ordered triples does such a plane exist?

Solution

Pyramid has square base congruent edges and and Let be the measure of the dihedral angle formed by faces and Given that where and

are integers, find

Solution

Let be the integer closest to Find

Solution

In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18. The two chords divide theinterior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of

them can be expressed uniquely in the form where and are positive integers and is not divisible by thesquare of any prime number. Find

Solution

Let be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 5 heads before oneencounters a run of 2 tails. Given that can be written in the form where and are relatively prime positive integers,find .

Solution

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems 4/4

Copyright © 2020 Art of Problem Solving

1995 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1995))

Preceded by1994 AIME Problems

Followed by1996 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems&oldid=97723"

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1996 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1996)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

In a magic square, the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of theentries of a magic square. Find .

Solution

For each real number , let denote the greatest integer that does not exceed . For how many positive integers is it truethat and that is a positive even integer?

Solution

Find the smallest positive integer for which the expansion of , after like terms have beencollected, has at least 1996 terms.

1996 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of a shadow, which doesnot include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed .

Solution

Suppose that the roots of are , , and , and that the roots of are , , and . Find .

Solution

In a five-team tournament, each team plays one game with every other team. Each team has a chance of winning any game

it plays. (There are no ties.) Let be the probability that the tournament will produce neither an undefeated team nor a winless

team, where and are relatively prime integers. Find .

Solution

Two squares of a checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if onecan be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible?

Solution

The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many orderedpairs of positive integers with is the harmonic mean of and equal to ?

Solution

A bored student walks down a hall that contains a row of closed lockers, numbered 1 to 1024. He opens the locker numbered 1,and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turnsaround and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening eachclosed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is thenumber of the last locker he opens?

Solution

Find the smallest positive integer solution to .

Solution

Let be the product of the roots of that have a positive imaginary part, and suppose that , where and . Find .

Solution

For each permutation of the integers , form the sum

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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.

The average value of all such sums can be written in the form , where and are relatively prime positive integers. Find .

Solution

In triangle , , , and . There is a point for which bisects , and is a right angle. The ratio

can be written in the form , where and are relatively prime positive integers. Find .

Solution

A rectangular solid is made by gluing together cubes. An internal diagonal of this solidpasses through the interiors of how many of the cubes?

Solution

In parallelogram let be the intersection of diagonals and . Angles and are each twice aslarge as angle and angle is times as large as angle . Find the greatest integer that does not exceed

.

Solution

1996 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1996))

Preceded by1995 AIME Problems

Followed by1997 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems&oldid=99643"

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems 1/4

1997 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1997)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegativeintegers?

Solution

The nine horizontal and nine vertical lines on an checkerboard form rectangles, of which are squares. The number can be written in the form where and are relatively prime positive integers. Find

Solution

Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placedthe two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times theproduct Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?

Solution

Circles of radii 5, 5, 8, and are mutually externally tangent, where and are relatively prime positive integers. Find

1997 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

The number can be expressed as a four-place decimal where and represent digits, any of which could bezero. It is desired to approximate by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such

fraction to is What is the number of possible values for ?

Solution

Point is in the exterior of the regular -sided polygon , and is an equilateral triangle. What is thelargest value of for which , , and are consecutive vertices of a regular polygon?

Solution

A car travels due east at miles per minute on a long, straight road. At the same time, a circular storm, whose radius is miles,

moves southeast at mile per minute. At time , the center of the storm is miles due north of the car. At time

minutes, the car enters the storm circle, and at time minutes, the car leaves the storm circle. Find .

Solution

How many different arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0and the sum of the entries in each column is 0?

Solution

Given a nonnegative real number , let denote the fractional part of ; that is, , where denotes the

greatest integer less than or equal to . Suppose that is positive, , and . Find the value of

.

Solution

Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, orgreen. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statementsare true:

i. Either each of the three cards has a different shape or all three of the card have the same shape.

ii. Either each of the three cards has a different color or all three of the cards have the same color.

iii. Either each of the three cards has a different shade or all three of the cards have the same shade.

How many different complementary three-card sets are there?

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Let . What is the greatest integer that does not exceed ?

Solution

The function defined by , where , , and are nonzero real numbers, has the properties ,

and for all values except . Find the unique number that is not in the range of .

Solution

Let be the set of points in the Cartesian plane that satisfy

If a model of were built from wire of negligible thickness, then the total length of wire required would be , where and are positive integers and is not divisible by the square of any prime number. Find .

Solution

Let and be distinct, randomly chosen roots of the equation . Let be the probability that

, where and are relatively prime positive integers. Find .

Solution

The sides of rectangle have lengths and . An equilateral triangle is drawn so that no point of the triangle liesoutside . The maximum possible area of such a triangle can be written in the form , where , , and arepositive integers, and is not divisible by the square of any prime number. Find .

Solution

1997 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1997))

Preceded by1996 AIME Problems

Followed by1998 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems 4/4

Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems&oldid=99928"

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1998 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1998)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

For how many values of is the least common multiple of the positive integers and , and ?

Solution

Find the number of ordered pairs of positive integers that satisfy and .

Solution

The graph of partitions the plane into several regions. What is the area of the bounded region?

Solution

Nine tiles are numbered respectively. Each of three players randomly selects and keeps three of the tiles, andsums those three values. The probability that all three players obtain an odd sum is where and are relatively primepositive integers. Find

Solution

1998 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Given that find

Solution

Let be a parallelogram. Extend through to a point and let meet at and at Given that and find

Solution

Let be the number of ordered quadruples of positive odd integers that satisfy Find

Solution

Except for the first two terms, each term of the sequence is obtained by subtracting the precedingterm from the one before that. The last term of the sequence is the first negative term encountered. What positive integer produces a sequence of maximum length?

Solution

Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9a.m. and 10 a.m., and stay for exactly minutes. The probability that either one arrives while the other is in the cafeteria is and where and are positive integers, and is not divisible by the square of any prime. Find

Solution

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are thevertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres.The radius of this last sphere is where and are positive integers, and is not divisible by the square of anyprime. Find .

Solution

Three of the edges of a cube are and and is an interior diagonal. Points and are on and respectively, so that and What is the area of the

polygon that is the intersection of plane and the cube?

Solution

Let be equilateral, and and be the midpoints of and respectively. There exist points and on and respectively, with the property that is on is on and is on The ratioof the area of triangle to the area of triangle is where and are integers, and is not divisibleby the square of any prime. What is ?

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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If is a set of real numbers, indexed so that its complex power sumis defined to be where Let be the sum of the complex power sums ofall nonempty subsets of Given that and where and are integers,find

Solution

An rectangular box has half the volume of an rectangular box, where and are integers, and What is the largest possible value of ?

Solution

Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos inwhich the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which

and do not both appear for any and . Let be the set of all dominos whose coordinates are no larger than 40.Find the length of the longest proper sequence of dominos that can be formed using the dominos of

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

1998 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1998))

Preceded by1997 AIME Problems

Followed by1999 AIME Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems&oldid=110749"

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems 1/4

1999 AIME (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=1999)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all four preceding terms also being prime.

Solution

Consider the parallelogram with vertices and A line through the origin cuts thisfigure into two congruent polygons. The slope of the line is where and are relatively prime positive integers. Find

Solution

Find the sum of all positive integers for which is a perfect square.

Solution

The two squares shown share the same center and have sides of length 1. The length of is and the area ofoctagon is where and are relatively prime positive integers. Find

1999 AIME Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

For any positive integer , let be the sum of the digits of , and let be For example, How many values of do not exceed 1999?

Solution

A transformation of the first quadrant of the coordinate plane maps each point to the point The vertices ofquadrilateral are and Let be the area of the region enclosed by the image of quadrilateral Find the greatest integer that does not exceed

Solution

There is a set of 1000 switches, each of which has four positions, called , and . When the position of any switchchanges, it is only from to , from to , from to , or from to . Initially each switch is in position . The switchesare labeled with the 1000 different integers , where , and take on the values . At step of a1000-step process, the -th switch is advanced one step, and so are all the other switches whose labels divide the label on the -thswitch. After step 1000 has been completed, how many switches will be in position ?

Solution

Let be the set of ordered triples of nonnegative real numbers that lie in the plane Let us saythat supports when exactly two of the following are true: Let consist of those

triples in that support The area of divided by the area of is where and are relatively prime

positive integers, find

Solution

A function is defined on the complex numbers by where and are positive numbers. This functionhas the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that

and that where and are relatively prime positive integers. Find

Solution

Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random,all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are amongthe ten given points is where and are relatively prime positive integers. Find

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Given that where angles are measured in degrees, and and are relatively prime positive integers

that satisfy find

Solution

The inscribed circle of triangle is tangent to at and its radius is 21. Given that and find the perimeter of the triangle.

Solution

Forty teams play a tournament in which every team plays every other( different opponents) team exactly once. No ties occur,and each team has a chance of winning any game it plays. The probability that no two teams win the same number ofgames is where and are relatively prime positive integers. Find

Solution

Point is located inside triangle so that angles and are all congruent. The sides of thetriangle have lengths and and the tangent of angle is where and are relatively prime positive integers. Find

Solution

Consider the paper triangle whose vertices are and The vertices of its midpoint triangle are themidpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is thevolume of this pyramid?

Solution

1999 AIME (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1999))

Preceded by1998 AIME Problems

Followed by2000 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems&oldid=107010"

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

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Copyright © 2020 Art of Problem Solving

5/3/2020 Art of Problem Solving

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2000 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2000)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find the least positive integer such that no matter how is expressed as the product of any two positive integers, at leastone of these two integers contains the digit .

Solution

Let and be integers satisfying . Let , let be the reflection of across the line , let be the reflection of across the y-axis, let be the reflection of across the x-axis, and let be the reflection of acrossthe y-axis. The area of pentagon is . Find .

Solution

In the expansion of where and are relatively prime positive integers, the coefficients of and are equal.Find .

Solution

The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height ofthe rectangle are relatively prime positive integers, find the perimeter of the rectangle.

2000 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is One marble is

taken out of each box randomly. The probability that both marbles are black is and the probability that both marbles are white

is where and are relatively prime positive integers. What is ?

Solution

For how many ordered pairs of integers is it true that and that the arithmetic mean of and isexactly more than the geometric mean of and ?

Solution

Suppose that and are three positive numbers that satisfy the equations and

Then where and are relatively prime positive integers. Find .

Solution

A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and itsbase horizontal, the height of the liquid is where and are positive integers and is not divisible by thecube of any prime number. Find .

Solution

The system of equations

has two solutions and . Find .

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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A sequence of numbers has the property that, for every integer between and inclusive, the

number is less than the sum of the other numbers. Given that , where and are relatively prime positive

integers, find .

Solution

Let be the sum of all numbers of the form , where and are relatively prime positive divisors of What is the greatest

integer that does not exceed ?

Solution

Given a function for which

holds for all real what is the largest number of different values that can appear in the list ?

Solution

In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at miles per hour along the highways and at miles per hour across the prairie. Consider the set of points that can be reached

by the firetruck within six minutes. The area of this region is square miles, where and are relatively prime positive

integers. Find .

Solution

In triangle it is given that angles and are congruent. Points and lie on and respectively, so that Angle is times as large as angle where is a positive real number. Find the

greatest integer that does not exceed .

Solution

A stack of cards is labelled with the integers from to with different integers on different cards. The cards in thestack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to thebottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, andthe next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards alreadyon the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is foundthat, reading from left to right, the labels on the cards are now in ascending order: In the originalstack of cards, how many cards were above the card labeled ?

Solution

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems 4/4

Copyright © 2020 Art of Problem Solving

2000 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2000))

Preceded by1999 AIME Problems

Followed by2000 AIME II Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems&oldid=117532"

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2000 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2000)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

The number

can be written as where and are relatively prime positive integers. Find .

Solution

A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola ?

Solution

A deck of forty cards consists of four 1's, four 2's,..., and four 10's. A matching pair (two cards with the same number) is removedfrom the deck. Given that these cards are not returned to the deck, let be the probability that two randomly selected cardsalso form a pair, where and are relatively prime positive integers. Find

Solution

2000 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution

Given eight distinguishable rings, let be the number of possible five-ring arrangements on the four fingers (not the thumb) of onehand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost threenonzero digits of .

Solution

One base of a trapezoid is units longer than the other base. The segment that joins the midpoints of the legs divides thetrapezoid into two regions whose areas are in the ratio . Let be the length of the segment joining the legs of the trapezoidthat is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does notexceed .

Solution

Given that

find the greatest integer that is less than .

Solution

In trapezoid , leg is perpendicular to bases and , and diagonals and are perpendicular.

Given that and , find .

Solution

Given that is a complex number such that , find the least integer that is greater than .

Solution

A circle is inscribed in quadrilateral , tangent to at and to at . Given that , , , and , find the square of the radius of the circle.

Solution

The coordinates of the vertices of isosceles trapezoid are all integers, with and .The trapezoid has no horizontal or vertical sides, and and are the only parallel sides. The sum of the absolute values ofall possible slopes for is , where and are relatively prime positive integers. Find .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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Copyright © 2020 Art of Problem Solving

The points , and lie on the surface of a sphere with center and radius . It is given that , ,

, and that the distance from to triangle is , where , , and are positive integers, and are

relatively prime, and is not divisible by the square of any prime. Find .

Solution

The equation has exactly two real roots, one of which is , where

, and are integers, and are relatively prime, and . Find .

Solution

Every positive integer has a unique factorial base expansion , meaning that , where each is an integer, , and . Given

that is the factorial base expansion of , find the value of

.

Solution

Find the least positive integer such that

Solution

2000 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2000))

Preceded by2000 AIME I Problems

Followed by2001 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems&oldid=103486"

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems 1/3

2001 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2001)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find the sum of all positive two-digit integers that are divisible by each of their digits.

Solution

A finite set of distinct real numbers has the following properties: the mean of is less than the mean of , and themean of is more than the mean of . Find the mean of .

Solution

Find the sum of the roots, real and non-real, of the equation , given that there are no multiple

roots.

Solution

2001 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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In triangle , angles and measure degrees and degrees, respectively. The bisector of angle intersects at , and . The area of triangle can be written in the form , where , , and are positiveintegers, and is not divisible by the square of any prime. Find .

Solution

An equilateral triangle is inscribed in the ellipse whose equation is . One vertex of the triangle is , one

altitude is contained in the y-axis, and the length of each side is , where and are relatively prime positive integers. Find

.

Solution

A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may beexpressed in the form , where and are relatively prime positive integers. Find .

Solution

Triangle has , and . Points and are located on and , respectively,such that is parallel to and contains the center of the inscribed circle of triangle . Then , where

and are relatively prime positive integers. Find .

Solution

Call a positive integer a if the digits of the base-7 representation of form a base-10 number that is twice .For example, is a 7-10 double because its base-7 representation is . What is the largest 7-10 double?

Solution

In triangle , , and . Point is on , is on , and is on . Let , , and , where , , and are positive and satisfy

and . The ratio of the area of triangle to the area of triangle can be written in the form , where and are relatively prime positive integers. Find .

Solution

Let be the set of points whose coordinates and are integers that satisfy and Two distinct points are randomly chosen from The probability that the midpoint of the segment they determine

also belongs to is where and are relatively prime positive integers. Find

Solution

In a rectangular array of points, with 5 rows and columns, the points are numbered consecutively from left to right beginningwith the top row. Thus the top row is numbered 1 through the second row is numbered through and so forth.Five points, and are selected so that each is in row Let be the number associated with Nowrenumber the array consecutively from top to bottom, beginning with the first column. Let be the number associated with after the renumbering. It is found that and Find the smallest possiblevalue of

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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A sphere is inscribed in the tetrahedron whose vertices are and The radius of the sphere is where and are relatively prime positive integers. Find

Solution

In a certain circle, the chord of a -degree arc is 22 centimeters long, and the chord of a -degree arc is 20 centimeters longerthan the chord of a -degree arc, where The length of the chord of a -degree arc is centimeters,where and are positive integers. Find

Solution

A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent housesever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How manydifferent patterns of mail delivery are possible?

Solution

The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are randomly written on the faces of a regular octahedron so that each face contains adifferent number. The probability that no two consecutive numbers, where 8 and 1 are considered to be consecutive, are written onfaces that share an edge is where and are relatively prime positive integers. Find

Solution

2001 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2001))

Preceded by2000 AIME II Problems

Followed by2001 AIME II Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems&oldid=107679"

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems 1/4

2001 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2001)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Let be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of forms a perfect square. What are the leftmost three digits of ?

Solution

Each of the 2001 students at a high school studies either Spanish or French, and some study both. The number who study Spanishis between 80 percent and 85 percent of the school population, and the number who study French is between 30 percent and 40percent. Let be the smallest number of students who could study both languages, and let be the largest number of studentswho could study both languages. Find .

Solution

Given that

2001 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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find the value of .

Solution

Let . The lines whose equations are and contain points and , respectively, such that

is the midpoint of . The length of equals , where and are relatively prime positive integers. Find .

Solution

A set of positive numbers has the if it has three distinct elements that are the lengths of the sides of atriangle whose area is positive. Consider sets of consecutive positive integers, all of whose ten-elementsubsets have the triangle property. What is the largest possible value of ?

Solution

Square is inscribed in a circle. Square has vertices and on and vertices and on the circle.

The ratio of the area of square to the area of square can be expressed as where and are relatively

prime positive integers and . Find .

Solution

Let be a right triangle with , , and . Let be the inscribed circle. Construct with on and on , such that is perpendicular to and tangent to . Construct with on and on such that is perpendicular to and tangent to . Let be the inscribed circle of and

the inscribed circle of . The distance between the centers of and can be written as . What is ?

Solution

A certain function has the properties that for all positive real values of , and that for . Find the smallest for which .

Solution

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be

used. The probability of obtaining a grid that does not have a 2-by-2 red square is , where and are relatively prime positive

integers. Find .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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How many positive integer multiples of 1001 can be expressed in the form , where and are integers and ?

Solution

Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that

Club Truncator will win, lose, or tie are each . The probability that Club Truncator will finish the season with more wins than

losses is , where and are relatively prime positive integers. Find .

Solution

Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra is definedrecursively as follows: is a regular tetrahedron whose volume is 1. To obtain , replace the midpoint triangle of every face

of by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of is , where and

are relatively prime positive integers. Find .

Solution

In quadrilateral , and , , , and .

The length may be written in the form , where and are relatively prime positive integers. Find .

Solution

There are complex numbers that satisfy both and . These numbers have the form , where and angles are measured in degrees. Find the

value of .

Solution

Let , , and be three adjacent square faces of a cube, for which , and let be theeighth vertex of the cube. Let , , and , be the points on , , and , respectively, so that

. A solid is obtained by drilling a tunnel through the cube. The sides of the tunnel are planesparallel to , and containing the edges, , , and . The surface area of , including the walls of the tunnel, is

, where , , and are positive integers and is not divisible by the square of any prime. Find .

Solution

2001 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2001))

Preceded by2001 AIME I Problems

Followed by2002 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems 4/4

Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems&oldid=103488"

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2002 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2002)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Giventhat each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least onepalindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is

, where and are relatively prime positive integers. Find .

Solution

The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to oneanother and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter

dimension can be written as , where and are positive integers. Find .

2002 AIME I Problems

Contents

Problem 1

Problem 2

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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems 2/4

Solution

Jane is 25 years old. Dick is older than Jane. In years, where is a positive integer, Dick's age and Jane's age will both be two-digit number and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let be Dick's presentage. How many ordered pairs of positive integers are possible?

Solution

Consider the sequence defined by for . Given that , for

positive integers and with , find .

Solution

Let be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagonhave at least two vertices in the set ?

Solution

The solutions to the system of equations

are and . Find .

Solution

The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers , , and with ,

What are the first three digits to the right of the decimal point in the decimal representation of ?

Solution

Find the smallest integer k for which the conditions

(1) is a nondecreasing sequence of positive integers

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

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(2) for all

(3)

are satisfied by more than one sequence.

Solution

Harold, Tanya, and Ulysses paint a very long picket fence. Harold starts with the first picket and paints every th picket; Tanyastarts with the second picket and paints every th picket; and Ulysses starts with the third picket and paints every th picket. Callthe positive integer when the triple of positive integers results in every picket beingpainted exactly once. Find the sum of all the paintable integers.

Solution

In the diagram below, angle is a right angle. Point is on , and bisects angle . Points and are on and , respectively, so that and . Given that and , find the integer

closest to the area of quadrilateral .

Solution

Let and be two faces of a cube with . A beam of light emanates from vertex and reflects offface at point , which is units from and units from . The beam continues to be reflected off the facesof the cube. The length of the light path from the time it leaves point until it next reaches a vertex of the cube is given by

, where and are integers and is not divisible by the square of any prime. Find .

Solution

Let for all complex numbers , and let for all positive integers . Given that

and , where and are real numbers, find .

Solution

In triangle the medians and have lengths 18 and 27, respectively, and . Extend to intersectthe circumcircle of at . The area of triangle is , where and are positive integers and is notdivisible by the square of any prime. Find .

Solution

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

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A set of distinct positive integers has the following property: for every integer in the arithmetic mean of the set of valuesobtained by deleting from is an integer. Given that 1 belongs to and that 2002 is the largest element of what is thegreatest number of elements that can have?

Solution

Polyhedron has six faces. Face is a square with face is a trapezoid with parallel to and and face has The other three

faces are and The distance from to face is 12. Given that where and are positive integers and is not divisible by the square of any prime, find

Solution

2002 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2002))

Preceded by2001 AIME II Problems

Followed by2002 AIME II Problems

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All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems&oldid=103489"

Problem 15

See also

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2002 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2002)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Given that and are both integers between and , inclusive; is the number formed by reversing the digits of ; and . How many distinct values of are possible?

Solution

Three vertices of a cube are , , and . What is the surface area of thecube?

Solution

It is given that , where , , and are positive integers that form an increasing geometricsequence and is the square of an integer. Find .

Solution

Patio blocks that are hexagons unit on a side are used to outline a garden by placing the blocks edge to edge with on eachside. The diagram indicates the path of blocks around the garden when .

2002 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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If , then the area of the garden enclosed by the path, not including the path itself, is square units, where

is a positive integer. Find the remainder when is divided by .

Solution

Find the sum of all positive integers where and are non-negative integers, for which is not a divisor of .

Solution

Find the integer that is closest to .

Solution

It is known that, for all positive integers ,

.

Find the smallest positive integer such that is a multiple of .

Solution

Find the least positive integer for which the equation has no integer solutions for . (The notation means

the greatest integer less than or equal to .)

Solution

Problem 5

Problem 6

Problem 7

Problem 8

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Let be the set Let be the number of sets of two non-empty disjoint subsets of . (Disjoint sets aredefined as sets that have no common elements.) Find the remainder obtained when is divided by .

Solution

While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correctangular mode. He was lucky to get the right answer. The two least positive real values of for which the sine of degrees is the

same as the sine of radians are and , where , , , and are positive integers. Find .

Solution

Two distinct, real, infinite geometric series each have a sum of and have the same second term. The third term of one of the

series is , and the second term of both series can be written in the form , where , , and are positive integers

and is not divisible by the square of any prime. Find .

Solution

A basketball player has a constant probability of of making any given shot, independent of previous shots. Let be the ratio ofshots made to shots attempted after shots. The probability that and for all such that isgiven to be where , , , and are primes, and , , and are positive integers. Find

.

Solution

In triangle , point is on with and , point is on with and , , and and intersect at . Points and lie on so that is parallel to and is parallel

to . It is given that the ratio of the area of triangle to the area of triangle is , where and arerelatively prime positive integers. Find .

Solution

The perimeter of triangle is , and the angle is a right angle. A circle of radius with center on isdrawn so that it is tangent to and . Given that where and are relatively prime positive integers,find .

Solution

Circles and intersect at two points, one of which is , and the product of the radii is . The x-axis and the line

, where , are tangent to both circles. It is given that can be written in the form , where , , and are positive integers, is not divisible by the square of any prime, and and are relatively prime. Find .

Solution

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

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Copyright © 2020 Art of Problem Solving

2002 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2002))

Preceded by2002 AIME I Problems

Followed by2003 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems&oldid=103490"

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m/Forum/resources.php?c=182&cid=45&year=2003)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Given that

where and are positive integers and is as large as possible, find

Solution

One hundred concentric circles with radii are drawn in a plane. The interior of the circle of radius 1 is coloredred, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color.The ratio of the total area of the green regions to the area of the circle of radius 100 can be expressed as where and are relatively prime positive integers. Find

Solution

Let the set Susan makes a list as follows: for each two-element subset of she writeson her list the greater of the set's two elements. Find the sum of the numbers on the list.

Solution

2003 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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Given that and that find

Solution

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units.

Given that the volume of this set is where and are positive integers, and and are relatively prime, find

Solution

The sum of the areas of all triangles whose vertices are also vertices of a 1 by 1 by 1 cube is where and are integers. Find

Solution

Point is on with and Point is not on so that and and areintegers. Let be the sum of all possible perimeters of Find

Solution

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form ageometric progression, and the first and fourth terms differ by Find the sum of the four terms.

Solution

An integer between and inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its tworightmost digits. How many balanced integers are there?

Solution

Triangle is isosceles with and Point is in the interior of the triangle so that and Find the number of degrees in

Solution

An angle is chosen at random from the interval Let be the probability that the numbers and are not the lengths of the sides of a triangle. Given that where is the number of degrees in

and and are positive integers with find

Solution

In convex quadrilateral and The perimeter of is640. Find (The notation means the greatest integer that is less than or equal to )

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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Let be the number of positive integers that are less than or equal to 2003 and whose base-2 representation has more 1's than0's. Find the remainder when is divided by 1000.

Solution

The decimal representation of where and are relatively prime positive integers and contains the digits 2, 5,and 1 consecutively, and in that order. Find the smallest value of for which this is possible.

Solution

In and Let be the midpoint of and let be the point on such that bisects angle Let be the point on such that Suppose that meets

at The ratio can be written in the form where and are relatively prime positive integers. Find

Solution

2003 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2003))

Preceded by2002 AIME II Problems

Followed by2003 AIME II Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems&oldid=109738"

Problem 13

Problem 14

Problem 15

See also

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2003 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2003)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

The product of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find the sum ofall possible values of .

Solution

Let be the greatest integer multiple of 8, whose digits are all different. What is the remainder when is divided by 1000?

Solution

Define a as a sequence of letters that consists only of the letters , , and - some of these letters may notappear in the sequence - and in which is never immediately followed by , is never immediately followed by , and isnever immediately followed by . How many seven-letter good words are there?

Solution

In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of thesmaller tetrahedron to that of the larger is , where and are relatively prime positive integers. Find .

Solution

2003 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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A cylindrical log has diameter inches. A wedge is cut from the log by making two planar cuts that go entirely through the log.The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a angle with the plane of the firstcut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedgecan be expressed as , where n is a positive integer. Find .

Solution

In triangle and point is the intersection of the medians. Points and are the images of and respectively, after a rotation about What is the area of the union of the tworegions enclosed by the triangles and

Solution

Find the area of rhombus given that the radii of the circles circumscribed around triangles and are and , respectively.

Solution

Find the eighth term of the sequence whose terms are formed by multiplying the correspondingterms of two arithmetic sequences.

Solution

Consider the polynomials and Given that and are the roots of find

Solution

Two positive integers differ by The sum of their square roots is the square root of an integer that is not a perfect square. Whatis the maximum possible sum of the two integers?

Solution

Triangle is a right triangle with and right angle at Point is the midpoint of and is on the same side of line as so that Given that the area of triangle may be expressed

as where and are positive integers, and are relatively prime, and is not divisible by the square of any

prime, find

Solution

The members of a distinguished committee were choosing a president, and each member gave one vote to one of the candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least than the number ofvotes for that candidate. What is the smallest possible number of members of the committee?

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currentlylocated, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex onits tenth move is where and are relatively prime positive integers, find

Solution

Let and be points on the coordinate plane. Let be a convex equilateral hexagon suchthat and the y-coordinates of its vertices are distinct

elements of the set The area of the hexagon can be written in the form where and arepositive integers and n is not divisible by the square of any prime. Find

Solution

Let . Let be the distinct zeros of and let for where and are real numbers. Let

where and are integers and is not divisible by the square of any prime. Find

Solution

2003 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2003))

Preceded by2003 AIME I Problems

Followed by2004 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems&oldid=103492"

Problem 14

Problem 15

See also

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2004 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2004)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See Also

The digits of a positive integer are four consecutive integers in decreasing order when read from left to right. What is the sum ofthe possible remainders when is divided by 37?

Solution

Set consists of consecutive integers whose sum is and set consists of consecutive integers whose sum is The absolute value of the difference between the greatest element of and the greatest element of is 99. Find

Solution

A convex polyhedron has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. Aspace diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many spacediagonals does have?

Solution

A square has sides of length 2. Set is the set of all line segments that have length 2 and whose endpoints are on adjacent sidesof the square. The midpoints of the line segments in set enclose a region whose area to the nearest hundredth is Find

2004 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attemptedquestions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 pointsout of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer scoreon each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less thanAlpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta couldachieve is where and are relatively prime positive integers. What is ?

Solution

An integer is called snakelike if its decimal representation satisfies if is odd and if is even. How many snakelike integers between 1000 and 9999 have four distinct digits?

Solution

Let be the coefficient of in the expansion of the product Find

Solution

Define a regular -pointed star to be the union of line segments such that

the points are coplanar and no three of them are collinear,each of the line segments intersects at least one of the other line segments at a point other than an endpoint,all of the angles at are congruent,all of the line segments are congruent, andthe path turns counterclockwise at an angle of less than 180 degrees at each vertex.

There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similarregular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?

Solution

Let be a triangle with sides 3, 4, and 5, and be a 6-by-7 rectangle. A segment is drawn to divide triangle into a triangle and a trapezoid and another segment is drawn to divide rectangle into a triangle and atrapezoid such that is similar to and is similar to The minimum value of the area of can be written in theform where and are relatively prime positive integers. Find

Solution

A circle of radius 1 is randomly placed in a 15-by-36 rectangle so that the circle lies completely within the rectangle.Given that the probability that the circle will not touch diagonal is where and are relatively prime positiveintegers, find

Solution

A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone,including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid

and a frustum-shaped solid in such a way that the ratio between the areas of the painted surfaces of and and the ratiobetween the volumes of and are both equal to Given that where and are relatively prime positiveintegers, find

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Let be the set of ordered pairs such that and and are

both even. Given that the area of the graph of is where and are relatively prime positive integers, find The notation denotes the greatest integer that is less than or equal to

Solution

The polynomial has 34 complex roots of the form with

and Given that where and are relatively prime positive integers, find

Solution

A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope isattached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the

rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower is feet, where

and are positive integers, and is prime. Find

Solution

For all positive integers , let

and define a sequence as follows: and for all positive integers . Let be the smallest suchthat . (For example, and .) Let be the number of positive integers such that

. Find the sum of the distinct prime factors of .

Solution

2004 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2004))

Preceded by2003 AIME II Problems

Followed by2004 AIME II Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Problem 12

Problem 13

Problem 14

Problem 15

See Also

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Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems&oldid=104638"

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m/Forum/resources.php?c=182&cid=45&year=2004)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions

into which the chord divides the circle to the smaller can be expressed in the form where and are

positive integers, and are relatively prime, and neither nor is divisible by the square of any prime. Find the remainder whenthe product is divided by 1000.

Solution

A jar has 10 red candies and 10 blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies atrandom. Given that the probability that they get the same color combination, irrespective of order, is where and arerelatively prime positive integers, find

Solution

A solid rectangular block is formed by gluing together congruent 1-cm cubes face to face. When the block is viewed so thatthree of its faces are visible, exactly 231 of the 1-cm cubes cannot be seen. Find the smallest possible value of

Solution

2004 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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How many positive integers less than 10,000 have at most two different digits?

Solution

In order to complete a large job, 1000 workers were hired, just enough to complete the job on schedule. All the workers stayed onthe job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then 100 workerswere laid off, so the second quarter of the work was completed behind schedule. Then an additional 100 workers were laid off, sothe third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is theminimum number of additional workers, beyond the 800 workers still on the job at the end of the third quarter, that must be hiredafter three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

Solution

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them,and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth ofthem, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananaswhenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the processare in the ratio what is the least possible total for the number of bananas?

Solution

is a rectangular sheet of paper that has been folded so that corner is matched with point on edge Thecrease is where is on and is on The dimensions and are given.The perimeter of rectangle is where and are relatively prime positive integers. Find

Solution

How many positive integer divisors of are divisible by exactly 2004 positive integers?

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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A sequence of positive integers with and is formed so that the first three terms are in geometricprogression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all the terms

are in geometric progression, and the terms and are in arithmetic progression. Let be the greatest term in this sequence that is less than 1000. Find

Solution

Let be the set of integers between 1 and whose binary expansions have exactly two 1's. If a number is chosen at randomfrom the probability that it is divisible by 9 is where and are relatively prime positive integers. Find

Solution

A right circular cone has a base with radius 600 and height A fly starts at a point on the surface of the cone whosedistance from the vertex of the cone is 125, and crawls along the surface of the cone to a point on the exact opposite side of the

cone whose distance from the vertex is Find the least distance that the fly could have crawled.

Solution

Let be an isosceles trapezoid, whose dimensions are and Draw circlesof radius 3 centered at and and circles of radius 2 centered at and A circle contained within the trapezoid is tangent

to all four of these circles. Its radius is where and are positive integers, is not divisible by the

square of any prime, and and are relatively prime. Find

Solution

Let be a convex pentagon with and Given that the ratio

between the area of triangle and the area of triangle is where and are relatively prime positiveintegers, find

Solution

Consider a string of 's, into which signs are inserted to produce an arithmetic expression. For example, could be obtained from eight 's in this way. For how many values of is it possible to

insert signs so that the resulting expression has value ?

Solution

A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in halfrepeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a512 by 1 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end,resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has becomea stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting fromthe left?

Solution

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

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Copyright © 2020 Art of Problem Solving

2004 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2004))

Preceded by2004 AIME I Problems

Followed by2005 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems&oldid=103605"

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m/Forum/resources.php?c=182&cid=45&year=2005)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See Also

Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent toa circle with radius 30. Let be the area of the region inside circle and outside of the six circles in the ring. Find

Solution

For each positive integer let denote the increasing arithmetic sequence of integers whose first term is 1 and whosecommon difference is For example, is the sequence For how many values of does contain theterm 2005?

Solution

How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is lessthan 50?

Solution

The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilledpositions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges thegroup in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this

2005 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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band can have.

Solution

Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side,but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face.Find the number of possible distinguishable arrangements of the 8 coins.

Solution

Let be the product of the nonreal roots of Find

Solution

In quadrilateral and Given that where and are positive integers, find

Solution

The equation has three real roots. Given that their sum is where and are

relatively prime positive integers, find

Solution

Twenty-seven unit cubes are painted orange on a set of four faces so that the two unpainted faces share an edge. The 27 cubesare then randomly arranged to form a cube. Given that the probability that the entire surface of the larger cube is

orange is where and are distinct primes and and are positive integers, find

Solution

Triangle lies in the Cartesian Plane and has an area of 70. The coordinates of and are and respectively, and the coordinates of are The line containing the median to side has slope Find the largestpossible value of

Solution

A semicircle with diameter is contained in a square whose sides have length 8. Given the maximum value of is find

Solution

For positive integers let denote the number of positive integer divisors of including 1 and For example, and Define by Let denote the number of

positive integers with odd, and let denote the number of positive integers with even.Find

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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A particle moves in the Cartesian Plane according to the following rules:

1. From any lattice point the particle may only move to or 2. There are no right angle turns in the particle's path.

How many different paths can the particle take from to ?

Solution

Consider the points and There is a unique square such that each of thefour points is on a different side of Let be the area of Find the remainder when is divided by 1000.

Solution

Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find

Solution

2005 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2005))

Preceded by2004 AIME II Problems

Followed by2005 AIME II Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and Solutions2005 AIME I Math Jam Transcript (http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=50)Mathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems&oldid=103607"

Problem 14

Problem 15

See Also

5/3/2020 Art of Problem Solving

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2005 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2005)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

A game uses a deck of different cards, where is an integer and The number of possible sets of 6 cards that can bedrawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find

Solution

A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut,cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from oneanother. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each

type is where and are relatively prime integers, find

Solution

An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the

sum of the original series. The common ratio of the original series is where and are relatively prime integers. Find

Solution

2005 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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Find the number of positive integers that are divisors of at least one of

Solution

Determine the number of ordered pairs of integers such that and

Solution

The cards in a stack of cards are numbered consecutively from 1 through from top to bottom. The top cards areremoved, kept in order, and form pile The remaining cards form pile The cards are then restacked by taking cardsalternately from the tops of pile and respectively. In this process, card number becomes the bottom card of thenew stack, card number 1 is on top of this card, and so on, until piles and are exhausted. If, after the restacking process, atleast one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. Findthe number of cards in the magical stack in which card number 131 retains its original position.

Solution

Let Find

Solution

Circles and are externally tangent, and they are both internally tangent to circle The radii of and are 4 and 10,respectively, and the centers of the three circles are all collinear. A chord of is also a common external tangent of and

Given that the length of the chord is where and are positive integers, and are relatively prime, and is not

divisible by the square of any prime, find

Solution

For how many positive integers less than or equal to 1000 is true for all real ?

Solution

Given that is a regular octahedron, that is the cube whose vertices are the centers of the faces of and that the ratio of the

volume of to that of is where and are relatively prime integers, find

Solution

Let be a positive integer, and let be a sequence of reals such that and

for Find

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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Copyright © 2020 Art of Problem Solving

Square has center and are on with and between and and Given that where and are positive integers and

is not divisible by the square of any prime, find

Solution

Let be a polynomial with integer coefficients that satisfies and Given that has two distinct integer solutions and find the product

Solution

In triangle and Point is on with Point is on

such that Given that where and are relatively prime positive integers, find

Solution

Let and denote the circles and respectively. Let be the smallest positive value of for which the line contains the center of a circle that is externally

tangent to and internally tangent to Given that where and are relatively prime integers, find

Solution

2005 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2005))

Preceded by2005 AIME I Problems

Followed by2006 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and Solutions2005 AIME II Math Jam Transcript (http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=51)Mathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems&oldid=118561"

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

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2006 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2006)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

In quadrilateral is a right angle, diagonal is perpendicular to and Find the perimeter of

Solution

Let set be a 90-element subset of and let be the sum of the elements of Find the number ofpossible values of

Solution

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is of the original integer.

Solution

Let be the number of consecutive 0's at the right end of the decimal representation of the product Find the remainder when is divided by 1000.

Solution

2006 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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The number can be written as where and are positive integers. Find

Solution

Let be the set of real numbers that can be represented as repeating decimals of the form where are distinctdigits. Find the sum of the elements of

Solution

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region to the area ofshaded region is 11/5. Find the ratio of shaded region to the area of shaded region

Solution

Hexagon is divided into five rhombuses, and as shown. Rhombuses and are

congruent, and each has area Let be the area of rhombus Given that is a positive integer, find the number ofpossible values for

Solution

The sequence is geometric with and common ratio where and are positive integers. Given that find the number of possible ordered pairs

Solution

Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region be the union of theeight circular regions. Line with slope 3, divides into two regions of equal area. Line 's equation can be expressed in the form

where and are positive integers whose greatest common divisor is 1. Find

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Solution

A collection of 8 cubes consists of one cube with edge-length for each integer A tower is to be built using all8 cubes according to the rules:

Any cube may be the bottom cube in the tower.The cube immediately on top of a cube with edge-length must have edge-length at most

Let be the number of different towers than can be constructed. What is the remainder when is divided by 1000?

Solution

Find the sum of the values of such that where is measured in degrees and

Solution

For each even positive integer let denote the greatest power of 2 that divides For example, and

For each positive integer let Find the greatest integer less than 1000 such that is

a perfect square.

Solution

A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the anglebetween any other pair, and the top of the tripod is 4 feet from the ground. In setting up the tripod, the lower 1 foot of one legbreaks off. Let be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then can be

written in the form where and are positive integers and is not divisible by the square of any prime. Find

(The notation denotes the greatest integer that is less than or equal to )

Solution

Given that a sequence satisfies and for all integers find the minimum possible value of

Solution

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems 4/4

Copyright © 2020 Art of Problem Solving

2006 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006))

Preceded by2005 AIME II Problems

Followed by2006 AIME II Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics ExaminationAIME Problems and Solutions2006 AIME I Math Jam Transcript (http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=144)Mathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems&oldid=103609"

5/3/2020 Art of Problem Solving

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2006 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2006)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

In convex hexagon , all six sides are congruent, and are right angles, and and

are congruent. The area of the hexagonal region is Find .

Solution

The lengths of the sides of a triangle with positive area are , , and , where is a positive integer.Find the number of possible values for .

Solution

Let be the product of the first 100 positive odd integers. Find the largest integer such that is divisible by .

Solution

Let be a permutation of for which

2006 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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An example of such a permutation is Find the number of such permutations.

Solution

When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face is greaterthan 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four facesis 1/6, and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is47/288. Given that the probability of obtaining face is where and are relatively prime positive integers, find

Solution

Square has sides of length 1. Points and are on and respectively, so that is equilateral. Asquare with vertex has sides that are parallel to those of and a vertex on The length of a side of this smaller

square is where and are positive integers and is not divisible by the square of any prime. Find

Solution

Find the number of ordered pairs of positive integers such that and neither nor has a zero digit.

Solution

There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the samecolor on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large trianglesare considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so thattheir corresponding small triangles are of the same color.

Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles maybe formed?

Solution

Circles and have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line is a commoninternal tangent to and and has a positive slope, and line is a common internal tangent to and and has a negativeslope. Given that lines and intersect at and that where and are positive integers and isnot divisible by the square of any prime, find

Solution

Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is

awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game ofthe tournament, team beats team The probability that team finishes with more points than team is where and are relatively prime positive integers. Find

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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A sequence is defined as follows and, for all positive integers

Given that and find the remainder when is divided

by 1000.

Solution

Equilateral is inscribed in a circle of radius 2. Extend through to point so that and extend through to point so that Through draw a line parallel to and through draw a line

parallel to Let be the intersection of and Let be the point on the circle that is collinear with and and

distinct from Given that the area of can be expressed in the form where and are positive integers,

and are relatively prime, and is not divisible by the square of any prime, find

Solution

How many integers less than 1000 can be written as the sum of consecutive positive odd integers from exactly 5 values of ?

Solution

Let be the sum of the reciprocals of the non-zero digits of the integers from to inclusive. Find the smallest positiveinteger n for which is an integer.

Solution

Given that and are real numbers that satisfy:

and that where and are positive integers and is not divisible by the square of any prime, find

Solution

2006 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006))

Preceded by2006 AIME I Problems

Followed by2007 AIME I Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

American Invitational Mathematics Examination

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

5/3/2020 Art of Problem Solving

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Copyright © 2020 Art of Problem Solving

AIME Problems and Solutions2006 AIME I Math Jam Transcript (http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=144)Mathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems&oldid=103610"

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2007 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2007)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

How many positive perfect squares less than are multiples of 24?

Solution

A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands.Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet persecond. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constantrate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, findthe distance in feet between the start of the walkway and the middle person.

Solution

The complex number is equal to , where is a positive real number and . Given that the imaginary parts of and are the same, what is equal to?

Solution

2007 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Theirperiods are , , and years. The three planets and the star are currently collinear. What is the fewest number of years fromnow that they will all be collinear again?

Solution

The formula for converting a Fahrenheit temperature to the corresponding Celsius temperature is An

integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and againrounded to the nearest integer.

For how many integer Fahrenheit temperatures between and inclusive does the original temperature equal the finaltemperature?

Solution

A frog is placed at the origin on the number line, and moves according to the following rule: in a given move, the frog advances toeither the closest point with a greater integer coordinate that is a multiple of , or to the closest point with a greater integercoordinate that is a multiple of . A move sequence is a sequence of coordinates which correspond to valid moves, beginningwith , and ending with . For example, is a move sequence. How many move sequences arepossible for the frog?

Solution

Let

Find the remainder when is divided by 1000. ( is the greatest integer less than or equal to , and is the least integergreater than or equal to .)

Solution

The polynomial is cubic. What is the largest value of for which the polynomials and are both factors of ?

Solution

In right triangle with right angle , and . Its legs and are extended beyond and . Points and lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center is tangent to the hypotenuse and to the extension of leg , the circle with center is tangent to the hypotenuse and to

the extension of leg , and the circles are externally tangent to each other. The length of the radius of either circle can beexpressed as , where and are relatively prime positive integers. Find .

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Solution

In a grid ( rows, columns), of the squares are to be shaded so that there are two shaded squares in each rowand three shaded squares in each column. Let be the number of shadings with this property. Find the remainder when isdivided by .

Solution

For each positive integer , let denote the unique positive integer such that . For example,

and . If find the remainder when is divided by 1000.

Solution

Problem 10

Problem 11

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In isosceles triangle , is located at the origin and is located at (20,0). Point is in the first quadrant with and angle . If triangle is rotated counterclockwise about point until the image of

lies on the positive -axis, the area of the region common to the original and the rotated triangle is in the form

, where are integers. Find .

Solution

A square pyramid with base and vertex has eight edges of length 4. A plane passes through the midpoints of , , and . The plane's intersection with the pyramid has an area that can be expressed as . Find .

Solution

A sequence is defined over non-negative integral indexes in the following way: , .

Find the greatest integer that does not exceed

Solution

Let be an equilateral triangle, and let and be points on sides and , respectively, with and

. Point lies on side such that angle . The area of triangle is . The twopossible values of the length of side are , where and are rational, and is an integer not divisible by thesquare of a prime. Find .

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 12

Problem 13

Problem 14

Problem 15

See also

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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems 5/5

Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems&oldid=74275"

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2007 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2007)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of fivecharacters chosen from the four letters in AIME and the four digits in . No character may appear in a sequence more timesthan it appears among the four letters in AIME or the four digits in . A set of plates in which each possible sequence appearsexactly once contains N license plates. Find N/10.

Solution

Find the number of ordered triples where , , and are positive integers, is a factor of , is a factor of , and .

Solution

Square has side length , and points and are exterior to the square such that and . Find .

2007 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers,but not necessarily equal for the two products. In one hour, workers can produce widgets and whoosits. In twohours, workers can produce widgets and whoosits. In three hours, workers can produce widgets and whoosits. Find .

Solution

The graph of the equation is drawn on graph paper with each square representing one unit in eachdirection. How many of the by graph paper squares have interiors lying entirely below the graph and entirely in the firstquadrant?

Solution

An integer is called parity-monotonic if its decimal representation satisfies if is odd, and if is even. How many four-digit parity-monotonic integers are there?

Solution

Given a real number let denote the greatest integer less than or equal to For a certain integer there are exactly positive integers such that and divides for all such that

Find the maximum value of for

Solution

A rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangledetermined by the intersections of some of these lines is called basic if

(i) all four sides of the rectangle are segments of drawn line segments, and(ii) no segments of drawn lines lie inside the rectangle.

Given that the total length of all lines drawn is exactly 2007 units, let be the maximum possible number of basic rectanglesdetermined. Find the remainder when is divided by 1000.

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

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Rectangle is given with and Points and lie on and respectively, such that The inscribed circle of triangle is tangent to at point and the inscribed circle of triangle

is tangent to at point Find

Solution

Let be a set with six elements. Let be the set of all subsets of Subsets and of , not necessarily distinct, are

chosen independently and at random from . The probability that is contained in at least one of or is where

, , and are positive integers, is prime, and and are relatively prime. Find (The set is the setof all elements of which are not in )

Solution

Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube hasradius and rolls along the surface toward the smaller tube, which has radius . It rolls over the smaller tube and continuesrolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one completerevolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance from whereit starts. The distance can be expressed in the form where and are integers and is not divisible by thesquare of any prime. Find

Solution

The increasing geometric sequence consists entirely of integral powers of Given that

and

find

Solution

A triangular array of squares has one square in the first row, two in the second, and in general, squares in the th row for With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated

in given diagram). In each square of the eleventh row, a or a is placed. Numbers are then placed into the other squares, with theentry for each square being the sum of the entries in the two squares below it. For how many initial distributions of 's and 's inthe bottom row is the number in the top square a multiple of ?

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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Solution

Let be a polynomial with real coefficients such that and for all , Find

Solution

Four circles and with the same radius are drawn in the interior of triangle such that is tangent tosides and , to and , to and , and is externally tangent to and . If the sides

of triangle are and the radius of can be represented in the form , where and are relatively prime

positive integers. Find

Solution

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Followed by2008 AIME I

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All AIME Problems and Solutions

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Mathematics Competitions (http://amc.maa.org).

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Problem 14

Problem 15

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m/Forum/resources.php?c=182&cid=45&year=2008)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

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1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Of the students attending a school party, of the students are girls, and of the students like to dance. After thesestudents are joined by more boy students, all of whom like to dance, the party is now girls. How many students now atthe party like to dance?

Solution

Square has sides of length units. Isosceles triangle has base , and the area common to triangle and square is square units. Find the length of the altitude to in .

Solution

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constantrates. Ed covers kilometers after biking for hours, jogging for hours, and swimming for hours, while Sue covers kilometers after jogging for hours, swimming for hours, and biking for hours. Their biking, jogging, and swimming rates areall whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution

2008 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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There exist unique positive integers and that satisfy the equation . Find .

Solution

A right circular cone has base radius and height . The cone lies on its side on a flat table. As the cone rolls on the surface of thetable without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertextouches the table. The cone first returns to its original position on the table after making complete rotations. The value of can be written in the form , where and are positive integers and is not divisible by the square of any prime. Find

.

Solution

A triangular array of numbers has a first row consisting of the odd integers in increasing order. Each row belowthe first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top rowequals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiplesof ?

Solution

Let be the set of all integers such that . For example, is the set . How many of the sets do not contain a perfect square?

Solution

Find the positive integer such that

Solution

Ten identical crates each of dimensions ft ft ft. The first crate is placed flat on the floor. Each of the remaining nine

crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let be the

probability that the stack of crates is exactly ft tall, where and are relatively prime positive integers. Find .

Solution

Let be an isosceles trapezoid with whose angle at the longer base is . The diagonals have length

, and point is at distances and from vertices and , respectively. Let be the foot of thealtitude from to . The distance can be expressed in the form , where and are positive integers and isnot divisible by the square of any prime. Find .

Solution

Consider sequences that consist entirely of 's and 's and that have the property that every run of consecutive 's has evenlength, and every run of consecutive 's has odd length. Examples of such sequences are , , and , while

is not such a sequence. How many such sequences have length 14?

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: thedistance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour ofspeed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the carin front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each caris 4 meters long and that the cars can travel at any speed, let be the maximum whole number of cars that can pass thephotoelectric eye in one hour. Find the quotient when is divided by 10.

Solution

Let

.

Suppose that

.

There is a point for which for all such polynomials, where , , and are positive integers, and

are relatively prime, and . Find .

Solution

Let be a diameter of circle . Extend through to . Point lies on so that line is tangent to . Point isthe foot of the perpendicular from to line . Suppose , and let denote the maximum possible length ofsegment . Find .

Solution

A square piece of paper has sides of length . From each corner a wedge is cut in the following manner: at each corner, the two

cuts for the wedge each start at distance from the corner, and they meet on the diagonal at an angle of (see the figurebelow). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, theyare taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, theperpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form ,where and are positive integers, , and is not divisible by the th power of any prime. Find .

Problem 12

Problem 13

Problem 14

Problem 15

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Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems&oldid=55234"

See also

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2008 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2008)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

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1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Let , where the additions and subtractionsalternate in pairs. Find the remainder when is divided by .

Solution

Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate whichis three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer andRudolph begin biking at the same time and arrive at the -mile mark at exactly the same time. How many minutes has it takenthem?

Solution

A block of cheese in the shape of a rectangular solid measures cm by cm by cm. Ten slices are cut from the cheese.Each slice has a width of cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel toeach other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off?

Solution

2008 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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There exist unique nonnegative integers and integers ( ) with each either or such that

Find .

Solution

In trapezoid with , let and . Let , , and

and be the midpoints of and , respectively. Find the length .

Solution

The sequence is defined by

The sequence is defined by

Find .

Solution

Let , , and be the three roots of the equation

Find .

Solution

Let . Find the smallest positive integer such that

is an integer.

Solution

A particle is located on the coordinate plane at . Define a move for the particle as a counterclockwise rotation of radians about the origin followed by a translation of units in the positive -direction. Given that the particle's position after

moves is , find the greatest integer less than or equal to .

Solution

The diagram below shows a rectangular array of points, each of which is unit away from its nearest neighbors.

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutivepoints of the sequence is strictly increasing. Let be the maximum possible number of points in a growing path, and let be thenumber of growing paths consisting of exactly points. Find .

Solution

In triangle , , and . Circle has radius and is tangent to and . Circle is externally tangent to circle and is tangent to and . No point of circle lies outside of . The radius of

circle can be expressed in the form ,where , , and are positive integers and is the product of distinctprimes. Find .

Solution

There are two distinguishable flagpoles, and there are flags, of which are identical blue flags, and are identical greenflags. Let be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag andno two green flags on either pole are adjacent. Find the remainder when is divided by .

Solution

A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is

parallel to the imaginary axis. Let be the region outside the hexagon, and let . Then the area of has the

form , where and are positive integers. Find .

Solution

Let and be positive real numbers with . Let be the maximum possible value of for which the system of equations

has a solution satisfying and . Then can be expressed as a fraction , where and

are relatively prime positive integers. Find .

Solution

Find the largest integer satisfying the following conditions:

(i) can be expressed as the difference of two consecutive cubes;(ii) is a perfect square.

Solution

American Invitational Mathematics ExaminationAIME Problems and Solutions

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

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Copyright © 2020 Art of Problem Solving

Mathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=2009)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Call a -digit number geometric if it has distinct digits which, when read from left to right, form a geometric sequence. Find thedifference between the largest and smallest geometric numbers.

Solution

There is a complex number with imaginary part and a positive integer such that

Find .

Solution

2009 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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A coin that comes up heads with probability and tails with probability independently on each flip is flipped

eight times. Suppose the probability of three heads and five tails is equal to of the probability of five heads and three tails. Let

, where and are relatively prime positive integers. Find .

Solution

In parallelogram , point is on so that and point is on so that .

Let be the point of intersection of and . Find .

Solution

Triangle has and . Points and are located on and respectively so that , and is the angle bisector of angle . Let be the point of intersection of and , and let be

the point on line for which is the midpoint of . If , find .

Solution

How many positive integers less than are there such that the equation has a solution for ? (The notation denotes the greatest integer that is less than or equal to .)

Solution

The sequence satisfies and for . Let be the least integer greater than

for which is an integer. Find .

Solution

Let . Consider all possible positive differences of pairs of elements of . Let be the sum ofall of these differences. Find the remainder when is divided by .

Solution

A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from to inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of thethree prices are given. On a particular day, the digits given were . Find the total number of possible guessesfor all three prizes consistent with the hint.

Solution

The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and fiveEarthlings. At meetings, committee members sit at a round table with chairs numbered from to in clockwise order.Committee rules state that a Martian must occupy chair and an Earthling must occupy chair . Furthermore, no Earthling cansit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sitimmediately to the left of an Earthling. The number of possible seating arrangements for the committee is . Find .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Consider the set of all triangles where is the origin and and are distinct points in the plane with nonnegativeinteger coordinates such that . Find the number of such distinct triangles whose area is a positiveinteger.

Solution

In right with hypotenuse , , , and is the altitude to . Let be the circle having as a diameter. Let be a point outside such that and are both tangent to circle . The ratio of the

perimeter of to the length can be expressed in the form , where and are relatively prime positive integers.

Find .

Solution

The terms of the sequence defined by for are positive integers. Find the minimum

possible value of .

Solution

For , define , where . If and , find the minimum

possible value for .

Solution

In triangle , , , and . Let be a point in the interior of . Let and denotethe incenters of triangles and , respectively. The circumcircles of triangles and meet at distinctpoints and . The maximum possible area of can be expressed in the form , where , , and arepositive integers and is not divisible by the square of any prime. Find .

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems&oldid=85153"

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

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2009 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2009)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Before starting to paint, Bill had ounces of blue paint, ounces of red paint, and ounces of white paint. Bill paintedfour equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red andwhite, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the totalnumber of ounces of paint Bill had left.

Solution

Suppose that , , and are positive real numbers such that , , and . Find

Solution

In rectangle , . Let be the midpoint of . Given that line and line are perpendicular, findthe greatest integer less than .

Solution

2009 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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A group of children held a grape-eating contest. When the contest was over, the winner had eaten grapes, and the child in -thplace had eaten grapes. The total number of grapes eaten in the contest was . Find the smallest possiblevalue of .

Solution

Equilateral triangle is inscribed in circle , which has radius . Circle with radius is internally tangent to circle at onevertex of . Circles and , both with radius , are internally tangent to circle at the other two vertices of . Circles , ,

and are all externally tangent to circle , which has radius , where and are relatively prime positive integers. Find

.

Solution

Let be the number of five-element subsets that can be chosen from the set of the first natural numbers so that at least twoof the five numbers are consecutive. Find the remainder when is divided by .

Solution

Define to be for odd and for even. When

is expressed as a fraction in lowest terms, its denominator is with odd. Find .

Solution

Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears

for the first time. Let and be relatively prime positive integers such that is the probability that the number of times Dave

rolls his die is equal to or within one of the number of times Linda rolls her die. Find .

Solution

Let be the number of solutions in positive integers to the equation , and let be the number ofsolutions in positive integers to the equation . Find the remainder when is divided by .

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Solution

Four lighthouses are located at points , , , and . The lighthouse at is kilometers from the lighthouse at , thelighthouse at is kilometers from the lighthouse at , and the lighthouse at is kilometers from the lighthouse at .To an observer at , the angle determined by the lights at and and the angle determined by the lights at and areequal. To an observer at , the angle determined by the lights at and and the angle determined by the lights at and

are equal. The number of kilometers from to is given by , where , , and are relatively prime positive integers, and

is not divisible by the square of any prime. Find .

Solution

For certain pairs of positive integers with there are exactly distinct positive integers such that . Find the sum of all possible values of the product .

Solution

From the set of integers , choose pairs with so that no two pairs have a commonelement. Suppose that all the sums are distinct and less than or equal to . Find the maximum possible value of .

Solution

Let and be the endpoints of a semicircular arc of radius . The arc is divided into seven congruent arcs by six equally spacedpoints . All chords of the form or are drawn. Let be the product of the lengths of these twelvechords. Find the remainder when is divided by .

Solution

The sequence satisfies and for . Find the greatest integer less than

or equal to .

Solution

Let be a diameter of a circle with diameter . Let and be points on one of the semicircular arcs determined by

such that is the midpoint of the semicircle and . Point lies on the other semicircular arc. Let be the length of

the line segment whose endpoints are the intersections of diameter with the chords and . The largest possiblevalue of can be written in the form , where , , and are positive integers and is not divisible by the square of anyprime. Find .

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

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https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems 4/4

Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=2010)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Maya lists all the positive divisors of . She then randomly selects two distinct divisors from this list. Let be the probability

that exactly one of the selected divisors is a perfect square. The probability can be expressed in the form , where and

are relatively prime positive integers. Find .

Solution

Find the remainder when is divided by .

Solution

Suppose that and . The quantity can be expressed as a rational number , where and are

relatively prime positive integers. Find .

Solution

2010 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Jackie and Phil have two fair coins and a third coin that comes up heads with probability . Jackie flips the three coins, and then

Phil flips the three coins. Let be the probability that Jackie gets the same number of heads as Phil, where and are

relatively prime positive integers. Find .

Solution

Positive integers , , , and satisfy , , and . Find the number of possible values of .

Solution

Let be a quadratic polynomial with real coefficients satisfying for allreal numbers , and suppose . Find .

Solution

Define an ordered triple of sets to be if and . For example, is a

minimally intersecting triple. Let be the number of minimally intersecting ordered triples of sets for which each set is a subsetof . Find the remainder when is divided by .

Note: represents the number of elements in the set .

Solution

For a real number , let denote the greatest integer less than or equal to . Let denote the region in the coordinate plane

consisting of points such that . The region is completely contained in a disk of radius (a disk

is the union of a circle and its interior). The minimum value of can be written as , where and are integers and is

not divisible by the square of any prime. Find .

Solution

Let be a real solution of the system of equations , , . The

greatest possible value of can be written in the form , where and are relatively prime positive integers.

Find .

Solution

Let be the number of ways to write in the form , where the 'sare integers, and . An example of such a representation is .Find .

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Let be the region consisting of the set of points in the coordinate plane that satisfy both and . When is revolved around the line whose equation is , the volume of the resulting solid is

, where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime.

Find .

Solution

Let be an integer and let . Find the smallest value of such that for every partition of into two subsets, at least one of the subsets contains integers , , and (not necessarily distinct) such that .

Note: a partition of is a pair of sets , such that , .

Solution

Rectangle and a semicircle with diameter are coplanar and have nonoverlapping interiors. Let denote theregion enclosed by the semicircle and the rectangle. Line meets the semicircle, segment , and segment at distinctpoints , , and , respectively. Line divides region into two regions with areas in the ratio . Suppose that

, , and . Then can be represented as , where and are positiveintegers and is not divisible by the square of any prime. Find .

Solution

For each positive integer n, let . Find the largest value of n for which .

Note: is the greatest integer less than or equal to .

Solution

In with , , and , let be a point on such that the incircles of

and have equal radii. Let and be positive relatively prime integers such that . Find .

Solution

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The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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Problem 12

Problem 13

Problem 14

Problem 15

See also

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2010 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2010)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Let be the greatest integer multiple of all of whose digits are even and no two of whose digits are the same. Find theremainder when is divided by .

Solution

A point is chosen at random in the interior of a unit square . Let denote the distance from to the closest side of .

The probability that is equal to , where and are relatively prime positive integers. Find .

Solution

Let be the product of all factors (not necessarily distinct) where and are integers satisfying . Find the greatest positive integer such that divides .

Solution

2010 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Dave arrives at an airport which has twelve gates arranged in a straight line with exactly feet between adjacent gates. Hisdeparture gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different

gate, again at random. Let the probability that Dave walks feet or less to the new gate be a fraction , where and are

relatively prime positive integers. Find .

Solution

Positive numbers , , and satisfy and . Find

.

Solution

Find the smallest positive integer with the property that the polynomial can be written as a product of twononconstant polynomials with integer coefficients.

Solution

Let , where , , and are real. There exists a complex number such that the three roots of are , , and , where . Find .

Solution

Let be the number of ordered pairs of nonempty sets and that have the following properties:

,

,The number of elements of is not an element of ,The number of elements of is not an element of .

Find .

Solution

Let be a regular hexagon. Let , , , , , and be the midpoints of sides , , , , ,and , respectively. The segments , , , , , and bound a smaller regular hexagon. Let the ratio of

the area of the smaller hexagon to the area of be expressed as a fraction where and are relatively prime

positive integers. Find .

Solution

Find the number of second-degree polynomials with integer coefficients and integer zeros for which .

Solution

Define a T-grid to be a matrix which satisfies the following two properties:

1. Exactly five of the entries are 's, and the remaining four entries are 's.2. Among the eight rows, columns, and long diagonals (the long diagonals are and ,

no more than one of the eight has all three entries equal.

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Find the number of distinct T-grids.

Solution

Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of thebases of the two triangles is . Find the minimum possible value of their common perimeter.

Solution

The cards in a deck are numbered . Alex, Blair, Corey, and Dylan each pick a card from the deck randomly andwithout replacement. The two people with lower numbered cards form a team, and the two people with higher numbered cardsform another team. Let be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards

and , and Dylan picks the other of these two cards. The minimum value of for which can be written as

, where and are relatively prime positive integers. Find .

Solution

Triangle with right angle at , and . Point on is chosen such that

and . The ratio can be represented in the form , where , , are positive

integers and is not divisible by the square of any prime. Find .

Solution

In triangle , , , and . Points and lie on with and . Points and lie on with and . Let be the point,

other than , of intersection of the circumcircles of and . Ray meets at . The ratio can

be written in the form , where and are relatively prime positive integers. Find .

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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Problem 12

Problem 13

Problem 14

Problem 15

See also

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2011 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2011)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Jar A contains four liters of a solution that is acid. Jar B contains five liters of a solution that is acid. Jar C contains

one liter of a solution that is acid. From jar C, liters of the solution is added to jar A, and the remainder of the solution in jar

C is added to jar B. At the end both jar A and jar B contain solutions that are acid. Given that and are relatively primepositive integers, find .

Solution

In rectangle , and . Points and lie inside rectangle so that , , , , and line intersects segment . The length can be expressed in the form

, where , , and are positive integers and is not divisible by the square of any prime. Find .

Solution

Let be the line with slope that contains the point , and let be the line perpendicular to line that

contains the point . The original coordinate axes are erased, and line is made the -axis and line the -axis.In the new coordinate system, point is on the positive -axis, and point is on the positive -axis. The point withcoordinates in the original system has coordinates in the new coordinate system. Find .

2011 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

In triangle , , , and . The angle bisector of angle intersects at point ,and the angle bisector of angle intersects at point . Let and be the feet of the perpendiculars from to and , respectively. Find .

Solution

The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of thenumbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to beindistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishableacceptable arrangements.

Solution

Suppose that a parabola has vertex and equation , where and is an

integer. The minimum possible value of can be written in the form , where and are relatively prime positive integers. Find

.

Solution

Find the number of positive integers for which there exist nonnegative integers , , , such that

Solution

In triangle , , , and . Points and are on with on , points and are on with on , and points and are on with on . In addition, the points are positioned so that

, , and . Right angle folds are then made along , , and . The resultingfigure is placed on a level floor to make a table with triangular legs. Let be the maximum possible height of a table constructed

from triangle whose top is parallel to the floor. Then can be written in the form , where and are relatively

prime positive integers and is a positive integer that is not divisible by the square of any prime. Find .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

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Suppose is in the interval and . Find .

Solution

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular -gon determine an

obtuse triangle is . Find the sum of all possible values of .

Solution

Let be the set of all possible remainders when a number of the form , a nonnegative integer, is divided by 1000. Let bethe sum of the elements in . Find the remainder when is divided by 1000.

Solution

Six men and some number of women stand in a line in random order. Let be the probability that a group of at least four menstand together in the line, given that every man stands next to at least one other man. Find the least number of women in the linesuch that does not exceed 1 percent.

Solution

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled . The three vertices adjacent to

vertex are at heights 10, 11, and 12 above the plane. The distance from vertex to the plane can be expressed as ,

where , , and are positive integers and . Find .

Solution

Let be a regular octagon. Let , , , and be the midpoints of sides , , , and , respectively. For , ray is constructed from towards the interior of the octagon such

that , , , and . Pairs of rays and , and , and , and and meet at , , , respectively. If , then can be written in the form

, where and are positive integers. Find .

Solution

For some integer , the polynomial has the three integer roots , , and . Find .

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

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Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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2011 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2011)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Gary purchased a large beverage, but only drank of it, where and are relatively prime positive integers. If he hadpurchased half as much and drunk twice as much, he would have wasted only as much beverage. Find .

Solution

On square , point lies on side and point lies on side , so that . Find thearea of the square .

Solution

The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Findthe degree measure of the smallest angle.

Solution

2011 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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In triangle , . The angle bisector of angle intersects at point , and point is the midpoint of

. Let be the point of intersection of and the line . The ratio of to can be expressed in the form ,

where and are relatively prime positive integers. Find .

Solution

The sum of the first terms of a geometric sequence is . The sum of the first terms is . Find the sum of thefirst terms.

Solution

Define an ordered quadruple of integers as interesting if , and .How many interesting ordered quadruples are there?

Solution

Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of thered ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to thenumber of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let bethe maximum number of red marbles for which such an arrangement is possible, and let be the number of ways he can arrangethe marbles to satisfy the requirement. Find the remainder when is divided by .

Solution

Let be the 12 zeroes of the polynomial . For each , let be one of or . Then the

maximum possible value of the real part of can be written as where and are positive integers. Find

.

Solution

Let , , , be nonnegative real numbers such that , and

. Let and be positive relatively prime integers such that is the maximum possible value of

. Find .

Solution

A circle with center has radius 25. Chord of length 30 and chord of length 14 intersect at point . The distance

between the midpoints of the two chords is 12. The quantity can be represented as , where and are relatively prime

positive integers. Find the remainder when is divided by 1000.

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Let be the matrix with entries as follows: for , ; for ,

; all other entries in are zero. Let be the determinant of matrix . Then

can be represented as , where and are relatively prime positive integers. Find .

Note: The determinant of the matrix is , and the determinant of the matrix ; for

, the determinant of an matrix with first row or first column is equal to , where is the determinant of the matrix

formed by eliminating the row and column containing .

Solution

Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the

probability that each delegate sits next to at least one delegate from another country be , where and are relatively prime

positive integers. Find .

Solution

Point lies on the diagonal of square with . Let and be the circumcenters of triangles

and , respectively. Given that and , then , where and are positive integers. Find .

Solution

There are permutations of such that for , divides for all integers with . Find the remainder when is divided by 1000.

Solution

Let . A real number is chosen at random from the interval . The probability that

is equal to , where , , , , and are positive integers. Find

.

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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Problem 12

Problem 13

Problem 14

Problem 15

See also

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Copyright © 2020 Art of Problem Solving

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2012 AIME I (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2012)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

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1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find the number of positive integers with three not necessarily distinct digits, , with and such that both and are multiples of .

Solution

The terms of an arithmetic sequence add to . The first term of the sequence is increased by , the second term is increasedby , the third term is increased by , and in general, the th term is increased by the th odd positive integer. The terms of thenew sequence add to . Find the sum of the first, last, and middle term of the original sequence.

Solution

Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chickenmeal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waitercould serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

Solution

2012 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse,Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that areconveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reachesSparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue inthis manner. Sparky, Butch, and Sundance walk at and miles per hour, respectively. The first time Butch and Sundancemeet at a milepost, they are miles from Dodge, and they have been traveling for minutes. Find .

Solution

Let be the set of all binary integers that can be written using exactly zeros and ones where leading zeros are allowed. If allpossible subtractions are performed in which one element of is subtracted from another, find the number of times the answer is obtained.

Solution

The complex numbers and satisfy and the imaginary part of is , for relatively prime

positive integers and with Find

Solution

At each of the sixteen circles in the network below stands a student. A total of coins are distributed among the sixteenstudents. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in thenetwork. After the trade, all students have the same number of coins as they started with. Find the number of coins the studentstanding at the center circle had originally.

Solution

Cube labeled as shown below, has edge length and is cut by a plane passing through vertex and themidpoints and of and respectively. The plane divides the cube into two solids. The volume of the larger of thetwo solids can be written in the form where and are relatively prime positive integers. Find

Problem 5

Problem 6

Problem 7

Problem 8

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Solution

Let and be positive real numbers that satisfy

The value of can be expressed in the form where and are relatively prime positive integers. Find

Solution

Let be the set of all perfect squares whose rightmost three digits in base are . Let be the set of all numbers of the

form , where is in . In other words, is the set of numbers that result when the last three digits of each number in

are truncated. Find the remainder when the tenth smallest element of is divided by .

Solution

A frog begins at and makes a sequence of jumps according to the following rule: from thefrog jumps to which may be any of the points or

There are points with that can be reached by a sequence of suchjumps. Find the remainder when is divided by

Solution

Let be a right triangle with right angle at Let and be points on with between and such that

and trisect If then can be written as where and are relatively prime

positive integers, and is a positive integer not divisible by the square of any prime. Find

Solution

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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Three concentric circles have radii and An equilateral triangle with one vertex on each circle has side length The largest

possible area of the triangle can be written as where and are positive integers, and are relatively prime,and is not divisible by the square of any prime. Find

Solution

Complex numbers and are zeros of a polynomial and Thepoints corresponding to and in the complex plane are the vertices of a right triangle with hypotenuse Find

Solution

There are mathematicians seated around a circular table with seats numbered in clockwise order. After abreak they again sit around the table. The mathematicians note that there is a positive integer such that

( ) for each the mathematician who was seated in seat before the break is seated in seat after the break (whereseat is seat );

( ) for every pair of mathematicians, the number of mathematicians sitting between them after the break, counting in boththe clockwise and the counterclockwise directions, is different from either of the number of mathematicians sittingbetween them before the break.

Find the number of possible values of with

Solution

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Mathematics Competitions (http://amc.maa.org).

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Problem 14

Problem 15

See also

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2012 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2012)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

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1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 1516 See also

Find the number of ordered pairs of positive integer solutions to the equation .

Solution

Two geometric sequences and have the same common ratio, with , , and . Find .

Solution

At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, andcomputer science. There are two male and two female professors in each department. A committee of six professors is to containthree men and three women and must also contain two professors from each of the three departments. Find the number ofpossible committees that can be formed subject to these requirements.

Solution

Ana, Bob, and Cao bike at constant rates of meters per second, meters per second, and meters per second,respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs duewest. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially

2012 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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heading south, and Cao bikes in a straight line across the field to a point on the south edge of the field. Cao arrives at point at the same time that Ana and Bob arrive at for the first time. The ratio of the field's length to the field's width to the distancefrom point to the southeast corner of the field can be represented as , where , , and are positive integers with and relatively prime. Find .

Solution

In the accompanying figure, the outer square has side length . A second square of side length is constructed inside with the same center as and with sides parallel to those of . From each midpoint of a side of , segments are drawn to thetwo closest vertices of . The result is a four-pointed starlike figure inscribed in . The star figure is cut out and then folded toform a pyramid with base . Find the volume of this pyramid.

Solution

Let be the complex number with and such that the distance between and ismaximized, and let . Find .

Solution

Let be the increasing sequence of positive integers whose binary representation has exactly ones. Let be the 1000thnumber in . Find the remainder when is divided by .

Solution

The complex numbers and satisfy the system

Find the smallest possible value of .

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Let and be real numbers such that and . The value of can be expressed in

the form , where and are relatively prime positive integers. Find .

Solution

Find the number of positive integers less than for which there exists a positive real number such that .

Note: is the greatest integer less than or equal to .

Solution

Let , and for , define . The value of that satisfies

can be expressed in the form , where and are relatively prime positive integers. Find .

Solution

For a positive integer , define the positive integer to be -safe if differs in absolute value by more than from all multiplesof . For example, the set of -safe numbers is . Find the number ofpositive integers less than or equal to which are simultaneously -safe, -safe, and -safe.

Solution

Equilateral has side length . There are four distinct triangles , , , and ,

each congruent to , with . Find .

Solution

In a group of nine people each person shakes hands with exactly two of the other people from the group. Let be the number ofways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shakehands under one arrangement do not shake hands under the other arrangement. Find the remainder when is divided by .

Solution

Triangle is inscribed in circle with , , and . The bisector of angle meets side at and circle at a second point . Let be the circle with diameter . Circles and meet at and a second point

. Then , where and are relatively prime positive integers. Find .

Solution

American Invitational Mathematics ExaminationAIME Problems and SolutionsMathematics competition resources

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

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Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=2013)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs atconstant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIMETriathlon in four and a quarter hours. How many minutes does he spend bicycling?

Solution

Find the number of five-digit positive integers, , that satisfy the following conditions:

(a) the number is divisible by

(b) the first and last digits of are equal, and

(c) the sum of the digits of is divisible by

Solution

Let be a square, and let and be points on and respectively. The line through parallel to andthe line through parallel to divide into two squares and two nonsquare rectangles. The sum of the areas of the

two squares is of the area of square Find

2013 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

In the array of squares shown below, squares are colored red, and the remaining squares are colored blue. If one of allpossible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated

around the central square is , where is a positive integer. Find .

Solution

The real root of the equation can be written in the form , where , , and

are positive integers. Find .

Solution

Melinda has three empty boxes and textbooks, three of which are mathematics textbooks. One box will hold any three of hertextbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into

these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as ,

where and are relatively prime positive integers. Find .

Solution

A rectangular box has width inches, length inches, and height inches, where and are relatively prime positive

integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a trianglewith an area of square inches. Find .

Solution

The domain of the function is a closed interval of length , where and are positive

integers and . Find the remainder when the smallest possible sum is divided by 1000.

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

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A paper equilateral triangle has side length . The paper triangle is folded so that vertex touches a point on side

a distance from point . The length of the line segment along which the triangle is folded can be written as , where ,

, and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .

Solution

There are nonzero integers , , , and such that the complex number is a zero of the polynomial . For each possible combination of and , let be the sum of the zeros of .

Find the sum of the 's for all possible combinations of and .

Solution

Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, N, of play blocks which satisfiesthe conditions:

(a) If 16, 15, or 14 students are present in the class, then in each case all the blocks can be distributed in equal numbers to eachstudent, and

(b) There are three integers such that when , , or students are present and the blocks aredistributed in equal numbers to each student, there are exactly three blocks left over.

Find the sum of the distinct prime divisors of the least possible value of N satisfying the above conditions.

Solution

Let be a triangle with and . A regular hexagon with side length 1 is drawninside so that side lies on , side lies on , and one of the remaining vertices lies on . There

are positive integers and such that the area of can be expressed in the form , where and are

relatively prime, and c is not divisible by the square of any prime. Find .

Solution

Triangle has side lengths , , and . For each positive integer , points and are located on and , respectively, creating three similar triangles

. The area of the union of all triangles for can be expressed as , where and are relatively prime positive integers. Find .

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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Solution

For , let

and

so that . Then where and are relatively prime positive integers. Find .

Solution

Let be the number of ordered triples of integers satisfying the conditions (a) , (b)there exist integers , , and , and prime where , (c) divides , , and , and(d) each ordered triple and each ordered triple form arithmetic sequences. Find .

Solution

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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Problem 15

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Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Suppose that the measurement of time during the day is converted to the metric system so that each day has metric hours,and each metric hour has metric minutes. Digital clocks would then be produced that would read just before midnight,

at midnight, at the former AM, and at the former PM. After the conversion, a person who wanted towake up at the equivalent of the former AM would set his new digital alarm clock for , where , , and are digits.Find .

Solution

Positive integers and satisfy the condition

Find the sum of all possible values of .

Solution

A large candle is centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approachesits bottom. Specifically, the candle takes seconds to burn down the first centimeter from the top, seconds to burn down thesecond centimeter, and seconds to burn down the -th centimeter. Suppose it takes seconds for the candle to burn downcompletely. Then seconds after it is lit, the candle's height in centimeters will be . Find .

2013 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

In the Cartesian plane let and . Equilateral triangle is constructed so that lies in the

first quadrant. Let be the center of . Then can be written as , where and are relativelyprime positive integers and is an integer that is not divisible by the square of any prime. Find .

Solution

In equilateral let points and trisect . Then can be expressed in the form , where

and are relatively prime positive integers, and is an integer that is not divisible by the square of any prime. Find .

Solution

Find the least positive integer such that the set of consecutive integers beginning with contains no squareof an integer.

Solution

A group of clerks is assigned the task of sorting files. Each clerk sorts at a constant rate of files per hour. At the end ofthe first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remainingclerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes thesorting in hours and minutes. Find the number of files sorted during the first one and a half hours of sorting.

Solution

A hexagon that is inscribed in a circle has side lengths , , , , , and in that order. The radius of the circle can bewritten as , where and are positive integers. Find .

Solution

A board is completely covered by tiles without overlap; each tile may cover any number of consecutive squares,and each tile lies completely on the board. Each tile is either red, blue, or green. Let be the number of tilings of the board in which all three colors are used at least once. For example, a red tile followed by a green tile, a green tile, a blue tile, and a green tile is a valid tiling. Note that if the blue tile is replaced by two bluetiles, this results in a different tiling. Find the remainder when is divided by .

Solution

Given a circle of radius , let be a point at a distance from the center of the circle. Let be the point on thecircle nearest to point . A line passing through the point intersects the circle at points and . The maximum possible area

for can be written in the form , where , , , and are positive integers, and are relatively prime, and

is not divisible by the square of any prime. Find .

Solution

Let , and let be the number of functions from set to set such that is aconstant function. Find the remainder when is divided by .

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Solution

Let be the set of all polynomials of the form , where , , and are integers. Find the number ofpolynomials in such that each of its roots satisfies either or .

Solution

In , , and point is on so that . Let be the midpoint of . Given that

and , the area of can be expressed in the form , where and are positive integersand is not divisible by the square of any prime. Find .

Solution

For positive integers and , let be the remainder when is divided by , and for let

. Find the remainder when is divided by .

Solution

Let be angles of an acute triangle with

There are positive integers , , , and for which

where and are relatively prime and is not divisible by the square of any prime. Find .

Solution

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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Problem 12

Problem 13

Problem 14

Problem 15

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m/Forum/resources.php?c=182&cid=45&year=2014)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has awidth of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between thevertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyeletsat the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace mustextend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.

Solution

An urn contains green balls and blue balls. A second urn contains green balls and blue balls. A single ball is drawn atrandom from each urn. The probability that both balls are of the same color is . Find .

2014 AIME I Problems

Contents

Problem 1

Problem 2

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Solution

Find the number of rational numbers , such that when is written as a fraction in lowest terms, the numerator andthe denominator have a sum of .

Solution

Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rideseast at miles per hour, and Steve rides west at miles per hour. Two trains of equal length, traveling in opposite directions atconstant but different speeds each pass the two riders. Each train takes exactly minute to go past Jon. The westbound traintakes times as long as the eastbound train to go past Steve. The length of each train is , where and are relatively primepositive integers. Find .

Solution

Let the set consist of the twelve vertices of a regular -gon. A subset of is called communalif there is a circle such that all points of are inside the circle, and all points of not in are outside of the circle. How manycommunal subsets are there? (Note that the empty set is a communal subset.)

Solution

The graphs and have y-intercepts of and , respectively, and eachgraph has two positive integer x-intercepts. Find .

Solution

Let and be complex numbers such that and . Let . The maximum possible value of

can be written as , where and are relatively prime positive integers. Find . (Note that , for w ,

denotes the measure of the angle that the ray from to makes with the positive real axis in the complex plane.

Solution

The positive integers and both end in the same sequence of four digits when written in base 10, where digit is notzero. Find the three-digit number .

Solution

Let be the three real roots of the equation . Find .

Solution

A disk with radius is externally tangent to a disk with radius . Let be the point where the disks are tangent, be the centerof the smaller disk, and be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to rollalong the outside of the larger disk until the smaller disk has turned through an angle of . That is, if the center of the smallerdisk has moved to the point , and the point on the smaller disk that began at has now moved to point , then is parallelto . Then , where and are relatively prime positive integers. Find .

Solution

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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A token starts at the point of an -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in adirection parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions andindependently of the other moves. The probability the token ends at a point on the graph of is , where and arerelatively prime positive integers. Find .

Solution

Let , and and be randomly chosen (not necessarily distinct) functions from to . The probability thatthe range of and the range of are disjoint is , where and are relatively prime positive integers. Find .

Solution

On square , points , and lie on sides and respectively, so that and . Segments and intersect at a point , and the areas of the quadrilaterals

and are in the ratio Find the area of square .

Solution

Let be the largest real solution to the equation

There are positive integers and such that . Find .

Solution

In and . Circle intersects at and at and and at

and . Given that and length where and are relatively prime positive

integers, and is a positive integer not divisible by the square of any prime. Find .

Solution

Problem 12

Problem 13

Problem 14

Problem 15

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Copyright © 2020 Art of Problem Solving

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=2014)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe beginsto paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room ispainted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abebegins for the three of them to finish painting the room.

Solution

Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of thethree factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the

third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is . The probability

that a man has none of the three risk factors given that he does not have risk factor A is , where and are relatively prime

positive integers. Find .

Solution

A rectangle has sides of length and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side oflength 36. The sides of length can be pressed toward each other keeping those two sides parallel so the rectangle becomes aconvex hexagon as shown. When the figure is a hexagon with the sides of length parallel and separated by a distance of 24, the

2014 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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hexagon has the same area as the original rectangle. Find .

Solution

The repeating decimals and satisfy

where , , and are (not necessarily distinct) digits. Find the three digit number .

Solution

Real numbers and are roots of , and and are roots of . Find the sum of all possible values of .

Solution

Charles has two six-sided dice. One of the die is fair, and the other die is biased so that it comes up six with probability and each

of the other five sides has probability . Charles chooses one of the two dice at random and rolls it three times. Given that the

first two rolls are both sixes, the probability that the third roll will also be a six is , where and are relatively prime positive

integers. Find .

Solution

Let . Find the sum of all positive integers for which

Solution

Circle with radius 2 has diameter . Circle is internally tangent to circle at . Circle is internally tangent to circle , externally tangent to circle , and tangent to . The radius of circle is three times the radius of circle , and can be

written in the form , where and are positive integers. Find .

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

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Solution

Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.

Solution

Let be a complex number with . Let be the polygon in the complex plane whose vertices are and every

such that . Then the area enclosed by can be written in the form , where is an integer. Find the

remainder when is divided by .

Solution

In , and . . Let be the midpoint of segment . Point lieson side such that . Extend segment through to point such that . Then

, where and are relatively prime positive integers, and is a positive integer. Find .

Solution

Suppose that the angles of satisfy . Two sides of the triangle havelengths 10 and 13. There is a positive integer so that the maximum possible length for the remaining side of is

. Find .

Solution

Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with aright shoe without regard to which shoes belong together. The probability that for every positive integer , no collection of

pairs made by the child contains the shoes from exactly of the adults is , where and are relatively prime positive

integers. Find .

Solution

In , , , and . Let , , and be points on line such that , , and . Point is the midpoint of segment , and point is on

ray such that . Then , where and are relatively prime positive integers. Find .

Solution

For any integer , let be the smallest prime which does not divide . Define the integer function to be theproduct of all primes less than if , and if . Let be the sequence defined by

, and for . Find the smallest positive integer such that .

Solution

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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Preceded by2014 AIME I

Followed by2015 AIME I

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=2015)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

The expressions = and = are obtained by writing multiplication and addition operators in an

alternating pattern between successive integers. Find the positive difference between integers and .

Solution

The nine delegates to the Economic Cooperation Conference include officials from Mexico, officials from Canada, and officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers

were determined randomly, the probability that exactly two of the sleepers are from the same country is , where and are

relatively prime positive integers. Find .

Solution

There is a prime number such that is the cube of a positive integer. Find .

Solution

2015 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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Point lies on line segment with and . Points and lie on the same side of line formingequilateral triangles and . Let be the midpoint of , and be the midpoint of . The area of

is . Find .

Solution

In a drawer Sandy has pairs of socks, each pair a different color. On Monday, Sandy selects two individual socks at random fromthe socks in the drawer. On Tuesday Sandy selects of the remaining socks at random, and on Wednesday two of the

remaining socks at random. The probability that Wednesday is the first day Sandy selects matching socks is , where and

are relatively prime positive integers. Find .

Solution

Point and are equally spaced on a minor arc of a circle. Points and are equally spaced ona minor arc of a second circle with center as shown in the figure below. The angle exceeds by . Findthe degree measure of .

Solution

In the diagram below, is a square. Point is the midpoint of . Points and lie on , and and lie on and , respectively, so that is a square. Points and lie on , and and lie on and ,

respectively, so that is a square. The area of is 99. Find the area of .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

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For positive integer , let denote the sum of the digits of . Find the smallest positive integer satisfying .

Solution

Let be the set of all ordered triple of integers with . Each ordered triple in generatesa sequence according to the rule for all . Find the number of such sequences forwhich for some .

Solution

Let be a third-degree polynomial with real coefficients satisfying

Find .

Solution

Triangle has positive integer side lengths with . Let be the intersection of the bisectors of and .Suppose . Find the smallest possible perimeter of .

Solution

Consider all 1000-element subsets of the set . From each such subset choose the least element. The

arithmetic mean of all of these least elements is , where and are relatively prime positive integers. Find .

Solution

With all angles measured in degrees, the product , where and are integers greater than 1.

Find .

Solution

For each integer , let be the area of the region in the coordinate plane defined by the inequalities and

, where is the greatest integer not exceeding . Find the number of values of with

for which is an integer.

Solution

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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A block of wood has the shape of a right circular cylinder with radius and height , and its entire surface has been painted blue.

Points and are chosen on the edge of one of the circular faces of the cylinder so that on that face measures . Theblock is then sliced in half along the plane that passes through point , point , and the center of the cylinder, revealing a flat,unpainted face on each half. The area of one of these unpainted faces is , where , , and are integers and isnot divisible by the square of any prime. Find .

Solution

2015 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2015))

Preceded by2014 AIME II

Followed by2015 AIME II

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=2015)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Let be the least positive integer that is both percent less than one integer and percent greater than another integer. Findthe remainder when is divided by .

Solution

In a new school, percent of the students are freshmen, percent are sophomores, percent are juniors, and percent areseniors. All freshmen are required to take Latin, and percent of sophomores, percent of the juniors, and percent of the

seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is , where and are

relatively prime positive integers. Find .

Solution

Let be the least positive integer divisible by whose digits sum to . Find .

Solution

In an isosceles trapezoid, the parallel bases have lengths and , and the altitude to these bases has length . The perimeter of the trapezoid can be written in the form , where and are positive integers. Find .

2015 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Solution

Two unit squares are selected at random without replacement from an grid of unit squares. Find the least positive integer

such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than .

Solution

Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form for some positive integers and . Can you tell me the values of and ?"

After some calculations, Jon says, "There is more than one such polynomial."

Steve says, "You're right. Here is the value of ." He writes down a positive integer and asks, "Can you tell me the value of ?"

Jon says, "There are still two possible values of ."

Find the sum of the two possible values of .

Solution

Triangle has side lengths , , and . Rectangle has vertex on ,vertex on , and vertices and on . In terms of the side length , the area of can beexpressed as the quadratic polynomial

Then the coefficient , where and are relatively prime positive integers. Find .

Solution

Let and be positive integers satisfying . The maximum possible value of is , where and are

relatively prime positive integers. Find .

Solution

A cylindrical barrel with radius feet and height feet is full of water. A solid cube with side length feet is set into the barrel sothat the diagonal of the cube is vertical. The volume of water thus displaced is cubic feet. Find .

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Solution

Call a permutation of the integers quasi-increasing if for each . For example, and are quasi-increasing permutations of the integers , but

is not. Find the number of quasi-increasing permutations of the integers .

Solution

The circumcircle of acute has center . The line passing through point perpendicular to intersects lines

and at and , respectively. Also , , , and , where and are relatively

prime positive integers. Find .

Solution

There are possible -letter strings in which each letter is either an A or a B. Find the number of such strings thatdo not have more than adjacent letters that are identical.

Solution

Define the sequence by , where represents radian measure. Find the index of the 100th

term for which .

Solution

Let and be real numbers satisfying and . Evaluate .

Solution

Circles and have radii and , respectively, and are externally tangent at point . Point is on and point is on such that is a common external tangent of the two circles. A line through intersects again at and intersects again at . Points and lie on the same side of , and the areas of and are equal. This common area

is , where and are relatively prime positive integers. Find .

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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Solution

2015 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2015))

Preceded by2015 AIME I

Followed by2016 AIME I

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems&oldid=99772"

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m/Forum/resources.php?c=182&cid=45&year=2016)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

For , let denote the sum of the geometric series

Let between and satisfy . Find .

Solution

Two dice appear to be normal dice with their faces numbered from to , but each die is weighted so that the probability of rolling

the number is directly proportional to . The probability of rolling a with this pair of dice is , where and are relatively

prime positive integers. Find .

Solution

A regular icosahedron is a -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. Theregular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices alladjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottomvertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part ofa path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.

2016 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

A right prism with height has bases that are regular hexagons with sides of length . A vertex of the prism and its threeadjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by theface of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain measures . Find .

Solution

Anh read a book. On the first day she read pages in minutes, where and are positive integers. On the second day Anh read pages in minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her

one more minute than on the previous day until she completely read the page book. It took her a total of minutes to readthe book. Find .

Solution

In let be the center of the inscribed circle, and let the bisector of intersect at . The line through

and intersects the circumscribed circle of at the two points and . If and , then

, where and are relatively prime positive integers. Find .

Solution

For integers and consider the complex number

Find the number of ordered pairs of integers such that this complex number is a real number.

Solution

For a permutation of the digits , let denote the sum of the three -digit numbers , , and . Let be the minimum value of subject to the condition that the units digit of is .

Let denote the number of permutations with . Find .

Solution

Triangle has and . This triangle is inscribed in rectangle with on

and on . Find the maximum possible area of .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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A strictly increasing sequence of positive integers , , , has the property that for every positive integer , thesubsequence , , is geometric and the subsequence , , is arithmetic. Suppose that

. Find .

Solution

Let be a nonzero polynomial such that for every real , and

. Then , where and are relatively prime positive integers. Find .

Solution

Find the least positive integer such that is a product of at least four not necessarily distinct primes.

Solution

Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line . A fence islocated at the horizontal line . On each jump Freddy randomly chooses a direction parallel to one of the coordinate axesand moves one unit in that direction. When he is at a point where , with equal likelihoods he chooses one of threedirections where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that wouldhave him cross over the fence to where . Freddy starts his search at the point and will stop once he reaches apoint on the river. Find the expected number of jumps it will take Freddy to reach the river.

Solution

Centered at each lattice point in the coordinate plane are a circle radius and a square with sides of length whose sides are

parallel to the coordinate axes. The line segment from to intersects of the squares and of the circles.Find .

Solution

Circles and intersect at points and . Line is tangent to and at and , respectively, with line closer topoint than to . Circle passes through and intersecting again at and intersecting again at .The three points , , are collinear, , , and . Find .

Solution

2016 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2016))

Preceded by2015 AIME II

Followed by2016 AIME II

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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2016 AIME II (Answer Key)| AoPS Contest Collections (http://www.artofproblemsolving.co

m/Forum/resources.php?c=182&cid=45&year=2016)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Initially Alex, Betty, and Charlie had a total of peanuts. Charlie had the most peanuts, and Alex had the least. The threenumbers of peanuts that each person had formed a geometric progression. Alex eats of his peanuts, Betty eats of her peanuts,and Charlie eats of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find thenumber of peanuts Alex had initially.

Solution

There is a chance of rain on Saturday and a chance of rain on Sunday. However, it is twice as likely to rain on Sunday if

it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is , where and

are relatively prime positive integers. Find .

Solution

Let and be real numbers satisfying the system

2016 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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Find the value of .

Solution

An rectangular box is built from unit cubes. Each unit cube is colored red, green, or yellow. Each of the layers of size parallel to the faces of the box contains exactly red cubes, exactly green cubes, andsome yellow cubes. Each of the layers of size parallel to the faces of the box contains exactly greencubes, exactly yellow cubes, and some red cubes. Find the smallest possible volume of the box.

Solution

Triangle has a right angle at . Its side lengths are pairwise relatively prime positive integers, and its perimeter is . Let be the foot of the altitude to , and for , let be the foot of the altitude to in . The

sum . Find .

Solution

For polynomial , define . Then

, where and are relatively prime positive integers. Find .

Solution

Squares and have a common center and . The area of is 2016, and the area of is a smaller positive integer. Square is constructed so that each of its vertices lies on a side of

and each vertex of lies on a side of . Find the difference between the largest and smallest positive integervalues for the area of .

Solution

Find the number of sets of three distinct positive integers with the property that the product of and is equal tothe product of and .

Solution

The sequences of positive integers and are an increasing arithmetic sequence and an increasinggeometric sequence, respectively. Let . There is an integer such that and .Find .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Triangle is inscribed in circle . Points and are on side with . Rays and meet again at and (other than ), respectively. If and , then

, where and are relatively prime positive integers. Find .

Solution

For positive integers and , define to be -nice if there exists a positive integer such that has exactly positivedivisors. Find the number of positive integers less than that are neither -nice nor -nice.

Solution

The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available andyou will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no twoadjacent sections can be painted with the same color.

Solution

Beatrix is going to place six rooks on a chessboard where both the rows and columns are labeled to ; the rooks areplaced so that no two rooks are in the same row or the same column. The of a square is the sum of its row number andcolumn number. The of an arrangement of rooks is the least value of any occupied square.The average score over all valid

configurations is , where and are relatively prime positive integers. Find .

Solution

Equilateral has side length . Points and lie outside the plane of and are on opposite sides of theplane. Furthermore, , and , and the planes of and form a

dihedral angle (the angle between the two planes). There is a point whose distance from each of and is . Find .

Solution

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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Copyright © 2020 Art of Problem Solving

For let and . Let be positive real numbers such that

and . The maximum possible value of , where and are

relatively prime positive integers. Find .

Solution

2016 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2016))

Preceded by2016 AIME I

Followed by2017 AIME I

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems&oldid=103716"

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m/Forum/resources.php?c=182&cid=45&year=2017)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Fifteen distinct points are designated on : the 3 vertices , , and ; other points on side ; other points onside ; and other points on side . Find the number of triangles with positive area whose vertices are among these points.

Solution

When each of , , and is divided by the positive integer , the remainder is always the positive integer . When eachof , , and is divided by the positive integer , the remainder is always the positive integer . Find

.

Solution

For a positive integer , let be the units digit of . Find the remainder when

is divided by .

Solution

2017 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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A pyramid has a triangular base with side lengths , , and . The three edges of the pyramid from the three corners of thebase to the fourth vertex of the pyramid all have length . The volume of the pyramid is , where and are positiveintegers, and is not divisible by the square of any prime. Find .

Solution

A rational number written in base eight is , where all digits are nonzero. The same number in base twelve is . Find thebase-ten number .

Solution

A circle circumscribes an isosceles triangle whose two congruent angles have degree measure . Two points are chosenindependently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects

the triangle is . Find the difference between the largest and smallest possible values of .

Solution

For nonnegative integers and with , let . Let denote the sum of all

, where and are nonnegative integers with . Find the remainder when is divided by .

Solution

Two real numbers and are chosen independently and uniformly at random from the interval . Let and be twopoints on the plane with . Let and be on the same side of line such that the degree measures of

and are and respectively, and and are both right angles. The probability that

is equal to , where and are relatively prime positive integers. Find .

Solution

Let , and for each integer let . Find the least such that is a multipleof .

Solution

Let , and where . Let be the unique complex number

with the properties that is a real number and the imaginary part of is the greatest possible. Find the real

part of .

Solution

Consider arrangements of the numbers in a array. For each such arrangement, let , , and bethe medians of the numbers in rows , , and respectively, and let be the median of . Let be the number ofarrangements for which . Find the remainder when is divided by .

Solution

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Call a set product-free if there do not exist (not necessarily distinct) such that . For example, the emptyset and the set are product-free, whereas the sets and are not product-free. Find the number ofproduct-free subsets of the set .

Solution

For every , let be the least positive integer with the following property: For every , there is always aperfect cube in the range . Find the remainder when

is divided by 1000.

Solution

Let and satisfy and . Findthe remainder when is divided by .

Solution

The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths , ,

and , as shown, is , where , , and are positive integers, and are relatively prime, and is not divisible by thesquare of any prime. Find .

Solution

2017 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2017))

Preceded by2016 AIME II

Followed by2017 AIME II

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Problem 12

Problem 13

Problem 14

Problem 15

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m/Forum/resources.php?c=182&cid=45&year=2017)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Find the number of subsets of that are subsets of neither nor .

Solution

Teams , , , and are in the playoffs. In the semifinal matches, plays , and plays . The winners of those twomatches will play each other in the final match to determine the champion. When plays , the probability that wins is

, and the outcomes of all the matches are independent. The probability that will be the champion is , where and

are relatively prime positive integers. Find .

Solution

A triangle has vertices , , and . The probability that a randomly chosen point inside the triangle is

closer to vertex than to either vertex or vertex can be written as , where and are relatively prime positive integers.

Find .

Solution

2017 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Find the number of positive integers less than or equal to whose base-three representation contains no digit equal to .

Solution

A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are , , , , , and . Find the greatest possible value of .

Solution

Find the sum of all positive integers such that is an integer.

Solution

Find the number of integer values of in the closed interval for which the equation has exactly one real solution.

Solution

Find the number of positive integers less than such that

is an integer.

Solution

A special deck of cards contains cards, each labeled with a number from to and colored with one of seven colors. Eachnumber-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given thatshe gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her

cards and have at least one card of each color and at least one card with each number is , where and are relatively

prime positive integers. Find .

Solution

Rectangle has side lengths and . Point is the midpoint of , point is the trisectionpoint of closer to , and point is the intersection of and . Point lies on the quadrilateral , and

bisects the area of . Find the area of .

Solution

Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of waysthere are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads(possibly passing through other towns on the way).

Solution

Circle has radius , and the point is a point on the circle. Circle has radius and is internally tangent to atpoint . Point lies on circle so that is located counterclockwise from on . Circle has radius and isinternally tangent to at point . In this way a sequence of circles and a sequence of points on the circles

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

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are constructed, where circle has radius and is internally tangent to circle at point , andpoint lies on counterclockwise from point , as shown in the figure below. There is one point inside all of

these circles. When , the distance from the center to is , where and are relatively prime positive integers.

Find .

Solution

For each integer , let be the number of -element subsets of the vertices of a regular -gon that are the vertices ofan isosceles triangle (including equilateral triangles). Find the sum of all values of such that .

Solution

A grid of points consists of all points in space of the form , where , , and are integers between and , inclusive. Find the number of different lines that contain exactly of these points.

Solution

Tetrahedron has , , and . For any point inspace, define . The least possible value of can be expressed as ,where and are positive integers, and is not divisible by the square of any prime. Find .

Solution

2017 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2017))

Preceded by2017 AIME I

Followed by2018 AIME I

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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Problem 13

Problem 14

Problem 15

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m/Forum/resources.php?c=182&cid=45&year=2018)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Let be the number of ordered pairs of integers with and such that the polynomial can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the

remainder when is divided by .

Solution

The number can be written in base as , can be written in base as , and can be written in base as , where . Find the base- representation of .

Solution

Kathy has red cards and green cards. She shuffles the cards and lays out of the cards in a row in a random order. She willbe happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders

RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is , where and

are relatively prime positive integers. Find .

Solution

2018 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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In and . Point lies strictly between and on and point lies strictly

between and on so that . Then can be expressed in the form , where and are

relatively prime positive integers. Find .

Solution

For each ordered pair of real numbers satisfying

there is a real number such that

Find the product of all possible values of .

Solution

Let be the number of complex numbers with the properties that and is a real number. Find theremainder when is divided by .

Solution

A right hexagonal prism has height . The bases are regular hexagons with side length . Any of the vertices determine atriangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Solution

Let be an equiangular hexagon such that , and . Denote the diameter of the largest circle that fits inside the hexagon. Find .

Solution

Find the number of four-element subsets of with the property that two distinct elements of a subset havea sum of , and two distinct elements of a subset have a sum of . For example, and aretwo such subsets.

Solution

The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bugwalks along the wheel, starting at point . At every step of the process, the bug walks from one labeled point to an adjacentlabeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug onlywalks in a clockwise direction. For example, the bug could travel along the path , which has steps.Let be the number of paths with steps that begin and end at point Find the remainder when is divided by .

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Solution

Find the least positive integer such that when is written in base , its two right-most digits in base are .

Solution

For every subset of , let be the sum of the elements of , with defined to be . If

is chosen at random among all subsets of , the probability that is divisible by is , where and are relatively

prime positive integers. Find .

Solution

Let have side lengths , , and . Point lies in the interior of , and points and are the incenters of and , respectively. Find the minimum possible area of as variesalong .

Solution

Let be a heptagon. A frog starts jumping at vertex . From any vertex of the heptagon except , the frogmay jump to either of the two adjacent vertices. When it reaches vertex , the frog stops and stays there. Find the number ofdistinct sequences of jumps of no more than jumps that end at .

Solution

David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, , which can each be inscribed in a circle with radius . Let denote the measure of the acute angle made by the

diagonals of quadrilateral , and define and similarly. Suppose that , , and

. All three quadrilaterals have the same area , which can be written in the form , where and are relatively

prime positive integers. Find .

Solution

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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2018 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2018))

Preceded by2017 AIME II

Followed by2018 AIME II

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

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m/Forum/resources.php?c=182&cid=45&year=2018)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Points , , and lie in that order along a straight path where the distance from to is meters. Ina runs twice asfast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at and runningtoward , Paul starting at and running toward , and Eve starting at and running toward . When Paul meets Eve, he turnsaround and runs toward . Paul and Ina both arrive at at the same time. Find the number of meters from to .

Solution

Let , , and , and for define recursively to be the remainder when is divided by . Find .

Solution

Find the sum of all positive integers such that the base- integer is a perfect square and the base- integer is a perfect cube.

Solution

2018 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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In equiangular octagon , and . The self-intersecting octagon enclosed six non-overlapping

triangular regions. Let be the area enclosed by , that is, the total area of the six triangular regions. Then

, where and are relatively prime positive integers. Find .

Solution

Suppose that , , and are complex numbers such that , , and , where . Then there are real numbers and such that . Find .

Solution

A real number is chosen randomly and uniformly from the interval . The probability that the roots of the polynomial

are all real can be written in the form , where and are relatively prime positive integers. Find .

Solution

Triangle has side lengths , , and . Points are on segment with between and for , and points

are on segment with between and for

. Furthermore, each segment , , is parallel to . The segments cut thetriangle into regions, consisting of trapezoids and triangle. Each of the regions has the same area. Find thenumber of segments , , that have rational length.

Solution

A frog is positioned at the origin of the coordinate plane. From the point , the frog can jump to any of the points , , , or . Find the number of distinct sequences of jumps in which the frog

begins at and ends at .

Solution

Octagon with side lengths and is formed by removing 6-8-10 triangles from the corners of a rectangle with

side on a short side of the rectangle, as shown. Let be the midpoint of , and partition the octagon into 7 triangles bydrawing segments , , , , , and . Find the area of the convex polygon whose vertices are the centroids ofthese 7 triangles.

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Solution

Find the number of functions from to that satisfy forall in .

Solution

Find the number of permutations of such that for each with , at least one of the first terms of thepermutation is greater than .

Solution

Let be a convex quadrilateral with , , and . Assume that thediagonals of intersect at point , and that the sum of the areas of triangles and equals the sum of theareas of triangles and . Find the area of quadrilateral .

Solution

Misha rolls a standard, fair six-sided die until she rolls 1-2-3 in that order on three consecutive rolls. The probability that she will roll

the die an odd number of times is where and are relatively prime positive integers. Find .

Solution

The incircle of triangle is tangent to at . Let be the other intersection of with . Points and

lie on and , respectively, so that is tangent to at . Assume that , , , and

, where and are relatively prime positive integers. Find .

Solution

Find the number of functions from to the integers such that , , and

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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for all and in .

Solution

2018 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2018))

Preceded by2018 AIME I

Followed by2019 AIME I

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems&oldid=94943"

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m/Forum/resources.php?c=182&cid=45&year=2019)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Consider the integer

Find the sum of the digits of .

Solution

Jenn randomly chooses a number from . Bela then randomly chooses a number from distinct from . The value of is at least with a probability that can be expressed in the form

, where and are relatively prime positive integers. Find .

Solution

In , , , and . Points and lie on , points and lie on , andpoints and lie on , with . Find the area of hexagon

.

Solution

2019 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

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A soccer team has available players. A fixed set of players starts the game, while the other are available as substitutes.During the game, the coach may make as many as substitutions, where any one of the players in the game is replaced by oneof the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may bereplaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter.Let be the number of ways the coach can make substitutions during the game (including the possibility of making nosubstitutions). Find the remainder when is divided by .

Solution

A moving particle starts at the point and moves until it hits one of the coordinate axes for the first time. When the particleis at the point , it moves at random to one of the points , , or , each withprobability , independently of its previous moves. The probability that it will hit the coordinate axes at is , where and

are positive integers. Find .

Solution

In convex quadrilateral side is perpendicular to diagonal , side is perpendicular to diagonal , , and . The line through perpendicular to side intersects diagonal at with

. Find .

Solution

There are positive integers and that satisfy the system of equations

Let be the number of (not necessarily distinct) prime factors in the prime factorization of , and let be the number of (notnecessarily distinct) prime factors in the prime factorization of . Find .

Solution

Let be a real number such that . Then where and arerelatively prime positive integers. Find .

Solution

Let denote the number of positive integer divisors of . Find the sum of the six least positive integers that are solutionsto .

Solution

For distinct complex numbers , the polynomial

can be expressed as , where is a polynomial with complex coefficients andwith degree at most . The value of

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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can be expressed in the form , where and are relatively prime positive integers. Find .

Solution

In , the sides have integer lengths and . Circle has its center at the incenter of . An excircleof is a circle in the exterior of that is tangent to one side of the triangle and tangent to the extensions of theother two sides. Suppose that the excircle tangent to is internally tangent to , and the other two excircles are bothexternally tangent to . Find the minimum possible value of the perimeter of .

Solution

Given , there are complex numbers with the property that , , and are the vertices of aright triangle in the complex plane with a right angle at . There are positive integers and such that one such value of is . Find .

Solution

Triangle has side lengths , , and . Points and are on ray with . The point is a point of intersection of the circumcircles of and

satisfying and . Then can be expressed as , where , , , and are positive integers suchthat and are relatively prime, and is not divisible by the square of any prime. Find .

Solution

Find the least odd prime factor of .

Solution

Let be a chord of a circle , and let be a point on the chord . Circle passes through and and is internallytangent to . Circle passes through and and is internally tangent to . Circles and intersect at points and .Line intersects at and . Assume that , , , and , where and arerelatively prime positive integers. Find .

Solution

2019 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2019))

Preceded by2018 AIME II

Followed by2019 AIME II

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All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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m/Forum/resources.php?c=182&cid=45&year=2019)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Two different points, and , lie on the same side of line so that and are congruent with , and . The intersection of these two triangular regions has area ,

where and are relatively prime positive integers. Find .

Solution

Lily pads lie in a row on a pond. A frog makes a sequence of jumps starting on pad . From any pad the frog jumpsto either pad or pad chosen randomly with probability and independently of other jumps. The probability that thefrog visits pad is , where and are relatively prime positive integers. Find .

Solution

Find the number of -tuples of positive integers that satisfy the following system of equations:

2019 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

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Solution

A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is ,where and are relatively prime positive integers. Find .

Solution

Four ambassadors and one advisor for each of them are to be seated at a round table with chairs numbered in order to .Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. Thereare ways for the people to be seated at the table under these conditions. Find the remainder when is divided by .

Solution

In a Martian civilization, all logarithms whose bases are not specified as assumed to be base , for some fixed . A Martianstudent writes down

and finds that this system of equations has a single real number solution . Find .

Solution

Triangle has side lengths , and . Lines , and are drawn parallel to , and , respectively, such that the intersections of , and with the interior of are segments of

lengths , and , respectively. Find the perimeter of the triangle whose sides lie on lines , and .

Solution

The polynomial has real coefficients not exceeding and

. Find the remainder when is divided by .

Solution

Call a positive integer -pretty if has exactly positive divisors and is divisible by . For example, is -pretty. Let bethe sum of the positive integers less than that are -pretty. Find .

Solution

There is a unique angle between and such that for nonnegative integers the value of is positive when is a multiple of , and negative otherwise. The degree measure of is , where and are relatively prime positive integers. Find

.

Solution

Triangle has side lengths and Circle passes through and is tangent to line at Circle passes through and is tangent to line at Let be the intersection of circles and not

equal to Then where and are relatively prime positive integers. Find

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Solution

For call a finite sequence of positive integers progressive if and divides for . Find the number of progressive sequences such that the sum of the terms in the sequence is equal to

Solution

Regular octagon is inscribed in a circle of area Point lies inside the circle so that the region

bounded by and the minor arc of the circle has area while the region bounded by and

the minor arc of the circle has area There is a positive integer such that the area of the region bounded by

and the minor arc of the circle is equal to Find

Solution

Find the sum of all positive integers such that, given an unlimited supply of stamps of denominations and cents, cents is the greatest postage that cannot be formed.

Solution

In acute triangle points and are the feet of the perpendiculars from to and from to , respectively.Line intersects the circumcircle of in two distinct points, and . Suppose , , and

. The value of can be written in the form where and are positive relatively prime integers.Find .

Solution

2019 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2019))

Preceded by2019 AIME I

Followed by2020 AIME I

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems&oldid=113134"

Problem 12

Problem 13

Problem 14

Problem 15

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m/Forum/resources.php?c=182&cid=45&year=2020)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

In with point lies strictly between and on side and point lies strictly between and on side such that The degree measure of is where and are

relatively prime positive integers. Find

Solution

There is a unique positive real number such that the three numbers and in that order, form ageometric progression with positive common ratio. The number can be written as where and are relatively primepositive integers. Find

Solution

A positive integer has base-eleven representation and base-eight representation where and represent(not necessarily distinct) digits. Find the least such expressed in base ten.

Solution

2020 AIME I Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Let be the set of positive integers with the property that the last four digits of are and when the last four digitsare removed, the result is a divisor of For example, is in because is a divisor of Find the sum of all thedigits of all the numbers in For example, the number contributes to this total.

Solution

Six cards numbered through are to be lined up in a row. Find the number of arrangements of these six cards where one of thecards can be removed leaving the remaining five cards in either ascending or descending order.

Solution

A flat board has a circular hole with radius and a circular hole with radius such that the distance between the centers of thetwo holes is . Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of theradius of the spheres is , where and are relatively prime positive integers. Find .

Solution

A club consisting of men and women needs to choose a committee from among its members so that the number ofwomen on the committee is one more than the number of men on the committee. The committee could have as few as memberor as many as members. Let be the number of such committees that can be formed. Find the sum of the prime numbersthat divide

Solution

A bug walks all day and sleeps all night. On the first day, it starts at point faces east, and walks a distance of units due east.Each night the bug rotates counterclockwise. Each day it walks in this new direction half as far as it walked the previous day.The bug gets arbitrarily close to the point Then where and are relatively prime positive integers. Find

Solution

Let be the set of positive integer divisors of Three numbers are chosen independently and at random with replacementfrom the set and labeled and in the order they are chosen. The probability that both divides and divides

is where and are relatively prime positive integers. Find

Solution

Let and be positive integers satisfying the conditions

is a multiple of and

is not a multiple of

Find the least possible value of

Solution

For integers and let and Find the number of ordered triples of integers with absolute values not exceeding for which there is an integer such that

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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Let be the least positive integer for which is divisible by Find the number of positive integerdivisors of

Solution

Point lies on side of so that bisects The perpendicular bisector of intersects thebisectors of and in points and respectively. Given that and the

area of can be written as where and are relatively prime positive integers, and is a positive integer notdivisible by the square of any prime. Find

Solution

Let be a quadratic polynomial with complex coefficients whose coefficient is Suppose the equation has four distinct solutions, Find the sum of all possible values of

Solution

Let be an acute triangle with circumcircle and let be the intersection of the altitudes of Suppose thetangent to the circumcircle of at intersects at points and with and The area of can be written as where and are positive integers, and is not divisible by the square of anyprime. Find

Solution

2020 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2020))

Preceded by2019 AIME II Problems

Followed by2020 AIME II Problems

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems&oldid=123366"

Problem 12

Problem 13

Problem 14

Problem 15

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m/Forum/resources.php?c=182&cid=45&year=2020)

Instructions

1. This is a 15-question, 3-hour examination. All answers are integersranging from to , inclusive. Your score will be the number ofcorrect answers; i.e., there is neither partial credit nor a penalty forwrong answers.

2. No aids other than scratch paper, graph paper, ruler, compass, andprotractor are permitted. In particular, calculators and computers are notpermitted.

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

1 Problem 12 Problem 23 Problem 34 Problem 45 Problem 56 Problem 67 Problem 78 Problem 89 Problem 910 Problem 1011 Problem 1112 Problem 1213 Problem 1314 Problem 1415 Problem 15

Find the number of ordered pairs of positive integers such that .

Solution

Let be a point chosen uniformly at random in the interior of the unit square with vertices at , and

. The probability that the slope of the line determined by and the point is greater than can be written as ,

where and are relatively prime positive integers. Find .

Solution

The value of that satisfies can be written as , where and are relatively prime positive

integers. Find .

Solution

2020 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

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Triangles and lie in the coordinate plane with vertices , , , , , . A rotation of degrees clockwise around the point where , will transform

to . Find .

Solution

For each positive integer , let be the sum of the digits in the base-four representation of and let be the sum of thedigits in the base-eight representation of . For example, , and

. Let be the least value of such that the base-sixteen representation of cannot be expressed using only the digits through . Find the remainder when is divided by .

Solution

Define a sequence recursively by , , and

for all . Then can be written as , where and are relatively prime positive integers. Find .

Solution

Two congruent right circular cones each with base radius and height have the axes of symmetry that intersect at right anglesat a point in the interior of the cones a distance from the base of each cone. A sphere with radius lies within both cones. The

maximum possible value of is , where and are relatively prime positive integers. Find .

Solution

Define a sequence recursively by and for integers . Find the leastvalue of such that the sum of the zeros of exceeds .

Solution

While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, theywent to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next toeach other before the break, then they did not sit next to each other after the break. Find the number of possible seating ordersthey could have chosen after the break.

Solution

Find the sum of all positive integers such that when is divided by , the remainder is .

Solution

Let , and let and be two quadratic polynomials also with the coefficient of equal to . David computes each of the three sums , , and and is surprised to find that each pair of these sums

has a common root, and these three common roots are distinct. If , then , where and are relatively

prime positive integers. Find .

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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Problem 11

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Let and be odd integers greater than An rectangle is made up of unit squares where the squares in the top roware numbered left to right with the integers through , those in the second row are numbered left to right with the integers

through , and so on. Square is in the top row, and square is in the bottom row. Find the number of orderedpairs of odd integers greater than with the property that, in the rectangle, the line through the centers ofsquares and intersects the interior of square .

Solution

Convex pentagon has side lengths , , and . Moreover, thepentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of .

Solution

For real number let be the greatest integer less than or equal to , and define to be the fractional partof . For example, and . Define , and let be the number of real-valuedsolutions to the equation for . Find the remainder when is divided by .

Solution

Let be an acute scalene triangle with circumcircle . The tangents to at and intersect at . Let and bethe projections of onto lines and , respectively. Suppose , , and

. Find .

Solution

2020 AIME II (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2020))

Preceded by2020 AIME I Problems

Followed by2021 AIME I Problems

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All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American

Mathematics Competitions (http://amc.maa.org).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems&oldid=125732"

Problem 12

Problem 13

Problem 14

Problem 15

AIME 2020 Enunciados y soluciones

Existe copia de este documento en formato Word: http://www.toomates.net/media/AIME2020.doc

AIME I 2020 Enunciados

1

En el triángulo ABC con ACAB , sea D un punto en el interior del lado AC y sea

E un punto en el interior del lado AB , de forma que BCDBEDAE . Determina

el ángulo ABC .

2

Existe un único número real positivo x tal que los tres números

)2(log8 x , x4log , x2log

en este orden, forman una progresión geométrica con razón positiva común. Determina

x.

3

Un entero positivo N se representa como cba en base 11, y se representa como acb1 ,

donde cba ,, representan dígitos, no necesariamente distintos. Determina el valor

mínimo de N expresado en base 10.

4

Sea S el conjunto de todos los enteros positivos N con la siguiente propiedad: Los

últimos cuatro dígitos de N son 2020, y cuando se eliminan estos cuatro dígitos, el

resultado es un divisor de N. Por ejemplo, 42020 pertenece a S porque 4 es divisor de

42020. Determina la suma de todos los dígitos de todos los números de S. Por ejemplo,

el número 42020 contribuye con 4+2+0+2+0=8 a este total.

5 Disponemos de seis cartas numeradas del 1 al 6 alineadas. Determina el número de

permutaciones de estas seis cartas tales que si se elimina una de ellas, las otras cinco

restantes quedan ordenadas, en orden descendente o ascendente.

6

En una mesa plana realizamos dos agujeros redondos de radios 1 y 2, respectivamente, y

cuyos centros están separados por una distancia igual a 7. Supongamos que dos esferas

del mismo radio 2R reposan cada una de ellas en un agujero, de forma que las dos

esferas son tangentes la una con la otra. Determina el cuadrado del radio de dichas

esferas.

7 Un club consistente en 11 hombres y 12 mujeres necesita escoger un comité entre sus

miembros, de forma que el número de mujeres del comité sea uno más que el número de

hombres de dicho comité. Este comité puede tener entre 1 y 23 miembros. Sea N el

número de comités que se pueden formar. Determina la suma de todos los números

primos que dividen N.

8

Un gusano se mueve durante todo el día y duerme durante toda la noche. Empieza en el

punto O y avanza una distancia de 5 unidades hacia el este. Cada noche, el gusano gira

60º en el sentido contrario al de las agujas del reloj, y al día siguiente avanza en esta

nueva dirección la mitad de la distancia del día anterior. Este gusano se acerca más y

más a un cierto punto P. Determina las coordenadas de dicho punto.

9

Sea S el conjunto de todos los enteros positivos divisores de 920 . Se toman tres

números de este conjunto, aleatoriamente y con reemplazamiento, y se etiquetan como

321 ,, aaa en el orden en que van apareciendo. Determina la probabilidad P de que 1a

divida a 2a y de que 2a divida a 3a .

10

Sean m y n enteros positivos satisfaciendo las siguientes condiciones:

- El máximo común divisor 1)210,( nm .

- mm es un múltiplo de nn .

- m no es un múltiplo de n .

Determina el menor valor posible de nm .

11

Para ciertos enteros dcba ,,, , definimos baxxxf 2)( y dcxxxg 2)( .

Determina el número de ternas ordenadas cba ,, de enteros, con valores absolutos no

mayores de 10, para los cuales existe un entero d tal que

042 fgfg

12

Sea n el mínimo entero para el cual nn 2149 es divisible entre 753 753 . Determina el

número de divisores positivos de n.

13

El punto D pertenece al lado BC del triángulo ABC de forma que AD es la bisectriz

de BAC . La mediatriz del segmento AD corta las bisectrices de ABC y ACB en

los puntos E y F, respectivamente. Dados 4AB , 5BC y 6CA , determina el área

de AEF .

14

Sea )(xP un polinomio cuadrático con coeficientes complejos cuyo coeficiente de 2x

es 1. Supongamos que la ecuación 0)( xPP tiene cuatro soluciones distintas:

bax ,,4,3 . Determina la suma de todos los posibles valores de 2)( ba .

15

Sea ABC un triángulo acutángulo con circuncírculo , y sea H su ortocentro.

Supongamos que la recta tangente a la circunferencia circunscrita de HBC en H corta

en los puntos YX , de forma que 3HA , 2HX y 6HY . Determina el área de

ABC .

AIME I 2020 Soluciones 1

Sea ACB

Por ser DBC isósceles, ACBBDC , y 2180DBC

Por ser BAC isósceles, ACBABC , y 2180BAC

Por ser DEA isósceles, 2180 BACEDA

Por ángulos suplementarios, )2180(180EDB

Por ser BDE isósceles, 2

180 EBD

Luego, finalmente,

7

5402180

2

1802´180

EBDABC grados.

2

)(log3

1)(log)2(log)(log)2(log)2(log 882888 3 xxxxa

xxxx

xb 88

2

2

8

8

84 log

2

3

3/2

log

2log

log

4log

loglog

3

xxxx

xc 88

2

8

8

82 log3

3/1

log

2log

log

2log

loglog

3

Luego, denotando )(log8 xk , llegamos a

ka 3

1 , kb

2

3 , kc 3

Por ser una sucesión geométrica:

3

44341291249

3

62

2

32

3

2

2

3

3

12

2

32

2/3

3

3/1

2/3

kkkkkk

kkk

kk

k

k

k

k

k

b

c

a

b

16

1

2

128)(log

3

44

43/4

8 xx

3

acbcbaN 88811111 232 , con 7,,0 cba .

512753120

1207535120

11118118810

111188810

223

223

cba

acb

acb

cbaacb

Luego

4120/512512512753120 acba

Con 4a nos quedamos cortos, luego 5a .

Supongamos que 5a . La ecuación queda de la forma:

cb

cb

cb

75388

512753600

512753600

Con 0b la ecuación resultante cc 7705388 no tiene solución entera.

Con 1b la ecuación resultante c715388 tiene solución entera:

573575388715388 cccc

Luego hemos encontrado la solución 5,1,4 cba , y como en todo momento hemos

trabajado con valores mínimos, esta será la solución mínima que nos pide el enunciado.

4

202010000 aN para cierto Na |

Luego akN para cierto entero k y por tanto

10152202010000202010000 2 kkaaak

Los posibles valores de a son:

1 1

2 2

5 5

101 2

4 4

10 1

202 4

505 10

20 2

404 8

1010 2

2020 4

Total 45

Por otro lado, los 12 números de S acaban todos en 2020, y por lo tanto tenemos que sumar

48412)0202(12

Finalmente, en total, tenemos 934845 .

5

Para este problema lo más práctico es enumerarlas una a una, en función de la carta que

eliminamos. Por cada permutación ascendente válida habrá una descendente, por lo que

contaremos solo las ascendentes y el resultado lo multiplicaremos por dos.

En las cuales vemos algunas repeticiones, que se han marcado en gris. En total hay

26546

Y como el orden puede ser también descendente, 52226

6 Realizamos un esquema transversal de la posición de las dos esferas en sus respectivos

agujeros:

En donde vemos claramente tres triángulos rectángulos diferentes en los que aplicar Pitágoras,

obteniendo un sistema de ecuaciones que tenemos que resolver:

222

222

222

7)(2

2

1

baR

bR

aR

4122412222

244224424

492414927)(24

22222

222

22222222

RRRRabR

abRabRR

abRRabbabaRR

13

16039480544224

544224442245

22412241

22222

22242224

2222222

RRRR

RRRRRR

RRRRRR

7 Consideremos un subconjunto cualquiera de 11 miembros del club.

Sea 110 i el número de hombres de dicho grupo.

Luego el número de mujeres será i11 .

Y por tanto el número de mujeres fuera del grupo es 11112 ii .

Luego si tomamos los hombres de dicho grupo y las mujeres que no están en dicho grupo,

obtenemos un comité que se adapta al enunciado.

Recíprocamente, si tenemos un comité de 1i mujeres y i hombres, con las ii 11112

mujeres que no están en el comité y los i hombres que sí están podemos formar un grupo de 11

personas.

Así pues, existe una biyección entre el número de comités posible y los subconjuntos de 11

personas:

11

23N

2713171923234567891011

1314151617181920212223

!12!11

!23

11

23

N

Y por tanto la suma pedida es 812713171923

Fuente de la solución: Solución oficial MAA en AoPS

8

Veamos en primer lugar como se comporta en vertical:

Paso 1 OA: 0

Paso 2 AB: 34

5

2

3º60sin

2/5 y

y

Paso 3 BC: 38

5

2

3º30cos

4/5 y

y

Paso 4 CD: 0

Paso 5 DE: 332

5

2

3º60sin

16/5 y

y

Paso 6 EF: 364

5

2

3º30cos

32/5 y

y

La variación vertical es:

...32

53

2

50

2

53

2

53

2

50

2

5

32

53

2

50

2

53

2

53

2

50

2

5

121110987

65432

Vemos por separado cada uno de estos bloques de seis pasos:

16

213

2

5

2

1

2

1

2

113

2

53

2

53

2

53

2

53

2

524326532

16

213

2

5

2

1

2

1

2

113

2

53

2

53

2

53

2

53

2

58438121198

Así pues, estamos sumando:

3

5

63

64

16

213

2

5

2

1

16

213

2

5...

2

1

2

1

2

11

16

213

2

5

...2

1

2

1

2

11

16

213

2

5...

16

213

2

5

16

213

2

5

16

213

2

5

16

213

2

5

20

62362662

181262201482

n

n

Veamos ahora su comportamiento en horizontal.

Paso 1 OA: 5

Paso 2 AB: 4

5

2

1º60cos

2/5 x

x

Paso 3 BC: 8

5

2

1º30sin

4/5

x

x

Paso 4 CD: 8

5x

Paso 5 DE: 32

5

2

1º60cos

16/5

x

x

Paso 6 EF: 64

5

2

1º30sin

32/5 x

x

Luego se ha producido un cambio total de 64

315

64

5

32

5

8

5

8

5

4

55

En el siguiente grupo de 6 pasos será 64

315

2

16

Y el resultado final será

06

563

64

64

315

64

315

2

1

n

n

Así pues, el gusano se acerca más y más al punto

3

5,5

9

918929 525220

Los divisores de 920 son todos los números de la forma cb52 con 180 b y 90 c

Luego ii cb

ia 52 con 180 ib , 90 ic , 30 i

La divisibilidad se convierte en un problema de orden:

ji

ji

jicc

bbaa |

Luego nuestro problema se reduce a determinar la probabilidad de que seis números

18,,0 321 bbb y 9,,0 321 ccc cumplan 321 bbb y 321 ccc .

El total de casos es 33 1019 T .

Contemos ahora el total de casos favorables. Aplicaremos la Proposición 5.5:

El número de casos de tomar 180 321 bbb es 13303

21

El número de casos de tomar 90 321 ccc es 2203

12

Y por tanto, los casos favorables son: 2201330 F , y la probabilidad es:

1805

77

1019

220133033

T

FP

10

La clave para resolver este problema es ver que la condición mn mn | implica que si p es un

factor primo de n , también lo será de m .

Por otro lado, la condición mn | implica 1n .

Otra clave importante es ver que la condición 1)210,( nm , con 7532210 implica

que si un primo p de un dígito (2,3,5 o 7) divide n, entonces también dividirá a m, y por tanto

dividirá nm , con lo que 1)210,( nm .

Así pues, los factores primos de n serán 11, 13, 17…

Si estos factores primos de n estuvieran elevados a 0 o 1 como máximo, por ser también

factores de m , entonces m tendría todos los factores de n elevados a sus potencias respectivas,

es decir, mn | , contradiciendo la tercera hipótesis del enunciado.

Así pues, la factorización de n debe contener algún exponente mayor o igual que 2.

El menor candidato posible de n es 121112 n , y por tanto 2421212 1111 .

Puesto que, por hipótesis, mm|11242 , necesitamos un m que sea múltiplo de 11, mayor o igual

que 242, que no sea divisible por 121 y que 1)210,( nm .

242m no cumple pues es divisible entre 121.

253m no cumple pues 374121253 nm es divisible entre 2.

264m no cumple pues 385121264 nm es divisible entre 5.

275m no cumple pues 396121275 nm es divisible entre 2.

Finalmente, 286m cumple las condiciones del enunciado:

3711407121286 nm , y por tanto 1)210,( nm

13112 m , mn | .

Luego la solución es 407.

Fuente: https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_10

11

4222)2( 2 babaf

16444)4( 2 babaf

Supongamos, en primer lugar, que )4()2( ff . Entonces:

6122016442)4()2( aababaff .

8412)4()2( bbff

Vemos que 6a es fijo, y que no hay ninguna restricción sobre b .

)8()8(

)8()8(0)8(0042

2

2

bcbd

dbcbbgfgfg

Luego, para 6a y cualquier pareja cb, , tomando )8()8( 2 bcbd se satisfarán las

condiciones del enunciado.

Observamos que en el caso )4()2( ff el enunciado no nos está diciendo que el polinomio

)(xg tenga una única solución doble, solo que una de sus dos soluciones es )4()2( ff .

Luego el número de ternas cba ,, con 10, cb son 4412121 .

Supongamos, en segundo lugar, que )4()2( ff (y por tanto 6a ).

Vemos que )2(f y )4(f son raíces del polinomio mónico )(xg , luego por las “Fórmulas de

Vieta” (ver AG/2.1) tenemos que

)4()2( ffc y )4()2( ffd .

Sobre el valor de d no hay ninguna condición. Pero se debe cumplir 10c .

1032202616442)4()2( babababaffc

5315510355103101032 babababa

Para resolver 53 ba determinamos la frontera 53 ba

50 ba , 3/50 ab , y el )0,0( no satisface la inecuación.

Para resolver ba 315 determinamos la frontera ba 315

150 ba , 53/150 ab , y el )0,0( sí satisface la inecuación.

En total hay 69 soluciones (recordemos que 6a ), que junto a las 441 del primer caso hacen

un total de 510.

12 Solución: https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_12

13

Sea M el punto medio del segmento AD . Trazamos el segmento DE , y EDAE pues EF es

la mediatriz de AD . Sean HG , los pies de las bisectrices por B y C respectivamente.

La clave de este problema es demostrar que 2/CFAM y 2/BEAM

Aplicando el Teorema del Seno a los triángulos ABE y EBD , y teniendo en cuenta que

EBDABE , tenemos:

BDEBAEBE

BDE

BE

BAE

DE

EBD

BE

BDE

BE

BAE

AE

ABE

sinsinsinsin

sinsin

sinsin

Puesto que ABBD , los triángulos ABE y EBD no son congruentes, luego

BDEBAE , y por tanto BAE y BDE son suplementarios.

Esto significa que AEDB es un cuadrilátero cíclico y que, por tanto:

BEBDDAE 2

1

De la misma forma se demuestra que CFCBFAD 2

1

Por 11.4.4a tenemos: 210

20

64

45

BD , 325 DC .

Por 11.4.4b tenemos: 23181862432642 ADAD

Y por tanto: 2

23AM

Por el Teorema del Coseno en ABC :

8

1coscos40251636cos542546 222 BBB

8

73

64

63sin

64

63

64

11

8

11cos1sin

2

22

BBB

Y de la misma manera deducimos que 4

3cos C y

4

7sin C .

Con aplicando la fórmula de la tangente del ángulo mitad:

3

7

9

73

8/9

8/73

8/11

8/73

cos1

sin

2tantan

B

BBDAE

7

7

4/7

4/7

4/31

4/7

cos1

sin

2tantan

C

CCDAF

Finalmente:

14

715

21

710

2

9

2

1

7

7

3

7

2

9

2

1tantan

2

1 2

DAFDAEAMAEF

Fuente: https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_13

14

El enunciado nos está diciendo que )(),(),4(),3( bPaPPP son raíces del polinomio )(xP , que

solo puede tener dos raíces diferentes, digamos nm, .

Por Vieta sabemos que podemos escribir mnxnmxxP )()( 2

Por lo tanto se pueden dar fundamentalmente dos casos:

a) )()(),4()3( bPaPnPPm

7)(416)(39

)(44)(33)4()3( 22

nmnmnm

mnnmmnnmPP

497)(7

))(())((

))(()()(

)()(

)()()()(

22

22

22

22

banmba

nmbababa

nmbanmbnmaba

nmbbnmaa

mnnmbbmnnmaabPaP

En donde hemos utilizado la hipótesis ba

a) )()4(),()3( bPPnaPPm

nmanmaaa

nmanmanma

nmaanm

mnnmaamnnmaPP

3))(3()3)(3(

))(3()()(39

)()(39

)()(33)()3(

2

2

22

La condición )()4( bPP nos lleva, con un razonamiento similar, a nmb 4

Pero, por otro lado

nmnnmbbmnnmbPPn

mmnnmaamnnmaPPm

)()(44)()4(

)()(33)()3(

22

22

Y restando ambas ecuaciones llegamos a

2/72777)(7 nnnnmnnmmnnm

mmm

mmmmnmnnm

7283237728832

77724322

7

2

7

2

7416)(442

Y por tanto

36)6()(62

7

2

5

2

74

2

734

2

53

2

733

2

baba

nmb

nma

Finalmente: 853649

15

Solucionaremos este problema mediante potencias.

Sea BCAHD , y AHP . Sabemos por 11.6.9a que DPHD .

Por potencia de H en ,

22326212 DPHDHDHDHAHPHAHYHX

Sea K la intersección de XY y BC. Por potencia de K en HBC y en , tenemos

3)6)(2(2 KHKHKHKYKXBCKBKH

Aplicando Pitágoras en KHD , 523 22222 HDKHKD

Sea BDx , CDy . Por potencia de D en , 102)23( DPADCDBDxy

De nuevo, por potencia de K en HBC y en , tenemos

yxyKDxKDKCKBKH 55))((9 2

Luego tenemos que resolver el sistema

5)(510)(55)(55559

10

xyyxxyyxyx

xy

5

14

5

14)(514

22 xyxyxy

5

116

5

396

5

14200

5

1440)(40)(

401044)()(2)(

2)(

22222

22

222

222

BCyxyxBC

xyyxyxxyyxyx

xyyxyx

Finalmente, 55352

1165

2

BCADABC

Fuente de esta versión: artofproblemsolving.com

AIME II 2020 Enunciados

1

Determina todas las parejas de enteros positivos ),( nm tales que 202 20nm .

2

Sea P un punto tomado aleatoriamente en el interior del cuadrado unidad de vértices

)0,0( , )0,1( , )1,1( y )1,0( . Consideramos la pendiente de la recta determinada por P y el

punto

8

3,

8

5. ¿Cuál es la probabilidad de que dicha pendiente sea mayor que

2

1 ?

3 Resuelve la siguiente ecuación:

2020

2

20

23log3log 3 xx

4

Sean ABC y ''' CBA triángulos en el plano cartesiano con vértices )0,0(A ,

)12,0(B , )0,16(C , )18,24('A , )18,36('B , )2,24('C . Una rotación de m

grados, con 1800 m , en el sentido de las agujas del reloj alrededor del punto ),( yx

transforma ABC en ''' CBA . Determina yxm ,, .

5

Dado un entero positivo n , sea )(nf la suma de los dígitos de la representación de n

en base cuatro, y sea )(ng la suma de los dígitos de la representación de )(nf en base

ocho.

Por ejemplo: 84 1210133210)2020( ff , y 321)2020( g .

Determina el valor mínimo de n de forma que la representación en base 16 de )(ng no

pueda ser representada usando solo los dígitos 0 a 9.

6

Definimos la siguiente secuencia recursiva: 201 t , 212 t , y 2

1

25

15

n

nn

t

tt .

Determina 2020t .

7 Dos conos rectos iguales cada uno de ellos con radio de la base 3 y altura 8 se cortan en

ángulo recto por sus ejes de simetría en el punto interior de los conos a distancia 3 de

sus respectivas bases. Determina el radio de la esfera más grande que podemos trazar en

la zona común de ambos conos.

8

Definimos recursivamente la siguiente sucesión de funciones:

1)(1 xxf

nxfxf nn 1)( para todo entero 1n .

Determina el valor mínimo de n para el que la suma de los ceros de )(xfn exceda

500000.

9 Mientras ven un espectáculo, Ayako, Billy, Carlos, Dahlia, Ehuang y Frank se sientan

en este orden en una fila de seis bancos. Durante el intermedio, salen a tomar algo, y al

volver se sienten en esos mismos asientos de manera que, si dos de ellos estaban

sentados juntos antes, ahora no lo están. ¿De cuantas formas posibles se han podido

sentar al volver del intermedio?

10

Determina la suma de todos los enteros positivos n tales que, cuando 3333 ...321 n se divide entre 5n , el residuo es 17.

11

Sea 73)( 2 xxxP , y sean )(xQ y )(xR dos polinomios cuadráticos que también

tienen el coeficiente de 2x igual a 1. David calcula cada una de las tres sumas QP ,

RP y RQ y comprueba sorprendido que cada pareja de estas sumas tiene una raíz

en común, y que dichas raíces son distintas. Si además sabemos que 2)0( Q ,

determina )0(R .

12

Sean m y n números enteros impares mayores que 1. Dibujamos un rectángulo de

nm casillas, todas ellas cuadrados de una unidad de lado, que numeraremos de forma

ordenada: En la fila superior, de izquierda a derecha, los números del 1 a n . En la

segunda fila, de izquierda a derecha, los números de 1n a n2 . Y así sucesivamente.

Supongamos que la casilla correspondiente al número 200 está en la fila superior, y que

la casilla correspondiente al número 2000 está en la fila inferior.

Determina el número de pares ordenados nm, de números impares mayores que 1

con la siguiente propiedad:

En el rectángulo nm generado con ellos de la forma anterior, la recta que une los

centros de las casillas correspondientes a los números 200 y 2000 pasa por el interior de

la casilla correspondiente al número 1099.

13

Sea un pentágono convexo ABCDE con lados 5AB , 6 DECDBC y 7EA .

Además, sabemos que dicho pentágono tiene una circunferencia inscrita (una

circunferencia tangente a todos los lados del pentágono). Determina el área de dicho

pentágono.

14

Para cada número real x sea x el mayor entero menor o igual que x , y definimos

xxx como la parte decimal de x . Por ejemplo, 03 y 56.056.4 . Si

definimos xxxf )( , determina el número de soluciones reales de la ecuación

17xfff con 20200 x .

15

Dado un triángulo acutángulo ABC , sean P y Q los pies de las perpendiculares de C

en AB y de B en AC , respectivamente. La recta PQ corta el circuncírculo de ABC en

los puntos X, Y. Supongamos que 10XP , 25PQ y 15QY . Determina el valor de

ACAB .

AIME II 2020 Soluciones

1

204020 5220 , y por tanto los enteros m y n deben ser de la forma bam 52 y dcn 52

Luego

202

4025252

52525252525220

204022

20402220402202

db

ca

nm

dbca

dcbadcba

La ecuación 402 ca tiene las siguientes soluciones:

0,20,...,36,2,38,1,40,0 cacacaca , hay 21 en total.

La ecuación 202 db tiene las siguientes soluciones:

0,10,...,16,2,18,1,20,0 dbdbdbdb , hay 11 en total.

Luego el total de parejas será 2311121 .

2

Determinamos la recta que pasa por el punto

8

3,

8

5P y tiene pendiente

2

1:

16

1

2

1

16

1

8

5

2

1

8

3

8

5

2

1

8

3

2

1

xybbbxy

La zona que buscamos son todos aquellos puntos del cuadrado unidad que estén por encima de

esta recta, a su derecha, o por debajo si están a su izquierda.

Determinamos sus puntos de corte con los bordes del cuadrado:

16

9,0

16

9

16

11

2

11

16

1,0

16

1

16

10

2

10

Byx

Ayx

Determinamos el área de esta zona en dos rectángulos y dos triángulos:

128

43

168

43

2168

86

16

7

8

3

2

1

16

3

8

3

2

1

8

5

16

5

16

1

8

5

F

Puesto que el área total es igual al área del cuadrado de lado 1,

112 T

Y la probabilidad es 128

43

1

128/43

T

F

3

Primera versión.

101201012020100202020 3333

luego

20

2

10120

2

2020

23log1013log3log 333 xxx

Y por tanto

20

2

20

220

2

10120

2

2020

2

20

2 3log

3log1013log1013log3log3log

3

333

x

x

xxxx

Ahora aplicamos la fórmula del cambio de base: bc

ca

b

a loglog

log

Y por tanto:

x

xxx

x

x 32log

3log

3log101 3

220

2

20

2

3

Esto último se puede justificar mediante el concepto de logaritmo: 3/)3(22

xxxx

Así pues, llegamos finalmente a la ecuación

100

331003101

3101

xxxx

x

x

Segunda versión.

Utilizando la identidad bn

mb a

m

an loglog

Esta ecuación se reduce a

100

3

3

202020

3

20203log3log3log

20 2020

2

20

22 3

xxxxx

xx

4

Primera versión.

La rotación genera circunferencias concéntricas en su centro O. Dicho centro queda

determinado por la intersección de dos radios, que serán perpendiculares a dos cuerdas por su

punto medio.

)9,12(2

)18,24()0,0(

2

'

AAM ,

)4,3()3,4()18,24()0,0()18,24('' vAAAA

El primer radio será la recta

7534273484)9(3)12(44

9

3

12

yxyxyx

yx

)15,18(2

)18,36()12,0(

2

'

BBN

)6,1()1,6()6,36()12,0()18,36('' wBBBB

El segundo radio será la recta

1236151086)15(1)18(66

15

1

18

yxyxyx

yx

El centro de la rotación será su punto de intersección:

32161232129414

7518369475)6123(3461231236

7534

yxx

xxxxxyyx

yx

Luego el centro de rotación es el punto )3,21(

Vemos que esta rotación transforma una recta horizontal (el lado AC) en una recta vertical (el

lado A'C'), luego será una rotación de º90m . Esto también se puede observar determinado,

por ejemplo, los vectores OA y 'OA y viendo que son perpendiculares.

Segunda versión.

Si en primer lugar nos percatamos de que la rotación es de 90º, viendo que transforma un

segmento horizontal en uno vertical, la determinación del centro ),( yxO es mucho más fácil,

pues sabemos que el segmento de (0,0) a ),( yx y el segmento ),( yx a ),( xyyx son

perpendiculares y tienen la misma longitud. Así pues, llegamos a

)3,21(),(18

24)18,24('),(

yx

xy

yxAxyyx

5

Analizando detenidamente la larga cadena de operaciones involucradas en este problema,

podemos especular que 10)( ng , el primer valor que no se puede representar en base 16 con

los dígitos 9,...,2,1,0 .

En base 8 disponemos de los dígitos del 0 al 7, y por lo tanto, el valor mínimo para que la suma

sea 10 será 3+7, es decir, el número 31783378 .

Finalmente, buscamos un número cuya suma de dígitos en su representación en base 4 sea 31.

En base 4 tenemos los dígitos 3,2,1,0 . 110331 , luego el número buscado será

14...44344343...43434131333333333 91011091011

4 n

Aplicando la fórmula de la serie geométrica:

3

14

41

4114...44

1111910

Para calcular 2211 24 a mano vamos calculando potencias de 2:

41943044224

1048576102410242

10242

202211

20

10

Con lo que, finalmente:

20971513

1434

1111

n

6

Primera versión.

250

53

2025

106

2025

12153

t

502125

103

502125

5053

2125

50

5053

2125

150

53

2125

1250

535

4

t

525

101

5215

101

5215

101

53

10

50215

10153

53

10

50215

5353

53

10

50215

50215103

53

101

50215

103

10

53

150215

103

250

5325

1502125

1035

5

t

20100221

10212

105

50212

103

5021

105

206

103

50211

105

101

5021

103

1105

101

502125

10325

1525

1015

6

t

21

21

101

101

525

10125

1205

525

10125

12056

t

Luego vemos que hay un bucle de longitud 5:

525

101,

502125

103,

250

53,21,20

Y puesto que 54042020 , 525

10152020 tt

Segunda versión.

Similar a la primera, para ahorrarnos cálculos podemos definir nn ts 5 , y por tanto

2

1

2

1

2

1 1

5

1

25

15

n

nn

n

n

n

nn

s

ss

s

s

t

tt

Con 1001 s y 1052 s . Ahora hacemos los cálculos:

50

533 s ,

50105

1034

s ,

105

1015 s , 1006 s y 1057 s .

Y observamos el bucle.

Tercera versión. (Oficial de la MAA)

En general, para una sucesión de la forma

at 1 , bt 2 , ak

bkt

23

1

Tenemos:

abk

kbka

bak

ka

bak

kb

bkbak

kb

bkbkak

kb

bkka

kb

bk

akkb

bk

akkbkt

33323

22222

2

4

1111

1111

11)/()1(1)/()1(

22

2

2

2

22

235

1

1

1

11

1

11

1

11

1

11

1

bk

ak

bkbk

abkkbka

bkbk

abk

bkbk

kbka

bk

a

bk

a

abk

kbka

bk

a

abk

kbka

bk

a

abk

kbkakt

Y finalmente at 6 y bt 7

Y observamos el bucle de longitud 5.

7 Dibujando la situación de los dos conos podemos especular que el radio que nos piden es la

distancia del centro a la apotema:

Que la podemos representar como la distancia del punto )3,3( a la recta que pasa por )6,0( y

)3,8( , es decir, la recta 0488368

3

yxxy

La distancia la podemos calcular por la conocida fórmula de la distancia de un punto a una

recta:

76.173

15

83

243833

22

d

8 Solución: https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_8

9 Este problema se puede resolver por "bashing", "machaqueo", es decir: generando todo el árbol

de posibilidades. Por simetría nos reduciremos a detallar todas las posibilidades que comienzan

por "A", "B" o "C", y multiplicaremos por 2:

En total hay 90245 posibilidades diferentes.

10

Primera versión.

4

1...321

223333

nnn

Queremos resolver la congruencia

)5(mod174

122

nnn

)5(mod681

68)5(4174)5(41

17)5(4

1)5(mod17

4

1

22

22

2222

nnn

nknknn

nknn

nnn

Pero observamos que

)5(mod41)5(mod41

)5(mod5)5(mod5

22

22

nnnn

nnnn

Y por tanto

)5(mod4001625)4()5(1 2222 nnn

Luego

832332|5

)5(mod0332)5(mod068400)5(mod68400

2

n

nnn

Con las condiciones del enunciado estamos suponiendo implícitamente que 175 n , luego

las únicas posibilidades son 327,161,78332,166,835 nn .

Comprobamos estas soluciones, y vemos que se cumple para 78 y para 161, pero no para 327:

)5(mod100

4

1327

22

nnn

n

Luego las soluciones son 161,78n .

Observación. Las comprobaciones exigen cálculo manual con números grandes. Por ejemplo:

107584)1(

106929327

2

2

n

nn

Segunda versión. Mediante aritmética modular.

682681|5

)5(mod0681

)5(mod681

)5(mod174

1

23422

22

22

22

nnnnnn

nnn

nnn

nnn

Realizando la división sintética tenemos que

5

33280163

5

682

332580163682

23234

23234

nnnn

n

nnn

nnnnnnn

Y por tanto

Zn

nnn

5

332681|5

22

Y se sigue igual que en la primera versión.

Tercera versión.

Con un cambio de variable:

41

55

mn

mnnm

Y por tanto

332|5332|mod0332mod068400mod68400

mod174400mod174

400mod17

4

)4()5(

mod174

)4()5()5(mod17

4

1

22

2222

nmmmm

mmm

mmm

nnn

Y se sigue igual que en la primera versión.

Fuente de estas soluciones: https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_10&oldid=125812

11

73)( 2 xxxP

2)( 2 xaxxQ

cxbxxR 2)(

5)3(2)()( 2 xaxxQxP

)7()3(2)()( 2 cxbxxRxP

)2()(2)()( 2 cxbaxxRxQ

Sea r la raíz común de )()( xQxP y )()( xRxP .

Sea s la raíz común de )()( xRxP y )()( xRxQ .

Sea t la raíz común de )()( xQxP y )()( xRxQ .

Entonces está claro que )()( xQxP tiene raíces r y t, )()( xRxP tiene raíces r y s, y

)()( xRxQ tiene raíces s y t.

Luego, aplicando Vieta:

tr

rta

trxrtxtxrxxaxxQxP

25

)(23

2)(22))((25)3(2)()( 22

rsc

srb

rsxsrxsxrxcxbxxRxP

27

)(23

2)(22))((2)7()3(2)()( 22

stc

tsba

stxtsxtxsxcxbaxxRxQ

22

)(2

2)(22))((2)2()(2)()( 22

Con lo que tenemos el siguiente sistema (del que solo nos interesa determinar c)

stc

tsba

rsc

srb

tr

rta

22

)(2

27

)(23

25

)(23

2

3064)(23)(23)(2

)(2

3)(2

3)(2

rrtssrrt

tsba

srb

rta

19/27

19/52

3

522

37

22

)(2

37

233

3/535

323

22

)(2

2

327

2

323

2

325

2

323

s

c

sc

sc

stc

tsba

sc

sb

tt

ta

stc

tsba

sc

sb

t

ta

La solución es 19

52)0( cR .

12 Solución: https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_12

13

Primera versión. Sin determinar el radio.

Sean TSRQP ,,,, los respectivos puntos de tangencia a los lados AB , BC , CD , DE , EF.

Sea PQPBx , CRQCy , DSRDz , ETESt , APATu

Se plantea el siguiente sistema que tiene solución única:

3

4

2

4

2

7

6

6

6

5

u

t

z

y

x

ut

tz

zy

yx

xu

Prolongamos la recta CD hasta cortar en N a AE y en M a AB

Por el criterio SSS de congruencia de triángulos,

CDEABCRIDSIDQIBQIBPIB

Y de la misma forma:

DEABCDEITEISCIRCIQ .

Luego, por triángulos suplementarios y el criterio ASA, DNEBMC , y por tanto

MN

Sea MCENa y MBNDb . Por tangentes concurrentes tenemos 24 ba ,

Es decir, 2 aMBND .

Aplicando el Teorema del Coseno en BMC :

)2(

162

)2(2

3242

)2(2

446

)2(2

)44(6

)2(2

)2(6

)2(2)2(6

22

222222222

222

aa

aa

aa

aa

aa

aaa

aa

aaa

aa

aaMCos

MCosaaaa

Aplicando el Teorema del Coseno en AMN :

7

4

)7(2

82

)7)(82(2

)82(

)82()7)(82(2

)7)(82(2)82(0

)7)(82(2)7()82()7(

)7)(62(2)7()62()52(

2

2

2

222

222

a

a

a

a

aa

aNCos

aNCosaa

NCosaaa

NCosaaaaa

NCosaaaaaaa

Luego:

8,

3

14

7

4

)2(

1622

aa

a

aa

aa

La única solución válida es la positiva 8a , y por tanto el triángulo ANM es un triángulo de

lados 15, 15, 24.

La altura de ANM por A tiene longitud 9811442251215 222 hh ,

Y por lo tanto el triángulo ANM tiene área 1089122

924

ANM .

Los triángulos END y BCM tienen lados 6,8,10, y por tanto sus áreas son, aplicando la

fórmula de Heron:

2424612)1012)(812)(612(12

122/)1086(

BCMEND

s

Y, finalmente, el área del pentágono será 602421082 ENDANM

Segunda versión. Determinando el radio.

Sea r el radio de la circunferencia inscrita, y sea R el punto de tangencia entre la circunferencia

y el lado CD . Como en la primera versión, hemos determinado los segmentos entre los puntos

de contacto y los vértices del pentágono:

Observamos que DB , EC , º90 DRICRI , AIAEIAB 2

1

La suma de los ángulos internos de los pentágonos ABCRI y AEDRI es la misma.

RIADRIDEIAERIACRICBIAB

(Nota: De aquí, en las soluciones oficiales “MAA 1” se deduce que º180RIA . Yo no veo

como se llega a este resultado.)

Luego los puntos R, I, A están alineados.

Ahora aplicamos Pitágoras a los triángulos rectángulos ARC y ARD

(*)1224424

222222222

222

222

ADACACAD

ACAR

ADAR

Por otro lado, podemos determinar 22 ADAC mediante trigonometría:

2

2

2

2

2

2

4

4

4/1

4/1

)2/(tan1

)2/(tan1cos

22tan

r

r

r

r

B

BB

rB

2

2

2

2

2

2

16

16

16/1

16/1

)2/(tan1

)2/(tan1cos

42tan

r

r

r

r

E

EE

rE

Y ahora, aplicando el Teorema del Coseno a los triángulos AED y ABC

2

222222

16

1667267)(2

r

rECosEDAEEDAEAD

2

222222

4

465265)(2

r

rBCosBCABBCABAC

Y por tanto

2474

4652

16

16672

16

1667267

4

465265

16

1667267

4

465265

2

2

2

2

2

2

222

2

222

2

222

2

22222

r

r

r

r

r

r

r

r

r

r

r

rADAC

Uniendo esta igualdad con (*) obtenemos la ecuación

222222

2

2

2

2

2

2

2

2

2

2

2

2

416316454167

34

45

16

167

364

4652

16

16672

12244

4652

16

16672

rrrrrr

r

r

r

r

r

r

r

r

r

r

r

r

Tenemos una ecuación bicuadrada en la que, con el cambio de variable 2rx , llegamos a

16

5/4064845 22 rxxx

La solución 5/42 r corresponde a una estrella de cinco puntas, que no es convexa. Además,

si 3r , entonces 2/tan D , 2/tan E y 2/tan C y por tanto D , E y C son

ángulos agudos, lo que no puede pasar en un pentágono convexo. Así pues,

4162 rr

Y el área es 60154 (el radio multiplicado por el semiperímetro del pentágono).

Fuente de estas versiones: Art of Problem Solving.

14 Solución: https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_14

15

Primera versión.

Sean APa , BPb , AQc , CQd .

Aplicando potencia de un punto:

525)2510(15

400)1525(10

cd

ab

Observamos que el cuadrilátero PQCB es cíclico, pues º90 BQCBPC .

Luego BAQP y CAPQ , y por tanto AQPAPQ , y en consecuencia

125

525400

22

2222

ca

cacdcabadccbaaba

c

dc

a

Por otro lado, aplicando el Teorema del Coseno en APQ :

Aacca cos225625 222

Y observamos que ba

cA

cos

Luego

525

12521252

400125

1252125

4002125

4002

4002225625

2

222

2

2222

2

2222

2

22222222

c

ccc

c

cccc

a

accc

a

aacca

aa

cacca

ba

cacca

Luego todo se reduce a solucionar la ecuación

35587587505252501500400525250250

400125525)525(250

125)525()525(250525

125)525(250

525

125250

525

12522500

525

12521252625

22222

224242

22222

2

2222

2

222

2

222

2

222

cccccc

cccccc

cccccc

cccc

c

ccc

c

ccc

c

ccc

Con este resultado se deducen el resto de incógnitas:

35

105

355

525525 dcd

10101000100012587512522 aca

10

40

1010

400400 bab

Finalmente,

145603510112

35810143533551041010

35

35105355

10

10401010

35

105355

10

401010))((

dcbaACAB

Segunda versión.

Sean APa , BPb , pero ahora definimos Ak cos .

Luego k

bABk

AB

b ,

k

aACk

AC

a

Como en la versión anterior, mediante Potencia de puntos tenemos

525)()1510(15525

400)()1525(10400

2

2

2

2

b

abkb

k

abb

k

abbACbCQb

a

abka

k

aba

k

baaABaBPa

De donde deducimos que

125525400525400

2222

22

baba

b

ab

a

abk

Por otro lado, aplicando el Teorema del Coseno en APQ :

abkba 225625 222

Para resolver este sistema de ecuaciones realizamos el siguiente cambio de variable:

525400 22 bau

Luego u

abk , 400 ua , 525 ub y por tanto:

875

10001400

)252)(400(775

)252)(400(221550

)252)(400(292522525400625

22

b

au

u

uuu

u

uuu

u

uuu

u

bauu

Y, finalmente, 145608571000

140014002

ab

u

u

ab

u

ab

ab

k

b

k

aACAB

Fuente de estas soluciones: www.artofproblemsolving.com

AIME 2019 Enunciados y soluciones

Existe copia de este documento en formato Word: http://www.toomates.net/media/AIME2019.doc

AIME I 2019 Enunciados

1

Consideremos el entero cifras

N321

99...99..9999999999

Calcula la suma de todas las cifras de N.

2

Jenn toma aleatoriamente un número J entre 1,2,3,...,19,20. Después Bela toma

aleatoriamente un número B entre 1, 2, 3,..., 19, 20, distinto de J. ¿Cuál es la

probabilidad de que JB sea como mínimo 2?

3

En el triángulo PQR , 15PR , 20QR y 25PQ . Sean los puntos A y B en PQ ,

C y D en QR y E y F en PR , con 5 PFRERDQCQBPA . Determina el

área del hexágono ABCDEF .

4

Un equipo de fútbol dispone de 22 jugadores. Un conjunto fijo de 11 jugadores empieza

el partido, mientras que los 11 restantes quedan como substitutos. Durante el partido, el

entrenador pude hacer hasta un máximo de 3 substituciones, en las que uno de 11

jugadores en el partido puede ser reemplazado por uno de los substitutos. Ningún

jugador que haya sido reemplazado puede volver a jugar, y un substituto que haya

entrado a jugar se puede reemplazar. No se pueden hacer dos subtitutituciones al mismo

tiempo. Determina el número total de posibles substituciones que puede hacer el

entrenador, sabiendo que los jugadores involucrados y el orden de las substituciones

importa.

5

Una partícula empieza en el punto )4,4( y se mueve hasta alcanzar uno de los ejes de

coordenadas. Cuando la partícula se encuentra en el punto ),( ba se puede mover

aleatoriamente hasta el punto ),1( ba , )1,( ba o )1,1( ba , en cada caso con una

probabilidad 1/3, independientemente de sus movimientos anteriores. Determina la

probabilidad de que alcance el punto )0,0( , representada como nm 3/ , con nm, enteros

positivos.

6

En un cuadrilátero convexo KLMN , el lado MN es perpendicular a la diagonal KM ,

el lado KL es perpendicular a la diagonal LN , 65MN y 28KL . La recta por L

perpendicular al lado KN corta la diagonal KM en O con 8KO . Determina MO .

7

Sean yx, enteros positivos satisfaciendo el siguiente sistema de ecuaciones:

570),(log2log

60),(log2log

1010

1010

yxmcmy

yxMcdx

8

Sea x un número real tal que 36

11cossin 1010 xx . Determina xx 1212 cossin .

9

Sea )(n el número de enteros positivos de n. Determina la suma de los seis enteros

positivos n más pequeños tales que

7)1()( nn

10

Supongamos que existen 67321 ...,,, zzz números complejos diferentes de forma que el

polinomio

3673

3

2

3

1 ... zxzxzx

se puede expresar como )(1920 201720182019 xgxxx , donde )(xg es un polinomio

con coeficientes complejos con grado menor o igual a 2016. Determina el valor de

6731 kj

kj zz

11

Sea ABC un triángulo con longitudes enteras y ACAB . Sea una circunferencia

cuyo centro sea el incentro de ABC . Llamamos excírculo de ABC a una

circunferencia en el exterior de ABC que es tangente a un lado del triángulo y

tangente a las extensiones de los otros dos lados. Supongamos que el excírculo tangente

a BC es internamente tangente a , y que los otros dos excírculos son ambos

externamente tangentes a . Detemina el valor mínimo posible del perímetro de

ABC .

12

Dada la función zzzf 19)( 2 , sabemos que existen números complejos z con la

propiedad de que z , )(zf y )(zff son los vértices de un triángulo rectángulo en el

plano complejo, con ángulo recto en )(zf . Existen enteros positivos m y n tales que

uno de estos valores de z es inm 11 . Determina nm .

13

Sea ABC un triángulo con 4AB , 5BC y 6CA . Los puntos D y E pertenecen a

la semirrecta AB con AEADAB . Sea el punto CF el punto de intersección de

los circuncírculos de ACD y EBC cumpliendo 2DF y 7EF . Determina la

longitud BE .

14

Determina el menor factor primo impar de 120198

15

Sea AB una cuerda de la circunferencia , y sea P un punto de dicha cuerda AB . La

circunferencia 1 pasa por A y P y es tangente interna a . La circunferencia 2 pasa

por B y P y es tangente interna a . Sean P y Q los puntos de corte de 1 y 2 y sean

X e Y los puntos de corte entre la recta PQ y . Suponiendo 5AP , 3PB ,

11XY , determinar PQ .

AIME I 2019 Soluciones

1

(*)32110...1010101...1110...101010

110...110110110

99...99..9999999999

321321

321

321321

321321

321

cifras

N

0111...111000...001..10001001010...101010321321

321321

0789111....1113210111...111(*)3183321321

Las cifras de este número suman 34298703181

2

Hacemos una tabla de casos. En columnas, la bola de Jenn, en filas, la bola de Bela:

El número total de casos es 1920 (pues descartamos la diagonal), y los casos favorables son

2

191812...15161718

Luego la probabilidad es 20

9

20

2/18

1920

2/1918

P

3

El área del triángulo PQR se puede calcular mediante la fórmula de Heron:

150225005101530

52530

102030

151530

302

252015

s

Sin embargo, se comprueba que PQR es rectángulo en R, por ejemplo, mediante el Teorema

de Pitágoras: 222 256251520

Para calcular el área del triángulo QCB tenemos varias alternativas.

Una de ellas es determinar Qcos mediante el teorema del coseno:

5

3sin

25

9

25

161cos1sin

25

16cos

5

4

20252

202515cos

cos20252202515

222222

222

QQQQQ

Q

Luego 2

15

5

355

2

1sin

2

1 QQCQBQBC

De la misma manera llegamos a 2

25RDE y 10PAF

Y por tanto 1202

25

2

15120 RDEPAFQBCPQRABCDEF

4

Ordenaremos los casos en función del número de substituciones hechas:

Primer caso: 0 substitutiones.

1 caso.

Segundo caso: 1 substitución.

Hay 11 jugadores potencialmente substituibles y 11 jugadores substitutos, luego en

total:

1211111 casos.

Tercer caso: 2 substituciones.

En la segunda substitución tenemos 10 posibles substitutos, luego el total es

1211111 casos para la primera substitución, y

1101011 para la segunda, 1331010111111 casos.

Cuarto caso: 3 substituciones.

Siguiendo la pauta anterior, cada vez tenemos un substituto menos, luego hay:

131769091110111111 casos

Luego el total es 13311221317690133101211 casos.

5

Sea nmP , la probabilidad de que alcance el punto )0,0( partiendo del punto ),( nm . Está claro

que, por simetría, mnnm PP ,, .

El problema se puede resolver calculando recursivamente las probabilidades, mediante

probabilidad condicional:

1,1,1,1,3

1

3

1

3

1 babababa PPPP

3/11,1 P

21,12,11,23

1

3

1

3

1

3

1 PPP

33332221,22,11,12,23

5

3

113

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

PPPP

321,23,11,33

1

3

1

3

1

3

1 PPP

244443231,32,12,23,22,33

1

3

9

3

1

3

3

3

5

3

1

3

1

3

1

3

1

3

5

3

1

3

1

3

1

3

1 PPPPP

442323,22,22,33,33

11

3

353

3

1

3

1

3

5

3

1

3

1

3

1

3

1

3

1

3

1

PPPP

431,31,43

1

3

1

3

1

3

1 PP

554321,41,32,32,43

13

3

139

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

PPPP

65242,42,33,33,43

73

3

13

3

1

3

1

3

1

3

11

3

1

3

1

3

1

3

1 PPPP

76464,33,33,44,43

245

3

73

3

1

3

11

3

1

3

73

3

1

3

1

3

1

3

1 PPPP

6

Primera versión. Por trigonometría.

Sea KNLOP .

Sean MKN y LNK .

Por ser KLN recto, tenemos KLP , y por tanto:

cos2sin7cos8sin28 KP

Por otro lado, aplicando la versión extendida del Teorema del seno,

sin28sin65sin

28

sin

652 KNR

Con las dos identidades anteriores llegamos a

98

65

cos

sintan

Y por ser KMN un ángulo recto, se deduce que

9089898

6565tan KMOMKM

KM

Segunda versión. Por semejanza de triángulos.

KNKNKN

KLKP

KN

KL

KL

KPLKNPKL

LKNPKL

KLNKPL 7842822

De nuevo por el criterio AA,

KN

KM

KN

KMKOKP

KN

KM

KO

KPKPOKMN

8

Con las dos igualdades anteriores llegamos a

90898988

7848784 KOKMMOKM

KN

KM

KN

Tercera versión. Mediante alturas y el ortocentro.

Prolongamos los lados KL y MN hasta encontrarse en el punto Q.

Sea H el punto de corte de las dos diagonales: LNKMH . Está claro que H es el

ortocentro del triángulo KQN .

Trazamos el segmento QH , que será paralelo a LO pues QH es altura de KQN .

kLQ

kOHQKHLKOHQLO

28

8// para cierto k.

Por el criterio AA,

908988)1(8

)1(28

2828

8

2828

88

88

28

2

OMOMk

k

k

OM

k

HMk

kKQ

KM

KH

KLKMQKLH

Cuarta versión. Mediante potencias.

El cuadrilátero KLMN está inscrito en la circunferencia de diámetro KN , luego

KNKPKMKO

PLN está inscrito en la circunferencia de diámetro LN , y KL es tangente a dicha

circunferencia, puesto que LNKL .

Luego KNKPKL 2

De todo lo anterior:

90898988

28288

222 OMKMKLKNKPKM

Quinta versión. Teorema de Pitágoras y machaqueo algebraico.

Sean PNa , KPb , ONc , OMd , LNg ,

Aplicando el Teorema de Pitágoras a los triángulos rectángulos que aparecen en el esquema

obtenemos el siguiente sistema de siete ecuaciones y cinco incógnitas:

222

222

222

222

222

222

222

)(28

)(

28)(

8

)(65)8(

65

bag

gfea

feb

eb

cea

bad

dc

Que es equivalente al siguiente:

2222

222

222

2222

222

)(28)(

28)(

8

65

)(65)8(

bafea

feb

eb

dea

bad

No hace falta resolverlo, solo determinar d.

De la primera ecuación:

222

22222222222

2222

22222222

282816

28228)()(816

65

265816)(65)8(

effd

effeafeabadea

dea

babaddbad

222222

2222222

222

8282282828228)(

8

feffef

fefebfeb

eb

De lo anterior llegamos a

90898908

8288288)828(216

28828282816

222222

222222

OMddd

effd

Fuente de estas soluciones: artofproblemsolving.com

7

Sea m el número de factores primos en la factorización de x (no necesariamente diferentes), y

sea n el número de factores primos en la factorización de y (no necesariamente diferentes).

Determina nm 23 .

2

101010 ),(log),(log2log60 yxMcdxyxMcdx

2

101010 ),(log),(log2log570 yxmcmyyxmcmy

Sumando las dos igualdades llegamos a

63022

22

10

2

10

2

10

10),(),(

),(),(log

),(log),(log630

yxmcmyyxMcdx

yxmcmyyxMcdx

yxmcmyyxMcdx

Utilizando ahora la identidad

yxyxmcmyxMcd ),(),(

Llegamos finalmente a

4205...552...221010210210

2106303322 nmxyyxyxyx

Sean bax 52 , dcy 52 .

Entonces ),min(),min( 52),( dbcayxMcd , ),max(),max( 52),( dbcayxmcm , y por tanto

60),min(2

60),min(25252

525252

5252

5210),(),(log60

6060),min(2),min(2

6060),min(2),min(2

6060),min(2),min(2

60606022

10

dbb

caa

x

yxMcdxyxMcdx

dbbcaa

dbcaba

dbca

570),max(2

570),max(25252

525252

10),(),(log570

570570),max(2),max(2

570570),max(),max(

57022

10

dbd

cac

yxmcmyyxmcmy

dbdcac

dbcadc

Supongamos ca , db

1905703

1905703

20603

20603

5702

5702

602

602

570),max(2

570),max(2

60),min(2

60),min(2

dd

cc

bb

aa

dd

cc

bb

aa

dbd

cac

dbb

caa

Y por tanto 202052x , 19019052y , xyxMcd ),( , yyxmcm ),( ,

En efecto:

604020220202 1010101010),( yxMcdx

57038028521901902 1010101010),( yxmcmy

Cualquier otra posibilidad nos lleva a una incompatibilidad.

Supongamos ca , db

150

190

360

20

5702

5702

602

602

570),max(2

570),max(2

60),min(2

60),min(2

d

c

b

a

bd

cc

db

aa

dbd

cac

dbb

caa

absurdo.

Supongamos ca , db

190

150

20

360

5702

5702

602

602

570),max(2

570),max(2

60),min(2

60),min(2

d

c

b

a

dd

ac

bb

ca

dbd

cac

dbb

caa

absurdo.

Supongamos ca , db

5702

570570

602

60

570),max(2

570),max(2

60),min(2

60),min(2

dd

aaac

bb

aa

dbd

cac

dbb

caa

absurdo.

La única solución posible es 4052 2020 mx , 38052 190190 ny , y por tanto

880380240323 nm

8

Sean xa sin y xb cos . Está claro que 122 ba .

882212121210221012221010 ))((136

11

36

11babababbabaababa

Luego todo se reduce a determinar 8822 baba

2266222266624426

02321222221232023223

3333

3

3

2

3

1

3

0

3)(11

bababababaababab

bababababa

Sea 22bay . La igualdad anterior se puede escribir como ybayba 3131 6666

2

266

1

224466221010

10284664821002521242

2232322242125202

5225

1031536

11

10536

11105

5101055

5

4

5

3

5

2

5

1

5

0

5

)(11

yyy

ybaybababababa

abababababbaba

babababa

ba

Resolvemos la ecuación resultante:

6/5

6/155

36

11110315

36

111 22

y

yyyyyy

La solución 6/5y la descartamos por extraña (???) y nos quedamos con:

6

122 yba

3

2

6

1212 2222244 bababa

18

7

18

1

9

4

36

12

9

4

6

12

3

22

22

4424488

bababa

Finalmente:

54

13

108

7

36

11

18

7

6

1

36

11

36

11 88221212 bababa

Fuente de la solución: https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8

9

Para 1n , 2)2(,1)1( y está claro que no cumple la condición. Luego 2n , y por tanto

2)1(,)( nn y en consecuencia 5)1(,)(7)1()( nnnn .

11)( nn y ya hemos visto que no cumple la igualdad del enunciado.

Luego las posibilidades que quedan son:

a) 5)1(,2)( nn

b) 4)1(,3)( nn

c) 3)1(,4)( nn

d) 2)1(,5)( nn

Teniendo en cuenta que:

4

2

5)(

3)(

2)(

pnn

pnn

pnn

3

,224)(

pn

qpqpnn

Tenemos

a) pn , 41 pn

b) 2pn , qpn 1

2pn , 31 pn

c) 3pn , 21 qn

qpn , 21 qn

d) 4pn primo y qn 1 primo.

Observamos que en todos los casos, uno de los dos números consecutivos es la potencia par de

un número primo, y con esto vamos probando casos:

2)5(,3)4(51,4

3)4(,2)3(41,3422

nn

nn ninguna cumple.

4)10(,3)9(101,9

3)9(,4)8(91,8932

nn

nn ambas cumplen la condición del enunciado.

2)17(,5)16(171,16

5)16(,4)15(161,151624

nn

nn cumple la segunda.

4)26(,3)25(261,25

3)25(,8)24(251,242552

nn

nn cumple la segunda.

6)50(,3)49(501,49

3)49(,10)48(491,484972

nn

nn ninguna cumple.

4)82(,5)81(821,81

5)81(,10)80(811,808134

nn

nn ninguna cumple.

4)122(,3)121(1221,121

3)121(,16)120(1211,120121112

nn

nn cumple la segunda.

8)170(,3)169(1701,169

3)169(,16)168(1691,168169132

nn

nn ninguna cumple.

8)290(,3)289(2901,289

3)289(,18)288(2891,288289172

nn

nn ninguna cumple.

4)362(,3)361(3621,361

3)361(,24)360(3611,360361192

nn

nn cumple la segunda.

Y paramos porque hemos conseguido las seis soluciones pedidas. El resultado es

8+9+16+25+121+361=540

Fuente de esta solución: www.artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_9

10

Podemos escribir

673673673222111

3

673

3

2

3

1

...

...)(

zxzxzxzxzxzxzxzxzx

zxzxzxxp

Con lo cual tenemos las 20196733 raíces de un polinomio de grado 2019, que se repiten en

grupos de tres:

673673673222111 ,,...,,,,,,, zzzzzzzzz

Podemos aplicar las fórmulas de Vieta (y observando detenidamente que los 2019 complejos

involucrados se van repitiendo en grupos de tres):

3

203...20

67316731

673673673222111

i

i

i

i zzzzzzzzzzz

ji

ji

i

i

ji

ji

i

i

ji

ji

i

i zzzzzzzzz 9193

1939319 222

Sea

6731 kj

kj zzS . Entonces Szi

i 9193

12

Por otro lado,

Szzzzzi

i

ji

ji

i

i

i

i 223

20

9

400

6731

2

6731

2

2

6731

Y por tanto

9

3432919

3

1

9

400 SSS

Y como nos piden su valor absoluto, el resultado es 9

343.

11 https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_11

12

Calculamos los “vectores de desplazamiento”:

)20(201919)( 222 zzzzzzzzzzzfza

)19)(1(20119

201919201919119)19(19

19)19(19)19()())((

22222

2222

zzazzzz

zzzzzzzzzzzz

zzzzzzzfzffb

Aplicando 20.7.5c tenemos:

iIRzziIRa

zzaiIR

a

bba

)19)(1(

)19)(1(

Por el enunciado, sabemos que ikz 11 , luego

ikkkkiikk

kikikkikik

zzzzzzz

ikz

kikikz

2219814018221981912118

19221211981819221119818

19181919)19)(1(

11181818

2211)11(

22

222

22

2222

Que será un imaginario puro si y solo si 2219014018 2 kkk

Por el enunciado, el complejo buscado será iz 112219 , y la respuesta es 2309221 .

13

Por el Teorema del Coseno, aplicado al triángulo ABC ,

4

3

60

253616coscos60253616

cos562564cos2 222222

CC

CCACABABACAB

Sabemos que ACBDFE . Esto es cierto pues

BCFACBACFFDADFEBEF y BCFBEF

Luego, de nuevo aplicando el Teorema del Coseno en el triángulo DEF :

2432

32214944

372272cos2 22222

DE

DEFEFDFEFDFDE

Sea CFABX , BXa , XDb .

Por PoP, abaababbbaXFCXba 2244244

a

XF

CX

bXBCXFEBEFBCF

5

724

Por otro lado, b

CX

XF

aXDFXCADFCCAD

4

2

6

Con estas ecuaciones ya podemos determinar a y b :

4

25

4

5

6

)4(2

5

7

)4(264

2

6

575

7

baaa

aXFXF

a

XFaa

XF

Y por tanto 4

221524

4

25

4

5 DEbaBE

Fuente: https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_13

14

Buscamos el menor número primo p tal que 12019| 8 p

)(mod12019

)(mod012019

12019|

8

8

8

p

p

p

Entonces, elevando al cuadrado ambos lados, )(mod1201916 p

Pero 16,8,4,2,1)2019()(mod1201916 pordp

Sin embargo, )(mod120198,4,2,1)2019( 8 pordp y no -1, como queríamos, luego

deducimos que 16)2019( pord .

Puesto que (p)|)2019( pord , (p) será un múltiplo de 16.

Puesto que por hipótesis p es primo, 11

1)(

p

ppp

Y por tanto )16(mod1p . Los dos primeros primos que cumplen )16(mod1p son 17 y 97.

Sin embargo, )17(mod120198 , pero )97(mod120198 , luego la solución es 97.

Fuente de esta solución: artofproblemsolving.com

15

Aplicando PoP, PXPYPBAP 3515 , y puesto que, además, 11PXPY ,

deducimos que 2

6111PX ,

2

6111PY .

La clave de este problema está en demostrar que Q es el punto medio del segmento XY. Pues

entonces

2

11

2

XYQY ,

2

61

2

6111

2

11

PYQYPQ

Sea Z el punto de corte de las dos tangentes a por A y B, respectivamente. AZ es también

tangente a 1 y BZ es tangente también a 2 . Sabemos que Z pertenece al eje radical de 1 y

2 , es decir, a PQ, y sabemos que PBZZAP .

Pero por 10.2.7 se cumple AQPZAP y PQBPBZ , así pues:

PQBPBZZAPAQP

De aquí ABZAQZ y por tanto AQBZ es un cuadrilátero cíclico.

Por otro lado, si O es el centro de la circunferencia , sabemos que OAZB es un cuadrilátero

cíclico de radio OZ (puesto que º90 ZAOZBO ), y por tanto tenemos que OAZBQ es

cíclico de diámetro OZ, luego, por Tales, º90OQZ , es decir, XYOQ , de donde se

deduce que Q es el punto medio de la cuerda XY, tal y como queríamos ver.

Fuente de esta solución: https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15

AIME II 2019 Enunciados

1

Sean C y D dos puntos distintos a un mismo lado de la recta AB, de forma que ABC y

BAD son congruentes con 9AB , 10 ADBC y 17 DBCA . Determina el

área de la región triangular intersección de estos dos triángulos.

2

Marcamos con 1, 2, 3, … unos nenúfares en fila en un estanque. Una rana genera una

secuencia de saltos empezando en el nenúfar 1. Desde el nenúfar k , la rana puede saltar

al nenúfar 1k o al nenúfar 2k al azar, con probabilidad 2/1 y con independencia

de los otros saltos. Determina la probabilidad de que la rana pase por el nenúfar 7.

3

Determina el número de 7-tuplas de números positivos ),,,,,,( gfedcba que satisfacen

el siguiente sistema de ecuaciones:

72

71

70

efg

cde

abc

4

Lanzamos un dado cuatro veces. ¿Cuál es la probabilidad de que el producto de los

cuatro lanzamientos sea un cuadrado perfecto?

5

Cuatro embajadores y sus respectivos cuatro consejeros se sientan en una mesa redonda

de 12 asientos, numerados del 1 al 12. Cada embajador debe sentarse en un asiento

numerado como par. Cada consejero debe sentarse en un asiento contiguo a su

embajador. ¿De cuantas formas posibles se pueden sentar?

6

Supongamos que estamos ante una civilización marciana en la que todo logaritmo cuya

base no está especificada se supone de base b , para cierto 2b fijo. Un estudiante

marciano escribe

56loglog3 xx

54)(loglog xx

y encuentra que este sistema de ecuaciones tiene como única solución cierto número

real 1x . Determina b .

7

Sea ABC el triángulo de lados 120AB , 220BC , 180AC . Trazamos las rectas

Al , Bl y Cl paralelas a los lados BC , AC y AB , respectivamente, de forma que las

intersecciones de Al , Bl y Cl con el interior de ABC sean segmentos de longitudes 55,

45 y 15, respectivamente. Determina el perímetro del triángulo cuyos lados pertenecen a

las rectas Al , Bl y Cl .

8

El polinomio 201620172018)( zczbzazf tiene coeficientes reales no superiores a

2019, y

ii

f 3201920152

31

Determina el residuo cuando )1(f se divide entre 1000.

9

Diremos que un entero positivo n es k-guapo si n tiene exactamente k divisores

positivos y es divisible entre k. Por ejemplo, 18 es 6-guapo. Sea S la suma de todos los

enteros positivos menores de 2019 que sean 20-guapos. Determina S/20.

10

Determina el único ángulo entre 0º y 360º tal que para todo entero no negativo n , el

valor de n2tan es positivo siempre que n sea múltiplo de 3, y negativo en caso

contrario.

11

Dado el triángulo ABC con 7AB , 8BC y 9CA . Sea 1 la circunferencia que

pasa por el punto B y es tangente a la recta AC en A, y sea 2 la circunferencia que

pasa por el punto C y es tangente a la recta AB en A. Sea AK el segundo punto de

intersección entre 1 y 2 . Determina AK.

12

Dado un 1n , diremos que una sucesión finita naaa ,...,, 21 de números enteros

positivos es progresiva si 1 ii aa y ia divide 1ia para todo 11 ni . Determian el

número de sucesiones progresivas cuya suma de términos sea igual a 360.

13

Sea un octágono regular 87654321 AAAAAAAA inscrito en un círculo de área 1. Sea P un

punto en su interior de forma que la región determinada por 1PA , 2PA y el arco menor

21AA de dicho círculo tiene área 1/7, y la región determinada por 3PA , 4PA y el arco

menor 43AA de dicho círculo tiene área 1/9. Determina el área de la región determinada

por 6PA , 7PA y el arco menor 76 AA .

14

Determina todos los posibles valores de enteros positivos n para los cuales 91 céntimos

es el mayor valor que no se puede formar disponiendo de una infinita cantidad de sellos

de 5, n y 1n céntimos.

15

Dado un triángulo acutángulo ABC , sean P y Q los pies de las perpendiculares de C

en AB y de B en AC , respectivamente. La recta PQ corta el circuncírculo de ABC en

los puntos X, Y. Supongamos que 10XP , 25PQ y 15QY . Determina el valor de

ACAB .

AIME II 2019 Soluciones

1

Calculamos el área del triángulo ABC mediante la fórmula de Heron:

3698118

99

810

117

182/10179

ABC

s

s

s

s

Sea P el punto de corte entre la altura de ABC por C y la recta AB.

892

136 PCPCABC

Por Pitágoras:

63664100810 22 BP

Sea E el punto de corte entre AC y BD . El triángulo ABE es isósceles en E puesto que

ABC y BAD son congruentes y por tanto CABDBA .

Sea Q el punto de corte entre la altura de ABE por E y el lado AB .

Por ser ABE isósceles, EP es mediana y por tanto 2

9

2

ABAQ

Por Tales, 5

12

69

8

2/9

EQ

EQ

AP

PC

AQ

EQ

Y finalmente, 5

54

5

129

2

1

2

1 EQABAEB

2

Generamos el árbol de posibilidades, hasta con 6 saltos:

Vemos que en total hay 62 caminos diferentes, de los cuales son favorables los siguientes:

1 camino en el salto 6.

5 caminos en el salto 5, en total 25 .

6 caminos en el salto 4, en total 226 .

1 caminos en el salto 3, en total 321

El total de caminos favorables es 432126251 32

La probabilidad es, por tanto, 64

43

2

436P

3

23 3272

7171

75270

efg

cde

abc

La clave está en la segunda ecuación: 71cde que es un número primo, luego solo puede

ocurrir uno de los tres casos siguientes:

a) 71,1,1 edc

Pero entonces 7271 fg , lo cual es imposible.

b) 1,1,71 edc

Pero entonces 7071ab , lo cual es imposible.

c) 1,71,1 edc

Tenemos el sistema

23 32

752

fg

ab

La primera ecuación permite los siguientes valores de a :

752,75,72,52,7,5,2,1 a , ocho casos diferentes.

La segunda ecuación permite los siguientes valores de f :

2332222232 32,32,32,32,32,32,3,3,2,2,2,1 f , doce casos diferentes.

Luego el total de casos es 96128 .

4

El número total de casos es 46 .

Sea nC el número de casos de obtener un cuadrado lanzando un dado dos veces, si ya tenemos

un n previo.

8)1,1( C : (1,1), (2,2),(3,3),(4,4),(5,5),(6,6),(1,4),(4,1)

6)1,2( C : (2,1), (1,2),(3,6),(6,3),(2,4),(4,2)

6)1,3( C : (1,3), (3,1),(3,4),(4,3),(2,6),(6,2)

8)1,4( C : (1,4), (4,1), (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)

4)1,5( C : (5,1), (1,5),(5,4),(4,5)

6)1,6( C : (1,6), (6,1),(2,3),(3,2),(4,6),(6,4)

8)2,2( C : (1,1), (2,2),(3,3),(4,4),(5,5),(6,6),(1,4),(4,1)

6)2,3( C : (2,3), (3,2) (1,6),(6,1),(4,6),(6,4)

6)2,4( C : (2,1), (1,2),(4,1),(1,4),(3,6),(6,3)

2)2,5( C : (5,2), (2,5)

6)2,6( C : (3,1), (1,3) (3,4),(4,3),(6,2),(2,6)

8)3,3( C : (1,1), (2,2),(3,3),(4,4),(5,5),(6,6),(1,4),(4,1)

6)3,4( C : (3,1), (1,3),(6,2),(2,6),(3,4),(4,3)

2)3,5( C : (5,3), (3,5)

6)3,6( C : (2,1), (1,2),(2,4),(4,2),(6,3),(3,6)

8)4,4( C : (1,1), (2,2),(3,3,),(4,4),(5,5),(6,6),(4,1),(1,4)

4)4,5( C : (5,1), (1,5),(5,4),(4,5)

6)4,6( C : (6,1), (1,6), (2,3),(3,2),(4,6),(6,4)

8)5,5( C : (1,1), (2,2),(3,3),(4,4),(5,5),(6,6),(4,1)(1,4)

2)5,6( C : (5,6), (6,5)

8)6,6( C : (1,1), (2,2),(3,3),(4,4),(5,5),(6,6),(1,4),(4,1)

Ahora escribimos todas las posibilidades en función de los dos dados primeros: 1,1 8

1,2 6

1,3 6

1,4 8

1,5 4

1,6 6 Total: 38

2,1 6

2,2 8

2,3 6

2,4 6

2,5 2

2,6 6 Total: 34

3,1 6

3,2 6

3,3 8

3,4 6

3,5 2

3,6 6 Total: 34

4,1 8

4,2 6

4,3 6

4,4 8

4,5 4

4,6 6 Total: 38

5,1 4

5,2 2

5,3 2

5,4 4

5,5 8

5,6 2 Total: 22

6,1 6

6,2 6

6,3 6

6,4 6

6,5 2

6,6 8 Total: 34

TOTAL: 38+34+34+38+22+34=200

162

25

6

2004P

Observación: En las soluciones oficiales se presentan varias alternativas, siempre teniendo en

cuenta esta o aquella propiedad. Como suele pasar, hay que valorar entre utilizar la fuerza bruta

del cómputo o aprovechar alguna propiedad para eliminar casos.

5

Lo primero que debemos observar en este problema es que los asientos están numerados, por lo

que no hay equivalencia por rotaciones.

Denotaremos por 4321 ,,, EEEE los embajadores y por 4321 ,,, CCCC sus respectivos cuatro

consejeros.

Aunque trabajamos con una mesa redonda, ponemos las posiciones en fila para mayor

comodidad. Marcamos en gris los asientos pares, los de los embajadores.

En primer lugar, colocamos a los embajadores. Hay que dejar dos sitios marcados como par

libres. Luego colocaremos los consejeros alrededor de estos.

Primer caso: Los dos sitios libres están juntos, y los cuatro embajadores juntos entre ellos.

Los huecos pueden ser (2,4), (4,6), (6,8), (8,10), (10,12) o (12,2), y en cada caso hay !4 formas

de colocar los embajadores, luego !46

Para colocar los consejeros:

1 2 3 4 5 6 7 8 9 10 11 12

E1 E2 E3 E4

C1 C2 C3 C4

C1 C2 C3 C4

C1 C2 C3 C4

C1 C2 C3 C4

C1 C2 C3 C4

Hay 5 configuraciones diferentes de consejeros. El total es 7205!46 combinaciones.

Segundo caso. Los asientos están separados por un embajador.

Pueden ser (2,6), (4,8), (6,10), (8,12), (10,2), (12,4) y para cada uno de ellos, hay !4 formas de

colocar los embajadores. En total: !46

Si los dos asientos libres están separados por algún consejero, digamos el "4" y el "8", las

posibilidades de colocar los consejeros son las siguientes:

1 2 3 4 5 6 7 8 9 10 11 12

E1 E2 E3 E4

C1 C2 C3 C4

C1 C2 C3 C4

C1 C2 C3 C4

C4 C1 C2 C3

C4 C1 C2 C3

C4 C1 C2 C3

C4 C1 C2 C3

C1 C2 C3 C4

Hay 8 formas posibles de situar a los consejeros.

Por tanto, el número total de posibilidades es 1152!468 configuraciones distintas.

Tercer caso.

Si hay dos embajadores entre los dos huecos, los huecos pueden ser: (2,8), (4,10), (6, 12), y de

nuevo hay !4 formas diferentes de poner los embajadores. En total: !43

Ahora, para colocar los consejersos, digamos que los huecos son el "4" y el "10":

1 2 3 4 5 6 7 8 9 10 11 12

E1 E2 E3 E4

C1 C2 C3 C4

C1 C2 C3 C4

C1 C2 C3 C4

C4 C1 C2 C3

C4 C1 C2 C3

C4 C1 C2 C3

C1 C2 C3 C4

C1 C2 C3 C4

C1 C2 C3 C4

Hay 9 posibilidades distintas. En total 6489!43

El número total de posibilidades es: 25206481152720

Fuente de esta solución: artofproblemsolving.com

6

Utilizando la fórmula del cambio de base: b

cc

a

ab

log

loglog

54

logloglog

loglog

log)(log54 log

xx

x

xxx

3614

27

3

56log

log27

14log

54

1

2

1

54

loglog

2

1

54

logloglogloglogloglog

3

56

3

56loglog56loglog3

x

xxx

x

xxxxxx

xxxx

Volviendo ahora a la segunda ecuación: 2754

36log 363654)(log54)(log xxxxx

2163636

3

236log36log28336log33636log356

2/33/2

2827

bb

7

Sean A’, B’, C’ los puntos de intersección de las rectas Al , Bl y Cl , y añadimos los puntos de

intersección D, E, F, G, H, I tal y como aparecen en el siguiente esquema:

Queremos determinar el perímetro del triángulo ''' CBA , que es semejante al triángulo

ABC , pues determina ángulos iguales por paralelismo de lados.

55180

22045

30180

12045

180

45

220120BG

HBBGHB

AC

HG

BC

BG

AB

HBABCHBG

2/45120

18015

2/55120

22015

120

15

180220CE

FCCEFC

BA

FE

CA

CE

BC

FCBCAFCE

30220

55120

45220

18055

220

55

120180AI

DAAIDA

BC

ID

AB

AI

AC

DACABDAI

Luego 603030120 HBAIABHI

Y por tanto

1102/5544'

902/4544'460

CFIC

ECHCEFHI

2/2752/5555220 CFBGBCFG

Y por tanto:

2/22555

452/275'

7555

302/275'

55

2/275

45

'

30

''''

GA

FAGAFA

ID

GF

AD

GA

AI

FAADIGFA

Finalmente, 2/49590452/225'''' HCGHGACA

8

11

180

2/495''

AC

CA que es la razón de proporcionalidad entre ''' CBA y ABC .

Luego el perímetro de ''' CBA será 8

11 veces el perímetro de ABC :

7151802201208

11

8

El número 2

31 ip

en notación exponencial es ie º60 , luego:

º60sinº60cosº120sinº120cosº1202 iiep i

1º1803 iep

1º3606 iep , y por tanto 11366366336632016 ppp

czbzazzczbzazf 22016201620172018)(

32019º60sin)(

2015º60cosº60sin)(º60cos

º60sinº60cos60sinº60cos

320192015 22016

ba

cabibacab

cbibaia

cpbpappfi

4038320192

3)(32019º60sin)( bababa

Puesto que, por hipótesis, 2019, ba , 20194038 baba

Finalmente, sustituyendo en la primera ecuación:

20152015º60cos ccab

Luego 6053201520192019111)1( 201620172018 cbacbaf , y su residuo al

dividir por 1000 es 53.

9

Sea n un número que cumpla las condiciones del enunciado.

n|20 , luego kn 20

2019n , luego 100k

kkn 522020 2

Aplicaremos detenidamente la fórmula 19.4b.

El exponente 2 del factor 2 genera un 3 en la función , luego será imposible obtener 20. Hay

que aumentar este exponente.

kk 5225220 32

Ahora tenemos un 824 , que nos impide obtener el 20.

Si aumentamos el exponente del 5 tampoco nos sirve, pues entonces

k 235220

Obtenemos un 1234 … y no son divisores de 20.

Así pues, el factor 2 tiene que tener exponente 4, luego:

kk 5225220 422

Ahora tenemos un factor 1025 y solo tenemos que multiplicarlo por 2, lo que sucede si y

solo si añadimos un primo p diferente de 2 y de 5. Luego serán todos los números de la forma

ppn 420524 , con 251004 pp , es decir: 23 19, 17, 13, 11, 7,,3p .

Su suma será 9352231917131173522352...1152352 44444

Otra combinación posible es 20005220 34 , luego la suma total será 344 529352

Y el resultado pedido es 4721003725293220

529352 222344

10 https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10

11

Primera versión. Mediante Triángulos semejantes.

Observamos que por ser AC tangente a 1 , por 10.2.7 se cumple KBAKAC , y por ser

AB tangente a 2 se cumple KCABAK .

Luego CKAAKB , y por tanto

KAKC

KAKB

KA

KC

KB

KA

7

9

9

7

7

9

Aplicando el Teorema del Coseno:

21

11

972

978)cos()cos(972978

222222

BACBAC

Y ahora observamos que BACBKA º180 . En efecto:

BAC

ABKBAKBKABAKABKBKA

º180

180180

Luego 21

11)cos(º180 BKABACBKA

Aplicando el Teorema del Coseno en el triángulo ABK :

2

9

4

81

81

19649

81

19649

21

11

9

72

81

4949

)cos(9

72

9

77

222222

2

22

AK

AKAKAKAKAKAK

BKAAKAKAKAK

Segunda versión. Mediante inversión.

Consideremos la inversión de centro A y radio AK . Sean *B y *C las imágenes respectivas

de B y C bajo esta inversión.

La imagen de 1 bajo esta inversión será una recta que pasará por K y *B , cumpliendo 22 *7* AKABAKABAB

La imagen de 2 bajo esta inversión será una recta que pasará por K y *C , cumpliendo 22 *9* AKACAKACAC

Observamos que **KCAB es un paralelogramo puesto que 1 y 2 son tangentes a AC y

AB, respetivamente. Luego ** ACKB y BACACBKAB 180**180* .

BACKABBACACBKAB cos*cos180**180*

Calculamos este coseno mediante el Teorema del Coseno como en la versión anterior:

21

11

972

978)cos()cos(972978

222222

BACBAC

Y aplicando el Teorema del Coseno llegamos a

2

9

132322

81491

21

11

972

97

)*cos(**2**

222

222

22

2

222

AKAKAKAK

AKAKAKAK

KABKBABKBABAK

Fuente de estas soluciones:

https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_11

12 https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_12

13 https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_13

14

El problema que se plantea es estudiar todas las combinaciones posibles

15 nznyx

con enteros 0,, zyx

Y determinar aquellos enteros positivos n para los cuales 91 no se puede formar pero sí se

pueden formar 92, 93, 94…

Está claro que, independientemente del valor n , se pueden formar todos los múltiplos de 5,

pues basta ir dando valores a x , con 0 zy .

Para un valor n cualquiera, está claro que podremos formar seguro todos los valores de la

forma n , 5n , 10n , 15n ,… También todos los valores de la forma 1n ,

651 nn , 11101 nn ,…

Todo esto nos indica que la clave para resolver este problema es pasar a módulo 5:

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

91 92 93 94 95

96 97 98 99 100

Aplicamos el Teorema “Chicken McNugget”, a las combinaciones 5 y n:

245591 nnn

Luego podemos garantizar que 24n .

Primer caso: 5mod1n

Supongamos que n se encuentra en la primera columna, es decir, que 5mod1n .

Vemos que 91 también está en la primera columna, luego nos obliga a 91n , es decir:

...,101,96n

Pero entonces no podríamos obtener 92 (ni 93, 94…) con lo que dicho número no es aceptable.

Segundo caso: 5mod2n

Si n se encuentra en la segunda columna, 1n se encontrará en la tercera, y por tanto n2 se

encontrará en la cuarta. En efecto:

5mod4410252225 kknkn

También vemos que 5mod12212 nn se puede obtener, y está en la primera

columna.

Luego 9112 n , y el primer candidato interesante sería el siguiente: 9612 n

479612 nn

Comprobamos que, efectivamente, 47n satisface las condiciones del enunciado:

Están todos los múltiplos de 47, en particular 5947454792 , todos los múltiplos de

48, 94472 y todos los múltiplos de 94, y observamos que 91 no se puede representar como

combinación de 47, 48 y 5, pero sí podemos representar 482)147(296 .

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

47 48

91 92 93 94 95

96 97 98 99 100

Tercer caso: 5mod3n

5mod1625mod3 nn , y por tanto n2 se encontrará en la primera columna. Como

en el caso anterior, deducimos que 912 n y el primer candidato interesante sería el siguiente:

48962 nn

Pero con este valor nos encontramos con un problema en la segunda columna: No se puede

representar 92, y por tanto no es válido.

En efecto, todo consiste en ir viendo las posibles combinaciones:

97494851,1,0 zyxzyx

102494851,1,1 zyxzyx

93494850,1,9 zyxzyx

…

Cuarto caso: 5mod4n

En este caso 5mod51n y es redundante, 5mod382 n , 5mod2123 n y

5mod14 n . Como en los casos anteriores, imponemos la condición

24964 nn

Y vemos que con este valor todas las cinco columnas están cubiertas, es decir, es una segunda

solución al problema.

Quinto caso: 5mod05 n

En este caso tendríamos 5mod11n , es decir, en la primera columna, y por tanto

92911 nn . El primer candidato sería 95n , pero con este valor no podríamos formar

92 (ni 93…), luego no es aceptable.

Así pues, las soluciones de este problema son 24n y 48n .

Fuente de esta solución: Vídeo “Momentum Learning” https://youtu.be/fTZP2e-_rjA

15

Primera versión.

Sean APa , BPb , AQc , CQd .

Aplicando potencia de un punto:

525)2510(15

400)1525(10

cd

ab

Observamos que el cuadrilátero PQCB es cíclico, pues º90 BQCBPC .

Luego BAQP y CAPQ , y por tanto AQPAPQ , y en consecuencia

125

525400

22

2222

ca

cacdcabadccbaaba

c

dc

a

Por otro lado, aplicando el Teorema del Coseno en APQ :

Aacca cos225625 222

Y observamos que ba

cA

cos

Luego

525

12521252

400125

1252125

4002125

4002

4002225625

2

222

2

2222

2

2222

2

22222222

c

ccc

c

cccc

a

accc

a

aacca

aa

cacca

ba

cacca

Luego todo se reduce a solucionar la ecuación

35587587505252501500400525250250

400125525)525(250

125)525()525(250525

125)525(250

525

125250

525

12522500

525

12521252625

22222

224242

22222

2

2222

2

222

2

222

2

222

cccccc

cccccc

cccccc

cccc

c

ccc

c

ccc

c

ccc

Con este resultado se deducen el resto de incógnitas:

35

105

355

525525 dcd

10101000100012587512522 aca

10

40

1010

400400 bab

Finalmente,

145603510112

35810143533551041010

35

35105355

10

10401010

35

105355

10

401010))((

dcbaACAB

Segunda versión.

Sean APa , BPb , pero ahora definimos Ak cos .

Luego k

bABk

AB

b ,

k

aACk

AC

a

Como en la versión anterior, mediante Potencia de puntos tenemos

525)()1510(15525

400)()1525(10400

2

2

2

2

b

abkb

k

abb

k

abbACbCQb

a

abka

k

aba

k

baaABaBPa

De donde deducimos que

125525400525400

2222

22

baba

b

ab

a

abk

Por otro lado, aplicando el Teorema del Coseno en APQ :

abkba 225625 222

Para resolver este sistema de ecuaciones realizamos el siguiente cambio de variable:

525400 22 bau

Luego u

abk , 400 ua , 525 ub y por tanto:

875

10001400

)252)(400(775

)252)(400(221550

)252)(400(292522525400625

22

b

au

u

uuu

u

uuu

u

uuu

u

bauu

Y, finalmente, 145608571000

140014002

ab

u

u

ab

u

ab

ab

k

b

k

aACAB

Fuente de estas soluciones: www.artofproblemsolving.com

AIME I 2018

1

Sea S el número de pares ordenados de enteros ),( ba , con 1001 a y 0b tales que

el polinomio baxx 2 se puede factorizar como un producto de dos factores lineales

no necesariamente diferentes con coeficientes enteros. Determina el residuo cuando S se

divide entre 1000.

2

El número n se escribe en base 14 como cba , se escribe en base 15 como bca y se

escribe en base 6 como caca , con 0a . Determina el número n en base 10.

3

Kathy tiene 5 cartas rojas y 5 cartas verdes. Las mezcla y toma 5 al azar, que coloca

alineadas. Digamos que Kathy quedará contenta si y solo si todas las cartas rojas

tomadas están han quedado adyacentes, y todas las cartas verdes han quedado

igualmente adyacentes. Por ejemplo, si sale “RRVVV”, “VVVVR” o “RRRRR”

quedará contenta, pero si sale “RRRVR” no lo será. ¿Cuál es la probabilidad de que

quede contenta?

4

Sea ABC un triángulo en el que 10 ACAB y 12BC . Sea D un punto situado en

el interior del lado AB y E un punto situado en el interior de AC tales que

ECDEAD . Determina AD .

5

Para cada par ordenado de números reales ),( yx satisfaciendo

22

42 7log2log yxyxyx

Existe un número real K tal que

22

93 43log3log Kyxyxyx

Determina el producto de todos los valores posibles de K.

6

Sea N la cantidad de números complejos z con la propiedad que 1z y !5!6 zz sea

un número real. Determina el residuo cuando N se divide entre 1000.

7

Sea un prisma hexagonal recto de altura 2, cuyas bases son hexágonos regulares de lado

1. Cualquier grupo de tres vértices del total de 12 determinan un triángulo. Determina el

número de esos triángulos que son isósceles (incluyendo triángulos equiláteros).

8

Sea ABCDEF un hexágono equiangular tal que 6AB , 8BC , 10CD , 12DE .

Denotamos por d el diámetro de la circunferencia más grande que podemos trazar en el

interior del hexágono. Determina 2d .

9

Determina el número de subconjuntos de cuatro elementos de 20,...,4,3,2,1 con la

siguiente propiedad: Contiene dos números distintos que suman 16 y dos números

distintos que suman 24. Por ejemplo: 19,13,5,3 y 18,20,10,6 son aceptables.

10

Disponemos de una rueda consistente en dos circunferencias y cinco radios, con una

etiqueta en cada punto de contacto entre los radios, tal y como aparece en la figura. Un

gusano avanza por la rueda, empezando en el punto A. Este gusano avanza en cada paso

desde un punto etiquetado a uno adyacente. Por la circunferencia interna este gusano

solo puede avanzar en el sentido contrario al de las agujas del reloj, mientras que por la

circunferencia exterior este gusano solo puede avanzar en el sentido de las agujas del

reloj. Por ejemplo, el gusano puede realizar el camino AJABCHCHIJA, que tiene 10

pasos. Sea n el número de caminos de 15 pasos que empiezan y acaban en A.

Determina el residuo cuando n se divide entre 1000.

11

Determina el menor entero positivo n tal que, cuando ese escribe n3 en base 143, sus

últimas dos cifras en dicha base son 01.

12

Para cada subconjunto T de 18,...,3,2,1U , sea Ts la suma de los elementos de T,

definiendo 0s . Si T se toma aleatoriamente entre todos los subconjuntos de U, la

probabilidad de que Ts sea divisible entre 3 es nm / , con m y n enteros positivos

coprimos. Determina m.

13

Sea ABC el triángulo con lados 30AB , 32BC , 34AC . Fijado un punto X en

el interior del lado BC , sean 1I , 2I los incentros de los triángulos ABX y ACX

respectivamente. Determina el área mínima del triángulo 21IAI cuando X se mueve en

el interior del lado BC .

14

Sea un heptágono 54321 PEPPPSP . Una rana empieza saltando desde el vértice S. Desde

cualquiera de los vértices del heptágono excepto E, la rana puede saltar a cualquiera de

sus dos vértices adyacentes. La rana se para cuando alcanza el vértice E. Determina el

número de secuencias diferentes de no más de 12 saltos que acaban en E.

15

David ha encontrado cuatro palitos de diferentes longitudes que se pueden usar para

formar tres cuadriláteros convexos cíclicos no congruentes entre ellos: A, B y C, que se

pueden inscribir en una circunferencia de radio 1. Denotaremos por A la medida del

ángulo agudo determinado por las diagonales del cuadrilátero A, y definimos B y C

de forma similar. Supongamos que 3/2sin A , 5/3sin B y 7/6sin C . Los tres

cuadriláteros tienen la misma área K, que se puede escribir de la forma nm / para

ciertos enteros positivos m y n coprimos. Determina nm .

Soluciones.

1

211221212211

2 xxxxbaxx

Luego

0,0

1,1

2121

1221

2121

b

a

Luego el problema se reduce a encontrar todas las parejas ),( ba de la forma

0,,1001 2112 a

Haciendo una tabla de doble entrada 21, , eliminado las parejas cuya suma excede 100 por la

derecha y eliminando las repeticiones que van produciendo por la izquierda:

El total es:

2601515010151

2

515025110125110121011...9799101

2

50

0

50

0

kk

kk

Pero debemos restar 1 de la combinación 021 que tampoco es aceptable.

Así pues, 2600S , y la respuesta correcta es 600.

2

Las condiciones del enunciado se corresponden con el siguiente sistema de ecuaciones

diofánticas:

cacan

bcan

cban

666

1515

1414

23

2

2

cumpliendo, además: 60 a , 60 c , 140 b

Simplificamos la tercera ecuación:

cacaacacan 37222)16()66(666 2323

y la sustituimos en las otras dos ecuaciones:

cbabcaca

cbacbacbaca

2230151537222

1871303614260141437222

2

2

Multiplicamos la segunda ecuación por 7 y le restamos la primera:

cacacba

cba434136340

1547210

187130

Luego ca 4

Puesto que 60 a , la única posibilidad es que 1c , y por tanto 4a , y

925378881374222 n

3

El número total de casos posibles, puesto que hay diez cartas, y las tomamos de 5 en 5, será

6789105

10 V . (Y no de 3225 como se podría suponer).

Para contabilizar los casos favorables, los vamos a ordenar en función del número de cartas

rojas que aparezcan.

a) 0 cartas rojas: “VVVVV”.

Hay 12012345 casos diferentes.

b) 1 carta roja: “RVVVV”.

Hay 60023455 casos, y puesto que la carta roja puede estar al principio o al final, hay

12002600 casos diferentes.

c) 2 cartas rojas: “RRVVV”.

Hay 120034545 casos, y puesto que las dos “RR” pueden estar al principio o al final,

hay 240021200 casos diferentes.

d) 3 cartas rojas: “RRRVV”

Hay 120045345 casos, y puesto que las dos “RRR” pueden estar al principio o al final,

hay 240021200 casos diferentes.

e) 4 cartas rojas: “RRRRV”.

Hay 60052345 casos, y puesto que la carta verde puede estar al principio o al final, hay

12002600 casos diferentes.

f) 5 cartas rojas: “RRRRR”.

Hay 12012345 casos diferentes.

En total: 74401201200240024001200120 casos favorables.

La probabilidad es, por tanto, 126

31

678910

7440

P

4

Aplicando el Teorema del Coseno al triángulo ABC :

25

7

10102

101012coscos10102101012

222222

AA

Sea ECDEADx . Aplicando el Teorema del Coseno al triángulo ADE , que es

isósceles en D, pues DEAD , y por tanto AAED ,

39

2500

25

1410

0010

25

7210)10(0

25

7)10(2)10(

cos2

222

222

xx

x

xx

xxx

xxxxx

AEDDEAEDEAEAD

La primera solución queda descartada pues suponemos que el punto está en el interior del

segmento, y por tanto la solución es 39

250

5

De la identidad ccb aba logloglog se deduce

2

loglogloglog2logloglog 222

2 cccccca a

aaaaaa

Que se puede aplicar a la primera ecuación:

22

2

2

2

22

22

22

222

2

22

42

7log2log7log2log2

2

7log7log7log2log 2

yxyxyxyxyxyx

yxyxyxyxyxyxyx

yx

yxyxyxyxyx

yxyxyxyxyxyxyxyx

yxyxyxyxyxyxyx

2220

236337440

74472

22

22222222

2222222

Y también a la segunda:

22222

3

2

3

22

33

22

3

22

3

22

93

43343log3log

43log3log22

43log

43log43log3log 2

KyxyxyxKyxyxyx

KyxyxyxKyxyx

KyxyxKyxyxyx

Si yx esta ecuación queda

2222222)7(16)7(434 xKxxKKxxxx

Puesto que 0 yx no es aceptable pues entonces 0log3log 33 yx y no existe el

logaritmo de cero, podemos suponer 0x y simplificar 2x para llegar a

9716716 KK

Si yx 2 , la segunda ecuación queda:

222

22222

222

)4(255

)4(8126

)2(4)2(323

yKyy

yKKyyyyy

Kyyyyyy

De nuevo, puesto que la opción 0 yx no es aceptable, podemos simplificar el factor 2y para llegar a

21425 KK

Así pues, la respuesta es 189921 .

6

Primera versión.

Pasando a forma exponencial: iez , iezz 720720!6 iezz 120120!5

!5!6!5!6 ImIm zzIRzz , y, puesto 1z , esto solo puede pasar si sucede una de las

dos condiciones siguientes:

a) !5!6 zz 0600120720120720 ii ee ,

ecuación que tiene 600 soluciones, puesto que estamos trabajando 2mod .

b) !5!6 , zz son reflexiones la una de la otra respecto del eje imaginario.

Luego 840120720120720 ii ee , ecuación que tiene 840 soluciones.

Un total de 1440840600 soluciones, y por tanto su residuo al dividirlo entre 1000 es 440.

Segunda versión.

Puesto que 1z , 1

1

z

z

Puesto que !5!6 zz es real, será igual a su conjugado, luego

120720

120720

120720

120720120720120720!5!6 1111

zzzz

zzzzzzzzzz

Multiplicando ambos lados por 720z obtenemos un polinomio de grado 1440 , que tendrá 1440

soluciones complejas.

Tercera versión.

0)120sin()720sin()120sin()720sin(ImIm !5!6!5!6 zzIRzz

Aplicando la identidad trigonométrica "Resta-A-Producto", tenemos

02

600sin

02

840cos

2

600sin

2

840cos2

2

600sin

2

840cos2

2

120720sin

2

120720cos2)120sin()720sin(0

La primera ecuación tiene 840 soluciones y la segunda 600, haciendo un total de 1440

soluciones.

Fuente: Soluciones #3, #4 y #2 respectivamente de la web "artofproblemsolving.com"

7

En el interior de las bases vemos que, fijado un vértice A, podemos formar los triángulos

isósceles ABF (6 triángulos) y ACE (dos triángulos) del siguiente esquema:

con un total de 1628 triángulos.

De una base a la otra, vemos que, fijado un vértice A, podemos formar los triángulos isósceles

ABC y ADE del siguiente esquema:

con un total de 24226 triángulos.

Pero, puesto que la longitud entre vértices opuestos del hexágono es 2, y coincide con la altura,

también los triángulos ABC del siguiente esquema:

con un total de 1226 triángulos.

En total podemos formar 52122416 triángulos isósceles.

8

Sabemos que la suma de ángulos internos de un polígono de n lados es igual a 2180 n ,

luego en nuestro caso , º720418026180 .

Puesto que es equiangular, cada uno de los ángulos internos medirá º1206

720 .

Podemos dibujar este hexágono mediante regla y transportador de ángulos, y observaremos que

sus lados opuestos son paralelos.

En efecto, prolongando los lados AB y CD hasta encontrarse en un punto G, vemos que

º60º60º60º180º60º120180 BGCBCGCBG

Y por tanto el triángulo BGC es un triángulo equilátero de lado 8.

Por otro lado, º60AGD , y es igual al ángulo suplementario de º120EDG , luego

AB//DE. Y con razonamientos similares se demuestra que EF//BD y AF//CD.

Así pues, cualquier circunferencia en el interior de este hexágono tendrá diámetro máximo la

distancia entre lados opuestos, y vemos que los lados más próximos son AF y CD. Así pues

solo nos queda encontrar la distancia d entre AF y CD.

Por trigonometría:

147349372

314

14sin

2

3 2 ddd

AG

dAGD

Segunda versión.

Un razonamiento alternativo podría ser el siguiente: Puesto que la figura es equiangular, se

puede completar hasta formar un triángulo equilátero GHI :

Los lados de este triángulo miden 246810 , y por tanto

2101224 EF , 162624 AF .

El triángulo GHI es semejante a CHD , que tiene altura

35510 22 h

Luego:

147373531235312

3510352424

35

10

35

24102

ddd

dddhh

9

Vamos a ordenarlos por subconjuntos que contengan una pareja que sume 16, y vamos a añadir

dos números más para que también haya dos que sumen 24.

},,15,1{ ba }4,20,15,1{ , 21}*,9,15,1{

},,14,2{ ba }4,20,14,2{ , 21}*,10,14,2{

},,13,3{ ba }4,20,13,3{ , 21}*,11,13,3{

},,12,4{ ba 21}*,20,12,4{

},,11,5{ ba }4,20,11,5{ , 21}*,19,11,5{ , 21}*,13,11,5{

},,10,6{ ba }4,20,10,6{ , 21}*,18,10,6{ , 21}*,14,10,6{

},,9,7{ ba }4,20,9,7{ , 21}*,17,9,7{ , 21}*,15,9,7{

Todos ellos suman 21662110 .

Pero vemos que hay alguna repetición, hay seis elementos que los hemos contado dos veces:

Los grupos 21}*,11,13,3{ y 21}*,13,11,5{ comparten el elemento }3,13,11,5{ .

Los grupos 21}*,10,14,2{ y 21}*,14,10,6{ comparten }2,14,10,6{

Los grupos 21}*,19,11,5{ y 21}*,13,11,5{ comparten }13,19,11,5{

Los grupos 21}*,18,10,6{ y 21}*,14,10,6{ comparten }18,14,10,6{

Los grupos 21}*,17,9,7{ y 21}*,15,9,7{ comparten }17,15,9,7{

Los grupos 21}*,9,15,1{ y 21}*,15,9,7{ comparten }1,15,9,7{

Por lo tanto hay 2106216 subconjuntos.

10

Primera versión: Mediante recursión pura.

Vamos a resolver este problema de forma recursiva, rellenando, paso a paso, una tabla en la

que las columnas son los puntos finales, y las filas son el número de pasos. En cada casilla

pondremos el número de caminos que hay para llegar al punto final (columna) mediante

caminos de n pasos (fila), siempre saliendo del punto A.

Por ejemplo, en la primera fila hay un 1 en B y en J, porque con caminos de 1 paso, y saliendo

de A, solo podemos llegar a B y a J, y de una sola manera.

En la segunda fila, hay un 1 en la columna A indicando que hay un camino posible de longitud

2 desde A, que es "AJA", un camino posible para llegar a C, que es "ABC", un camino posible

para llegar a F, que es "AJF", y un camino posible para llegar a I, que es "ABI".

Vemos claramente la pauta:

11 nnn JEA , 11 nnn IAB , 11 nnn HBC , 11 nnn GCD , 11 nnn FDE

11 nnn JEF , 11 nnn FDG , 11 nnn GCH , 11 nnn HBI , 11 nnn IAJ

n A B C D E F G H I J

1 0 1 0 0 0 0 0 0 0 1

2 1 0 1 0 0 1 0 0 1 0

3 0 2 0 1 1 0 1 1 0 2

...

14

15

Está claro que este problema, así planteado, requiere un trabajo más propio de una hoja de

cálculo que de un humano:

En todo caso llegamos a 300415 A y por tanto la respuesta correcta es 41000mod3004 .

Segunda versión: Determinando una fórmula recursiva.

Observamos un dato interesante: En cada paso, el gusano tiene siempre dos caminos posibles

para continuar, luego el número total de caminos es n2 :

n

nnnnnnnnn IHGEDCBAS 2

Con este dato podemos optimizar el proceso para llegar a 15A . En efecto:

3323333

222211

nnnnnnn

nnnnnnn

HBAJEGC

IAFDJEA

3213

323

3333

3333333333332

nnnn

nnn

nnnn

nnnnnnnnnnnnn

IJAS

IES

IFDS

IFDJIHGFEDCBA

31

3

313

33313

33313

2

nn

n

nnn

nnnnn

nnnnn

AA

AAS

IIAAS

IIAAS

Calculando directamente los primeros valores:

00 A , 01 A , 12 A , 03 A , 34 A

Y utilizando la fórmula recursiva anterior se llega a 300415 A .

Fuentes: https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_10

https://youtu.be/JD51G1wUKEs

11

Sabemos que decir que la expresión de n3 en base 143 es 143123 01... ddd es equivalente a

2

21

23

2

2

1 143mod13...1431431...143143143013 nn dddd

Así pues, en este problema nos piden resolver la congruencia 2143mod13 n con solución n

mínima, es decir, determinar el orden de 3 módulo 2143 .

Es importante remarcar que nos piden n mínimo, pues el Teorema de Euler (13.4) nos

garantiza que 2)143( 143mod132

puesto que 1143,3 2 , pero en este caso 171601432

y veremos que no es el valor mínimo posible.

222 13111431311143 , y puesto que 113,11 22 , para resolver la congruencia

22 1311mod13 n será suficiente resolver las congruencias 211mod13 a y

213mod13 b por separado.

Paso 1. Resolvemos la congruencia: 211mod13 a

Esta es sencilla. Se puede encontrar por tanteo:

121mod3331 121mod9932

121mod272733 121mod818134

121mod11121224335

El valor mínimo es 5a .

Paso 2. Resolvemos la congruencia: 213mod13 b . Esta es mucho más complicada.

Paso 2.1. Por tanteo ("bash"). Si lo hacemos por tanteo nos vamos a pasar un buen rato

calculando potencias hasta llegar al exponente 39 para encontrar 239 13mod13 :

1→3 2→9 3→27 4→81 5→74

6→53 7→159 8→139 9→79 10→68

11→35 12→105 13→146 14→100 15→131

16→55 17→165 18→157 19→133 20→61

21→14 22→42 23→126 24→40 25→120

26→22 27→66 28→29 29→87 30→92

31→107 32→152 33→118 34→16 35→48

36→144 37→94 38→113 39→1

Paso 2.2. Aplicando el Teorema de Euler. Una indicación nos la puede dar la función Phi de

Euler: 1332132 , luego el orden de 3 módulo 156 será un divisor de 1332156 .

Probando divisores encontramos la solución 13339 .

Paso 2.3. Aplicando el Teorema del binomio.

Observamos que 13mod1313mod13 2 bb .

En efecto: 13mod1311313113313mod13 22 bbb kk

Y la congruencia 13mod13 b tiene fácil solución: 13mod111322733 . Así pues,

cualquier solución de 13mod13 b será múltiple de 3.

Llegados a este punto podríamos aplicar la estrategia 2.2 anterior para llegar a la solución, pero

en su lugar vamos a aplicar el desarrollo binomial.

El número b buscado será múltiplo de 3: cb 3 para cierto entero c , con lo que la

congruencia se transforma en:

22323 13mod12713mod1313mod13 ccc

Ahora aplicamos el desarrollo binomial:

cccc

cc

c

c

c

ccc11321132

1...1132

11132

0

113227113227

011110

Y observamos que, trabajando módulo 213 , cualquier potencia c13 con 2c será (congruente

con) cero.

Así pues, a todos los efectos prácticos:

126113211321

27011

c

c

c

c

cccc

Y por tanto, la congruencia exponencial 213mod127 c se convierte en la congruencia lineal

213mod1126 c

Que se resuelve fácilmente:

222 13mod013213mod02613mod1126 ccc

Para que esta última congruencia se cumpla, es necesario y suficiente que c sea múltiplo de 13.

Así pues, finalmente llegamos a un resultado múltiplo de 3 y múltiplo de 13, y el valor mínimo

posible es 39133 c .

Paso 3. Juntamos las dos congruencias:

Hemos obtenido 25 11mod13 y 239 13mod13 , luego tomando el mínimo común múltiplo

de ambos exponentes: 19539539,5 , tendremos, aplicando 7.5d:

22195

22

25539539195

239395395195

1311mod13

113,11

13mod11333

11mod11333

Que será el valor mínimo posible pues en todos los pasos hemos obtenido los valores mínimos

posibles.

La solución del problema es 159.

Fuentes:

https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11

https://youtu.be/e7JGgykuK3w Analyzing the Expression in Mod 143² (2018 AIME I Prob 11)

"LetsSolveMathProblems"

https://youtu.be/b_Z_OGfyJyw 2018 AIME I Problem #11 "Osman Nal"

12

Una observación clave para solucionar este problema es que podemos pasar los números de U a

módulo 3:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0

Otra observación clave es que podemos dejar aparcados todos los números que sean

congruentes con 0 módulo 3: 18,15,12,9,6,3R , porque cualquier combinación de números

será múltiple de 3 si y solo si lo es añadiendo cualquier subconjunto de R. Luego nos

olvidamos de los elementos de R y deberemos multiplicar el resultado final por 62R .

Así pues, disponemos del conjunto

1 2 4 5 7 8 10 11 13 14 16 17

1 2 1 2 1 2 1 2 1 2 1 2

Y debemos contar todos los subconjuntos cuya suma sea múltiple de 3. Organizaremos el

recuento por número de elementos de dicho subconjunto:

0 elementos:

→ 1

1 elemento:

(no hay ningún caso)

2 elementos:

“12” → 3666

3 elementos:

“111” → 203

6

“222” → 203

6

4 elementos:

“2211” → 22515152

6

2

6

5 elementos:

“21111” → 901564

6

1

6

“22221” → 906151

6

4

6

6 elementos:

“222222” → 16

6

“111222” → 40020203

6

3

6

“111111” → 16

6

7 elementos:

“1122222” → 906155

6

2

6

“1111122” → 901562

6

5

6

8 elementos:

“11112222” → 22515154

6

4

6

9 elementos:

“111222222” → 201206

6

3

6

“111111222” → 202013

6

6

6

10 elementos:

“1111122222” → 36665

6

5

6

11 elementos:

(no hay ningún caso)

12 elementos:

“111111222222” → 1116

6

6

6

Total: 1+36+20+20+225+90+90+1+400+1+90+90+225+20+20+36+1=1366

Así pues, la probabilidad es 1118

6

2

683

2

21366

P , y la respuesta correcta es 683.

Fuente de esta solución: https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12

13

(Solución oficial MAA).

Sea AXB .

Sabemos por (11.4.11f) que 2

º901

BAI , luego 2/cossin 1 BAI .

Por otro lado, vemos que

222221212

ACABXABCAXXABCAXXAIAXIAII

Aplicando el Teorema del Seno al triángulo 1ABI tenemos:

)2/cos(

)2/sin(

sin

sin

sinsin 1

11

11

1

Bc

BAI

ABIABAI

BAI

AB

ABI

AI

Y de la misma manera tenemos:

)2/sin(

)2/sin(2

CbAI

Luego:

)2/sin()2/sin(2/sin

)sin(

)2/sin()2/sin(2/sin

)2/sin()2/cos(2

)2/sin()2/sin(2/sin

2/sin2

1

sin2

1

21

122121

CBAbcCBA

bc

CBAbc

AAIAI

AIIAIAIIAI

Y la igualdad acontece cuando º901)sin( , es decir, cuando la ceviana AX es la

altura, y en este caso el área buscada es

)2/sin()2/sin(2/sin CBAbc

Y calculamos los factores de este producto:

bc

cbacba

bc

acbbc

bc

acbbcbc

acb

AA

4

))((

4

2

4

2

2

21

2

cos1

2sin

222

222

222

Para llegar finalmente a:

a

bacacbcba

abc

bacacbcbabc

CBAbc

8

))()((

8

))()((

)2/sin()2/sin(2/sin

En nuestro caso particular:

126328

)323034)(303432)(343230(21

IAI

Fuente de esta solución: https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13

14

Primera versión: Mediante recursión pura.

Como en el problema anterior, sea n

kP el número de secuencias de n saltos que empiezan en S y

acaban en el vértice kP , y sean nE y nS el número de secuencias de n saltos que empiezan en

S y acaban en el vértice E y en el vértice S, respectivamente.

Está claro que: 1

5

1

1

nnn PPS , 1

2

1

1

nnn PSP , 1

3

1

12

nnn PPP , 1

23

nn PP , 1

4

1

3

nnn PPE ,

1

54

nn PP , 11

45

nnn SPP ,

Y se nos pide calcular 1221 ... EEE .

n nS nP1 nP2 nP3 nE nP4 nP5

1 0 1 0 0 0 0 1

2 2 0 1 0 0 1 0

3

...

12

Igual que con el problema anterior, el problema, así planteado, es tarea propia de una hoja de

cálculo:

Y la respuesta al problema es: 351155894728149431100

Segunda versión: Determinando una fórmula recursiva.

Para determinar la fórmula recursiva de nE vamos a utilizar la siguiente propiedad:

44

1

4

4

4

3

44

4

4

3

4

1

4

4

4

3

3

5

3

2

4

4

4

3

2

4

2

3

31

nnnnnnnn

nnnnnnnnnn

SPPPSPPP

PPPPPPPPEE

Y ahora vamos a buscar la fórmula recursiva:

3

1

3233

1

2

33

4

3

3

3

1

2

5

2

2

1

4

1

3

nnnnnnn

nnnn

nn

nnn

PSEESPE

SPPP

PP

PPE

Por otro lado:

231

3

3

3

4

31

4

2

4

5

31

4

2

4

5

4

1

4

4

2

44

5

4

1

3

1

3

nnn

nnnn

nnnn

nnnn

nnnn

nn

EEE

PPEE

PPEE

PPPS

PSPP

PS

En donde hemos aplicado la propiedad demostrada anteriormente. Así pues:

231

2312

2

nnnn

nnnnn

EEEE

EEEEE

Ahora calculamos directamente los primeros términos de la sucesión:

00 E , 01 E , 02 E , 13 E , 14 E

Y mediante la fórmula recursiva vamos determinando los siguientes términos:

35 E , 46 E , 97 E , 148 E , 289 E , 4710 E , 8911 E , 15512 E

Y, finalmente: 351155894728149431100 .

Fuentes:

https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_14

https://youtu.be/11utulhU590

15

Para resolver este problema nos vamos a basar en los siguientes resultados geométricos:

1) El área de un cuadrilátero (no necesariamente cíclico) ABCD se puede calcular sabiendo las

longitudes de las diagonales y el ángulo que estas determinan (ver GA/9.2.4):

sin2

1 BDACABCD

2) El Teorema de Ptolomeo: En todo cuadrilátero cíclico ABCD, el producto de las diagonales

es igual a la suma de los productos de los lados opuestos (ver GA/10.5.7):

ADBCCDABBDAC

3) La fórmula de Parameshavara: El radio de la circunferencia circunscrita se puede determinar

mediante los lados del cuadrilátero inscrito:

ABCD

bcadbdaccdabR

4

))()((

Donde a, b, c, d son los lados consecutivos del cuadrilátero cíclico (ver GA/10.5.15).

Con estos tres resultados procedemos a la resolución del problema.

Sean dcba ,,, las longitudes de los cuatro lados que forman el cuadrilátero.

Tenemos tres configuraciones posibles:

A) dcba ,,, , con a y c lados opuestos.

B) cdba ,,, , con a y d lados opuestos.

C) dbca ,,, , con a y b lados opuestos.

Mediante los resultados 1 y 2 llegamos a la siguiente conclusión:

AbdacABCDK sin)(2

1

BbcadABCDK sin)(2

1

CcdabABCDK sin)(2

1

Y multiplicando estas tres igualdades llegamos a:

))()((70

3

7

6

5

3

3

2))()((

8

1

sinsinsin))()((8

1

3

3

3

bcadbdaccdabK

bcadbdaccdabK

bcadbdaccdabK CBA

Ahora aplicamos la fórmula de Parameshavara, teniendo en cuenta que en nuestro caso 1R :

))()((16

))()((44

))()((

2 bcadbdaccdabK

bcadbdaccdabKK

bcadbdaccdabR

Así pues,

35

2416

70

316

70

3 23 KKK

Y la respuesta correcta es 593524 .

Fuente de esta solución: https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_15

AIME II 2018 Enunciados

1

Sean los puntos A, B y C alineados en este orden a lo largo de un camino recto donde la

distancia de A a C es de 1800 metros. Ina corre el doble de rápido que Eve, y Paul corre

el doble de rápido que Ina. Los tres corredores empiezan a correr al mismo tiempo, Ina

empezando en A y corriendo hacia C, Paul empezando en B y corriendo hacia C, y Eve

empezando en C y corriendo hacia A. Cuando Paul se encuentra con Eve, da media

vuelta y corre hacia A. Sabiendo que Paul e Ina llegan a B al mismo tiempo, determina

la longitud, en metros, de A a B.

2

Sean los puntos A, B y C alineados en este orden a lo largo de un camino recto donde la

distancia de A a C es de 1800 metros. Ina corre el doble de rápido que Eve, y Paul corre

el doble de rápido que Ina. Los tres corredores empiezan a correr al mismo tiempo, Ina

empezando en A y corriendo hacia C, Paul empezando en B y corriendo hacia C, y Eve

empezando en C y corriendo hacia A. Cuando Paul se encuentra con Eve, da media

vuelta y corre hacia A. Sabiendo que Paul e Ina llegan a B al mismo tiempo, determina

la longitud, en metros, de A a B.

3

Determina la suma de todos los números enteros positivos 1000b tal que el número

b36 (escrito en base b) es un cuadrado perfecto y el número b27 (también escrito en

base b) es un cubo perfecto.

4

Se toma aleatoriamente y de forma uniforme un número real a del intervalo 18,20 .

La probabilidad de que todas las raíces del polinomio

234222 234 xaxaaxx

sean reales se puede escribir de la forma nm / , donde m y n son enteros coprimos.

Determina nm .

5

Sean zyx ,, números complejos tales que ixy 32080 , 60yz , izx 2496 ,

donde 1i . Sean ba, números reales tales que biazyx . Determina 22 ba .

6

Se toma aleatoriamente y de forma uniforme un número real a del intervalo 18,20 .

La probabilidad de que todas las raíces del polinomio

234222 234 xaxaaxx

sean reales se puede escribir de la forma nm / , donde m y n son enteros coprimos.

Determina nm .

7

Sea el triángulo ABC de lados 9AB , 35BC y 12AC . Marcamos los puntos

BPPPPA 2450210 ,...,,, en el segmento AB de forma que kP se encuentra entre 1kP

y 1kP para todo 2449,...,2,1k , y marcamos los puntos CQQQQA 2450210 ,...,,, en

el segmento AC de forma que kQ se encuentra entre 1kQ y 1kQ para todo

2449,...,2,1k . Además, todo segmento kkQP , con 2449,...,2,1k , es paralelo a BC .

Estos segmentos cortan el triángulo en 2450 regiones, consistiendo en 2449 trapecios y

un triángulo. Todas estas 2450 regiones tienen el mismo área. Determina el número de

segmentos kkQP , con 2450,...,2,1k cuya longitud es racional.

8

Una rana está posicionada en el origen del plano coordenado. Desde el punto ),( yx

puede saltar a los puntos ),1( yx , ),2( yx , )1,( yx , )2,( yx . Determina el

número de secuencias de salto diferentes en los que la rana, saliendo de )0,0( , acaba en

)4,4( .

9

Sea ABCDEFGH el octágono con lados de longitud 10 GHEFCDAB y

11 HAFGDEBC que se ha formado eliminado los triángulos 6-8-10 de las

esquinas de un rectángulo 2723 , con lado AH en el lado corto del rectángulo, tal y

como se indica en la figura. Sea J el punto medio de AH , y dividimos este octágono en

7 triángulos trazando los segmentos JB , JC , JD , JE , JF y JG . Determina el área

del polígono convexo cuyos vértices son los baricentros de estos 7 triángulos.

10

Determina el número de funciones )(xf de 5,4,3,2,1 en 5,4,3,2,1 que satisfacen

)))((())(( xfffxff para todo x de 5,4,3,2,1 .

11

Determina el número de permutaciones de 1, 2, 3, 4, 5, 6 tales que para cada k con

51 k , al menos uno de los primeros k términos de la permutación es mayor que k .

12

Sea ABCD un cuadrilátero convexo con 10CDAB , 14BC y 652AD .

Supongamos que las diagonales de ABCD se cortan en un punto P, y que la suma de las

áreas de los triángulos APB y CPD es igual a la suma de las áreas de los triángulos

BPC y APD . Determina el área del cuadrilátero ABCD.

13

Misha lanza un dado convencional de seis caras hasta que obtiene la secuencia 1 – 2 – 3

en este orden en tres lanzamientos consecutivos. Determina la probabilidad de que el

número de lanzamientos sea impar.

14

Supongamos que el incírculo de un triángulo ABC es tangente a BC en X. Sea

XY el otro punto de intersección de AX con . Los puntos P y Q pertenecen a AB

y AC , respectivamente, de forma que PQ es tangente a en Y. Suponiendo además

que 3AP , 4PB , 8AC , entonces n

mAQ , con m y n enteros positivos

coprimos. Determina nm .

15

Determina el número de funciones f de 6,5,4,3,2,1,0 en los enteros tales que

0)0( f , 12)6( f y

yxyfxfyx 3)()(

para todo x , y en 6,5,4,3,2,1,0 .

AIME II 2018 Soluciones

1

Paul corre cuatro veces la velocidad de Eve, luego recorrerá cuatro veces la distancia de Eve,

digamos 4p y 1p. En el momento en que se encuentran Paul y Eve, Ina habrá recorrido la mitad

de camino que Paul, es decir, 2p, y Paul tendrá que recorrer 4p, y Ina 2p, luego en total, los

1800 metros se descomponen en

mpppppp 2009/180094221800

Y por tanto de A a B hay 4p, es decir, 800 m.

2

Paul corre cuatro veces la velocidad de Eve, luego recorrerá cuatro veces la distancia de Eve,

digamos 4p y 1p. En el momento en que se encuentran Paul y Eve, Ina habrá recorrido la mitad

de camino que Paul, es decir, 2p, y Paul tendrá que recorrer 4p, y Ina 2p, luego en total, los

1800 metros se descomponen en

mpppppp 2009/180094221800

Y por tanto de A a B hay 4p, es decir, 800 m.

3

Primera versión.

7227 bb es un cubo perfecto, es decir, 372 mb para cierto entero m.

2007721000 3 bmb

Puesto que hay menos cubos perfectos que cuadrados perfectos, listamos todos los cubos:

172812...,,644,273,82 3333

Puesto que 72 b es siempre impar, podemos eliminar de la lista todos los cubos pares, y

quedarnos solo con 33333 11,9,7,5,3

6336 bb es un cuadrado perfecto, es decir, 263 nb para cierto entero n.

22222 9)3(3|3|32363 kknknnnbbn

Luego nos podemos quedar solo con aquellos cubos divisibles entre 9: 33 9,3

Ya solo nos queda comprobar si estos candidatos se adaptan a nuestras condiciones: 23 636631072273 bbb , luego la base 10 es aceptable.

23 33108963613361727299 bb

Luego el resultado es 37110361 .

Segunda versión. 2)2(36336 nbbb es un cuadrado perfecto, luego 22 93 knkn .

23329)2(3 222 kbkbkb

El valor de k está limitado superiormente: 181000231000 2 kkb

Sustituyendo en la segunda ecuación:

12336746723272 22223 kkkkbm

y vemos que 3 es divisor de 3m , y por tanto 3 es divisor de m .

También vemos que es impar:

imparkimparkpark 123122 222

Y que está acotado superiormente:

2007710002721000 3 bmb

y 2197132 , luego 13m

Los cubos que cumplen las condiciones anteriores son dos: 33 9,3 , y como en la versión

anterior, solo nos queda comprobar que, efectivamente, satisfacen las condiciones del

enunciado.

Fuente de estas versiones: www.artofproblemsolving.com

4

Probando mediante Ruffini encontramos dos soluciones reales (El Teorema de las Raíces

racionales reduce la búsqueda a 2,2,1,1 ):

1 a2 22 a 34 a 2

1 -1 12 a 14 a 2

1 12 a 14 a 2 0

2 2 24 a 2

1 12 a 1 0

Luego

1)12()2)(1(234222 2234 xaxxxxaxaaxx

Así pues, este polinomio tendrá las cuatro raíces reales si y solo si

1)12(2 xax

Tiene sus dos raíces reales, es decir, cuando

2/1212

2/321224)12(0114)12( 222

aa

aaaa

Luego el conjunto de valores de a aceptables es 18,2/32/1,20 , cuya longitud es

362

72

2

336401

2

3

2

36

2

40

2

1

2

318)20(

2

1

El segmento total mide 382018 unidades de largo, luego la probabilidad es

19

18

38

36P

y la respuesta es 371918 .

5

Primera versión. Resolviendo el sistema.

)4(242496

60

)41(8032080

iizx

yz

iixy

)41(3

4)41(

3

4)41(

60

80)41(80

60

)41(80

6060

izxiiz

xi

zx

ixy

zyyz

i

i

iziizzizx 18

)41(3

4

)4(24)4(24)41(

3

4)4(24 2

Las raíces cuadradas de i son i 12

1, luego las raíces cuadradas de i18 son

13 iz

Sea 13 iz . Entonces:

iiz

y 1010)1(3

6060

ii

ixixy 1220

1010

)41(80)41(80

Finalmente:

742549575,757 2222 babaizyx

Nota: las condiciones del enunciado se cumplen tanto para

ixiyiz 1220,1010,13 como para ixiyiz 1220,1010,13 .

Segunda versión. Mediante notación exponencial.

Pasamos a notación exponencial: 1

1

ierx , 2

2

iery , 3

3

ierz .

17803208021 irrxy

606032 rryz

1724249631 irrxz

Este sistema tiene por solución: 3441 r , 2102 r y 233 r

)4arctan(178032080 21

)4arctan()(

2121

iieierrxy

2332

0)(

32 0606032

iieerryz

)4/1arctan(17242496 31

)4/1arctan()(

3131 ii

eierrxz

4/2/)4/1arctan()4arctan(2)4/1arctan(

)4arctan(22

21

21

Y por tanto 4/23 .

De todo lo anterior deducimos que iey i 1010210 4/ , iez i 3323 4/

y utilizando ixy 32080 , deducimos que ix 1220 .

Como en la primera versión, basta sumar zyx para llegar al resultado 74.

Fuente de esta versión: https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5

6

Probando mediante Ruffini encontramos dos soluciones reales (El Teorema de las Raíces

racionales reduce la búsqueda a 2,2,1,1 ):

1 a2 22 a 34 a 2

1 -1 12 a 14 a 2 1 12 a 14 a 2 0

2 2 24 a 2

1 12 a 1 0

Luego

1)12()2)(1(234222 2234 xaxxxxaxaaxx

Así pues, este polinomio tendrá las cuatro raíces reales si y solo si

1)12(2 xax

Tiene sus dos raíces reales, es decir, cuando

2/1212

2/321224)12(0114)12( 222

aa

aaaa

Luego el conjunto de valores de a aceptables es 18,2/32/1,20 , cuya longitud es

362

72

2

336401

2

3

2

36

2

40

2

1

2

318)20(

2

1

El segmento total mide 382018 unidades de largo, luego la probabilidad es

19

18

38

36P

y la respuesta es 371918 .

7

Fijando un 2450,...,2,1k , los triangulos ABC y kkQAP son semejantes por Tales, ya que

BCQP kk // .

Las regiones separadas por segmentos tienen la misma área

2450

ABC.

Los trapecios se pueden unir de forma que 2450

ABCkQAP kk

Es decir, la razón de proporcionalidad entre las áreas de los triángulos es:

22 7522450

kk

ABC

QAP kk

Sabemos que las áreas están en proporción el cuadrado de la razón de semejanza de las

longitudes de los lados, luego la razón de proporcionalidad de las longitudes es:

75

2/

752 22

kk

Y por tanto

67

3

23

33

7

1

2

3

7

135

75

2/35

kkkkBCQP kk

que será un número racional

si y solo si 6

k es racional, es decir, para todos los k tales que 6/k sea un cuadrado perfecto:

22 66/ pkpk

6161 2 kp , 24262 2 kp , ... , 240020620 2 kp ,

Con 264621621 2 kp ya nos pasamos luego hay 20 valores de k aceptables.

8

Llamaremos H1, H2, V1 y V2 a los saltos respectivos ),1( yx , ),2( yx , )1,( yx , )2,( yx .

Está claro que la rana ha de realizar cuatro pasos hacia la derecha y cuatro pasos hacia arriba.

Estos pasos se pueden realizar en cualquier orden.

Con ocho saltos, la única secuencia posible es

H1 H1 H1 H1 V1 V1 V1 V1 y todas sus permutaciones posibles.

70572234

5678

!4!4

!84,4

8

P

Con siete saltos:

H2 H1 H1 V1 V1 V1 V1 y todas sus permutaciones posibles.

H1 H1 H1 H1 V2 V1 V1 y todas sus permutaciones posibles.

21021051055372

567

!4!2

!74,2

7

P

Con seis saltos:

H2 H2 V1 V1 V1 V1 y todas sus permutaciones posibles.

H1 H1 H1 H1 V2 V2 y todas sus permutaciones posibles.

30215152

56

!4!2

!64,2

6

P

H2 H1 H1 V2 V1 V1 y todas sus permutaciones posibles.

18032562

3456

!2!2

!62,2

6

P

Con cinco saltos:

H2 H2 V2 V1 V1 y todas sus permutaciones posibles.

V2 V2 H2 H1 H1 y todas sus permutaciones posibles.

60230302

345

!2!2

!52,2

5

P

Con cuatro saltos:

H2 H2 V2 V2 y todas sus permutaciones posibles.

62

34

!2!2

!42,2

4

P

En total: 5566601803021070 secuencias posibles.

9

Dibujamos los puntos medios de los lados y con ellos trazamos las medianas y localizamos los

baricentros. Al hacer el dibujo vemos que, por simetría, podemos reducir nuestro estudio a la

parte superior:

Determinamos las coordenadas cartesianas de los puntos:

)0,0(J , )5.5,0(H , )5.11,8(G , )5.11,19(F , )5.5,27(E

Y aplicamos la fórmula de las coordenadas del baricentro (ver GA/20.9.2)

)3/17,3/8(3

)17,8(

3

)5.11,8()5.5,0()0,0(

3

GHJP

)3/23,9(3

)23,27(

3

)5.11,19()5.11,8()0,0(

3

FGJQ

)3/17,3/46(3

)17,46(

3

)5.5,27()5.11,19()0,0(

3

EFJR

)0,18()0,27(3

2S

Descomponiendo la figura en un rectángulo y dos triángulos obtenemos su área:

923

17

3

8

2

12

3

38

2

1

3

17

3

38

3

17

3

4618

2

1

3

17

3

23

3

8

3

46

2

1

3

17

3

8

3

46

Así pues, el área del polígono del enunciado es 184292 .

Observación. Una vez determinados los puntos, el área se podría haber determinado también

mediante la fórmula “Shoelace” (ver GA/18.6.5).

10

La condición del enunciado equivale a decir que todo elemento de 5,4,3,2,1 tiene un ciclo de

longitud igual o menor que 3:

Longitud 1:

xxfxffxfff

xxfxffxxf

)())(()))(((

)())(()(

Longitud 2:

yyfyffxfff

yyfxffyyfxyxf

)())(()))(((

)())(()(,)(

Longitud 3:

zzfyffxfff

zyfxffzzfyxzyfxyxf

)())(()))(((

)())(()(,,)(,)(

Sin embargo, podemos observar que la condición del enunciado no se cumpliría para longitudes

superiores a 3.

Sea a el número de elementos con ciclo de longitud 1, sea b el número de elementos con ciclo

de longitud 2 y sea c el número de elementos con ciclos de longitud 3.

Está claro que 5 cba , y que si 4a entonces 0b , puesto que si existe un elemento

con ciclo igual a 3, xxf )(3 , entonces xxfxff )())(( 32 y por tanto )(xf será un

elemento cuyo ciclo tendrá longitud 2.

Está claro también que 0a .

Ordenaremos los casos empezando por a , y luego por b :

1. 0,0,5 cba

1 caso.

2. 1,0,4 cba

No se puede dar.

3. 0,1,4 cba

2045 casos.

4. 2,0,3 cba

No se puede dar.

5. 1,1,3 cba

102

20

!2!3

!5

3

5

grupos de 3 elementos de ciclo 1,

603210 casos.

6. 0,2,3 cba

102

20

!2!3

!5

3

5

grupos de 3, y los otros dos pueden ir a cualquiera de los 3

elementos de ciclo 1, por tanto 903310 casos.

7. 3,0,2 cba

No se puede dar.

8. 2,1,2 cba

102

20

!3!2

!5

2

5

grupos de 2 elementos que tendrán asociados ciclos de longitud 1,

602310 casos.

9. 1,2,2 cba

240222310 casos.

10. 0,3,2 cba

8022210 casos.

11. 4,0,1 cba

No se puede dar.

12. 3,1,1 cba

Tenemos 5 candidatos para el elemento de ciclo 1.

Después, 4 candidatos para el elemento de ciclo 2.

Los tres que quedan han de ir todos al candidato de ciclo 2.

En total, 2045 casos.

13. 2,2,1 cba

Tenemos 5 candidatos para el elemento de ciclo 1.

De los 4 restantes, seleccionamos 2 para tener ciclo de longitud 2.

120222

34522

!2!2

!4522

2

45

casos.

14. 1,3,1 cba

603453!1!3

!453

3

45

casos.

15. 0,4,1 cba

Hay 5 casos.

En total, tendremos 7565601202080240609060201 casos.

Observaciones. En la solución "oficial" de AoSP, encontramos la fórmula general para

determinar el número de elementos en cada caso:

cb bab

a

a

55

También se indica que este problema apareció en la prueba "Standford Math Tournament,

Advanced Topics Test" del 2011, y en "Mock AIME 2" 2010 (problema 7)

Fuente de la solución: https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_10

11

El número total de permutaciones es 720!66 P , vamos a contar el número de permutaciones

que no cumplen la condición del enunciado, es decir, que para cada k con 51 k , todos los

primeros k términos de la permutación son menores o iguales que k .

Caso 1: 1k

Son todas las permutaciones que empiezan con 1: 1*****, es decir: 120!55 P .

Caso 2: 2k .

Son todas las permutaciones que empiezan por todas las permutaciones posibles de 12, es decir:

12**** y 21****. Las primeras ya están contadas en el caso 1, luego son todas las que

empiezan por 21: 24!44 P .

Caso 3: 3k

Son todas las permutaciones que empiezan por todas las permutaciones posibles de 123, es

decir:

123*** Ya contadas en el caso 1 132*** Ya contadas en el caso 1

213*** Ya contadas en el caso 2 231***

312*** 321***

Luego hay 18!333 3 P

Caso 4: 4k

Son todas las permutaciones que empiezan por todas las permutaciones posibles de 1234, es

decir:

1234** Ya contadas en el caso 1 1243** Ya contadas en el caso 1

1324** Ya contadas en el caso 1 1342** Ya contadas en el caso 1

1423** Ya contadas en el caso 1 1432** Ya contadas en el caso 1

2134** Ya contadas en el caso 2 2143** Ya contadas en el caso 2

2314** Ya contadas en el caso 3 2341**

2413** 2431**

3124** Ya contadas en el caso 3 3142**

3214** Ya contadas en el caso 3 3241**

3412** 3421**

4123** 4132**

4213** 4231**

4312** 4321**

Luego hay 26!21313 2 P

Caso 5: 5k

Son todas las permutaciones que empiezan por todas las permutaciones posibles de 12345, es

decir:

1***** Ya contadas en el caso 1

21**** Ya contadas en el caso 2 231*** Ya contadas en el caso 3

23415* Ya contadas en el caso 4 23451*

23514* 23541*

24135* Ya contadas en el caso 4 24153*

24315* Ya contadas en el caso 4 24351*

24513* 24531*

25134* 25143*

25314* 25341*

11 Casos que empiezan por 2

31245* Ya contadas en el caso 3 31254* Ya contadas en el caso 3

31425* Ya contadas en el caso 4 31452*

31524* 31542*

32145* Ya contadas en el caso 3 32154* Ya contadas en el caso 3

32415* Ya contadas en el caso 4 32451*

32514* 32541*

34125* Ya contadas en el caso 4 34152*

34215* Ya contadas en el caso 4 34251*

34512* 34521*

35124* 35142*

35214* 35241*

35412* 35421*

16 Casos que empiezan por 3

41235* Ya contadas en el caso 4 41253*

41325* 41352*

41523* 41532*

42135* Ya contadas en el caso 4 42153*

42315* Ya contadas en el caso 4 42351*

42513* 42531*

43125* Ya contadas en el caso 4 43152*

43215* Ya contadas en el caso 4 43251*

43512* 43521*

45123* 45132*

45213* 45231*

45312* 45321*

20 Casos que empiezan por 4

5***** 244 P Casos que empiezan por 5

En total hay 7124201611 casos.

Así pues, que no cumplan la condición del enunciado hay

25971261824120

Y finalmente, casos que sí cumplan la condición hay 461259720 .

Observación: En https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11

Se pueden encontrar otras soluciones alternativas, sin pasar al complementario, (tal vez) más elegantes que esta.

12

Utilitzando que )180sin(sin tenemos:

0))((

0

2

sin

2

sin

2

sin

2

sin

PDBPCPAP

CPBPDPAPCPDPBPAP

CPBPDPAPCPDPBPAP

BPCCPBPAPDDPAPDPCCPDPAPBBPAP

APDBPCCPDAPB

Luego el punto P de corte de las dos diagonales es el punto medio de una de ellas. Podemos

suponer, sin pérdida de generalidad, que CPAP .

Y por tanto CPDAPD y BPCAPB y por consiguiente

)1(sin49sin65

sin7sin65

sin1014sin65210

2

sin

2

sin

22 CA

CA

CA

CCDBCAADAB

ABDBDC

APDAPBCPDBPC

APDBPCCPDAPB

Por otro lado, aplicando el Teorema del Coseno:

CBD

ABD

cos141021410

cos65210265210

222

222

Y por tanto:

CACA

CA

CA

CA

cos35cos6558cos140cos652032

cos280cos6540196260

cos280196cos6540260

cos141021410cos65210265410 222

22

2

2

2

2

cos75

8

65

sin49165

cos75

8sin165

cos75

8cos65cos7cos65

5

8

CC

CA

CACA

5

4

25

16

25

91cos1sin

5

3coscos7

5

82

5

816

49cos75

82

5

865

cossin49cos75

82

5

865

cos49cos75

82

5

8sin4965

2

2

2

22

2

2

2

2

CCCC

C

CCC

CCC

Y por tanto, finalmente:

1125

4140

2

sin101422

CBDCABCD .

Fuente de esta solución: https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12

13

Primera versión.

Sea nf la probabilidad de que el número de lanzamientos sea n.

Observamos que nf es una función de densidad, es decir:

11

i

if

Está claro que 0210 fff puesto que al menos tiene que lanzar el dado 3 veces.

También está claro que 216

1

6

133 f .

Nosotros queremos determinar la probabilidad de un lanzamiento impar:

1

12531 ...i

iffffP

Para determinar nf observamos que “sacar 123 en la tirada n y no antes” equivale a “sacar 123

en la tirada n” y “no sacar 123 en las tiradas anteriores hasta n-3”, es decir:

3

1

1216

1 n

i

in ff

Por otro lado, vemos que

2

3

1

2

1

2

1

3

1

2

1

3

1

1

216

1

216

1

11216

11

216

11

216

1

n

n

i

i

n

i

i

n

i

i

n

i

i

n

i

i

n

i

inn

fff

ffffff

Así pues, 21216

1 nnn fff

Con esta igualdad ya podemos lanzarnos a calcular la probabilidad de P:

1

32

1

2

1

32

1

2

1

322

1

212112

1

12

216

1

216

1

216

1

216

1

i

i

i

i

i

i

i

i

i

ii

i

ii

i

i

ffff

fffffP

En esta última igualdad observamos que

1

2

i

if es la probabilidad de que aparezca la

combinación en una tirada par, es decir, el complementario de P:

Pfi

i

11

2

Y que

Pfi

i

1

32

Con lo que llegamos a la ecuación

PPP216

1)1(

que se resuelve fácilmente:

431

2161

216

4311

216

121

216

12

216

112

PPPPPPP

Segunda versión (no rigurosa).

Sean parP y imparP las probabilidades de un número par o impar de lanzamientos.

Está claro que 1 imparpar PP .

Podemos interpretar este problema en términos de probabilidad condicionada.

La probabilidad de obtener “123” en la primera vez, es 1/216, y si no, las probabilidades se

invierten (esto habría que justificarlo convenientemente). Luego, mediante la fórmula de la

probabilidad total:

impar

imparparimpar

P

PPP

1216

215

216

1

1216

111

216

1

216

111

216

1

Llegando a la ecuación equivalente a la primera versión.

Fuente de estas soluciones:

https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13

14

Primera versión (Mediante geometría proyectiva).

Sean M y N los puntos de contacto entre el circuncírculo y los lados respectivos AB y AC .

Aplicando el Teorema de Brianchon (ver GA/12.7.5) a los hexágonos tangenciales QNCBMP y

PYQCXB deducimos que las rectas MN, CP, BQ y XY son concurrentes en un punto al que

llamaremos O.

Además, por GA/12.5.3 sabemos que 1,;, XYOA

Por otro lado, sea BCPQZ . La polar de Z es XY, y el punto A pertenece a XY, luego

aplicando el Teorema de La Hire (GA/13.5.2), Z pertenecerá a la polar de A, que es MN, luego

las rectas PQ, MN y BC son concurrentes en un mismo punto Z.

Observamos que tenemos una perspectividad con centro Z, que conservará razón doble, luego:

CQNABPMAXYOA ,;,,;,,;,1

De la primera igualdad, y trabajando con valor absoluto ya que nuestros segmentos no están

orientados:

5

21

)3(7

)7(3,;,1

AM

AM

AM

MPAB

MBAPBPMA

Y de la misma manera, con la segunda igualdad, y teniendo en cuenta que ANAM por

GA/11.4.10,

59

168

)5/21(8

)5/218(,;,1

AQ

AQ

AQ

NQAC

NCAQCQNA

Y la respuesta correcta es 22759168 .

Segunda versión. (Mediante Teorema del Seno).

Como en la versión anterior, sean M y N los puntos de contacto entre el circuncírculo y los

lados respectivos AB y AC .

Sea BAX , y sea QYXPYA .

Puesto que QYX y YXC subtienden el mismo arco de , tenemos

YXCQYXPYA

Por otro lado, puesto que PYMP por ser ambas tangentes, aplicando el Teorema del Seno al

triángulo APY :

sin

sin111

AP

PY

AP

PM

AP

PMAP

AP

AM

Y ahora aplicando el Teorema del Seno al triángulo ABX :

sin

sin111

AB

BX

AB

BM

AB

BMAB

AB

AM

Y sumando las dos igualdades anteriores llegamos a

5

21

372 AM

AMAM

AP

AM

AB

AM

Aplicando el mismo argumento al triángulo AQY obtenemos la ecuación

59

168

8

11

5

212

AQ

AQAC

AM

AQ

AM

AC

AN

AQ

AN

Fuente de estas dos versiones:

https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_14

(en donde podemos encontrar otra tercera versión, mediante una combinación del Teorema del Seno y del

Teorema del Coseno).

15 Nota preliminar: La resolución que expongo no es nada elegante. A posteriori, más bien parece una tentativa más

o menos exitosa de sobrevivir en un mar de casos y condiciones. Pero las soluciones digamos oficiales de

https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_15

tampoco las he encontrado nada esclarecedoras.

3)()(

13)()(

13)()(

yx

yfxf

yx

yfxfyxyfxfyx

Por lo tanto podemos interpretar este problema en términos de pendiente de la función: La

variación vertical respecto de la variación horizontal.

31

x

y

Para 1x , está claro que 31 y y por lo tanto, de un valor x al siguiente 1x , los

incrementos verticales admitidos son seis: 32,1,1,2,3 .

Así pues, de momento, 3,2,1,1,2,3)1( f , 9)5( f , 6)4( f , 3)3( f , 0)2( f .

Para 2x , tenemos

6232

1

yy

Completando una tabla "Primer paso vs Segundo paso" y sombreando los valores que no

cumplen esta condición nos queda:

y +3 +2 +1 -1 -2 -3

+3 6 5 4 2 1 0

+2 5 4 3 1 0 -1

+1 4 3 2 0 -1 -2

-1 2 1 0 -2 -3 -4

-2 1 0 -1 -3 -4 -5

-3 0 -1 -2 -4 -5 -6

Luego los únicos casos aceptables para )2(f son: 6,5,4,3,2)2( f

Para 3x , tenemos

9333

1

yy

Completando una tabla "Primeros dos pasos vs Tercer paso" y sombreando los valores que no

cumplen esta condición nos queda:

y +3 +2 +1 -1 -2 -3

+6 9 8 7 5 4 3

+5 8 7 6 4 3 2

+4 7 6 5 3 2 1

+3 6 5 4 2 1 0

+2 5 4 3 1 0 -1

-2 1 0 -1 -3 -4 -5

-3 0 -1 -2 -4 -5 -6

-4 -1 -2 -3 -5 -6 -7

-5 -2 -3 -4 -6 -6 -8

-6 -3 -4 -5 -7 -8 -9

Así pues, puesto que necesito 12)6( f , y puesto que 6)2( f , los únicos valores aceptables

para )3(f son: 3,4,5,6,7,8,9)3( f

Eliminando algunos otros casos, los valores que puede tomar la función son los siguientes:

Que son lo suficientemente reducidos como para poder separar por casos:

A) 3)3( f .

Hacia delante: 1 camino:

Hacia atrás: 3 caminos:

Total: 3 caminos.

B) 4)3( f

Hacia delante: 3 caminos:

Hacia atrás: 5 caminos:

Total: 15 caminos.

C) 5)3( f

Hacia delante: 6 caminos:

Hacia atrás: 9 caminos:

Total: 54 caminos.

D) 6)3( f

Hacia delante: 7 caminos.

Hacia atrás: 7 caminos.

Total: 49 caminos.

E) 7)3( f

Hacia delante: 9 caminos.

Hacia atrás: 6 caminos.

Total: 54 caminos.

F) 8)3( f

Hacia delante: 5 caminos.

Hacia atrás: 3 caminos.

Total: 15 caminos.

G) 9)3( f

Hacia delante: 3 caminos.

Hacia atrás: 1 camino.

Total: 3 caminos.

Además, debemos descontar las combinaciones que hacen incompatible el

grupo )4(,)3(,)2( fff que son:

6)4(,3)3(,6)2( fff , 6)4(,5)3(,6)2( fff ,

6)4(,4)3(,6)2( fff , 6)4(,3)3(,6)2( fff ,

6)4(,7)3(,6)2( fff , 6)4(,8)3(,6)2( fff ,

6)4(,9)3(,6)2( fff , 7)4(,8)3(,6)2( fff .

Con lo que llegamos finalmente a: 1858315544954153 funciones.

AIME (1983-1999)

1983

1984

1985

1986

1987

1988

1989

1990

1991

1992

1993

1994

1995

1996

1997

1998

1999

1 60 93 384 337 300 770 869 528 146 400 728 63 255 200 750 25 29

2 15 592 26 104 137 169 968 828 840 502 580 312 25 340 125 480 118

3 20 144 198 150 182 27 750 117 166 164 943 561 67 44 126 800 38

4 26 649 32 181 480 20 675 13 159 62 870 312 224 166 17 17 185

5 4 512 986 890 588 634 283 432 128 660 763 103 51 23 417 40 223

6 35 24 315 33 193 142 160 840 743 156 495 660 589 49 42 308 314

7 57 997 757 981 70 110 925 89 383 320 5 72 27 300 198 196 650

8 61 160 61 141 112 364 334 560 10 819 365 315 85 799 90 618 25

9 12 20 49 306 33 192 144 73 44 164 118 394 616 342 233 87 259

10 432 119 600 358 120 840 994 144 532 572 250 450 215 159 117 152 489

11 288 106 85 816 486 163 947 23 135 945 93 465 40 276 241 525 177

12 65 401 182 61 19 441 137 720 677 792 344 702 5 58 58 83 345

13 448 15 401 560 931 987 905 184 990 820 163 850 400 65 66 368 742

14 130 38 25 750 373 84 490 594 384 94 448 71 378 768 582 130 463

15 175 36 864 400 462 704 108 20 12 396 997 597 37 777 554 761 408

AIME I (2000-2020)

2000

2001

2002

2003

2004

2005

2006

2007

2008

2009

2010

2011

2012

2013

2014

2015

2016

2017

2018

2019

2020

1 8 630 59 839 217 942 84 83 252 840 107 85 40 150 790 722 336 390 600 342 547

2 21 651 154 301 201 12 901 52 25 697 109 36 195 200 144 139 71 62 925 29 17

3 667 500 25 484 241 109 725 15 314 11 529 31 216 18 200 307 810 69 157 120 621

4 260 291 840 12 86 294 124 105 80 177 515 56 279 429 49 507 108 803 289 122 93

5 26 937 183 505 849 630 936 539 14 72 501 144 330 98 134 341 53 321 189 252 52

6 997 79 12 348 882 45 360 169 17 412 406 11 71 47 36 58 13 48 440 90 173

7 5 923 428 380 588 150 408 477 708 41 760 16 280 41 100 539 103 564 52 880 81

8 52 315 748 129 199 113 89 30 47 398 132 318 89 371 937 695 162 41 147 67 103

9 25 61 757 615 35 74 46 737 190 420 158 192 49 113 2 494 744 45 210 540 77

10 173 200 148 83 817 47 65 860 32 346 202 503 170 80 58 72 504 56 4 352 407

11 248 149 230 92 512 544 458 955 172 600 365 7 373 148 391 108 109 360 195 20 510

12 177 5 275 777 14 25 906 875 375 11 243 594 18 21 453 431 132 252 683 230 270

13 731 174 63 155 482 83 899 80 40 90 69 330 41 961 850 91 273 59 126 32 36

14 571 351 30 127 813 936 183 224 432 905 109 37 375 36 263 483 574 896 351 97 85

15 927 85 163 289 511 38 27 989 871 150 45 98 332 272 41 53 270 145 59 65 58

AIME II (2000-2020)

2000

2001

2002

2003

2004

2005

2006

2007

2008

2009

2010

2011

2012

2013

2014

2015

2016

2017

2018

2019

2020

1 7 816 9 336 592 13 46 372 100 114 640 37 34 275 334 131 108 196 800 59 231

2 98 298 294 120 441 79 893 200 620 469 281 810 363 881 76 25 107 781 112 107 171

3 758 898 111 192 384 802 49 578 729 141 150 143 88 350 720 476 265 409 371 96 103

4 180 67 803 28 927 435 462 450 21 89 52 51 61 40 447 18 180 222 23 187 108

5 376 253 42 216 766 54 29 888 504 32 75 542 750 20 420 90 182 791 74 520 151

6 181 251 521 112 408 392 12 640 561 750 8 80 125 282 167 440 275 195 37 216 626

7 137 725 112 400 293 125 738 553 753 401 136 3 32 945 21 161 840 501 20 715 298

8 110 429 49 348 54 405 336 896 251 41 772 784 40 272 254 36 728 134 556 53 101

9 0 929 501 6 973 250 27 259 19 0 11 559 107 106 581 384 262 13 184 472 90

10 647 784 900 156 913 11 831 710 240 96 163 57 496 146 147 486 43 546 756 547 239

11 131 341 518 578 625 889 834 179 254 125 68 73 8 399 56 23 749 544 461 11 71

12 118 101 660 134 134 307 865 91 310 803 676 97 958 540 399 548 732 110 112 47 248

13 200 69 901 683 484 418 15 640 29 672 263 96 677 10 28 628 371 245 647 504 60

14 495 840 98 51 108 463 63 676 7 983 7 440 16 512 77 89 450 168 227 71 10

15 1 417 282 15 593 169 9 389 181 14 218 850 919 222 149 129 863 682 185 574 717