Embed Size (px)
Citation preview
Math Team
Skills forOctober Rounds
Round1
Round2
Round3
Round4
Round5
Round6
Round 1 – Geometry: Volumes & Surfaces
Volume: Volume is measured in cubic units.
Surface Area (SA): Surface area is measured in square units. It is important to know how to derive surface area formulas.
Rectangular Solid CubeRight Cylinder
Lateral Surface Area (LA): the total surface area (SA) minus the area of the base(s).
2 2 2
V lwh
SA lw lh lh
3
26
V s
SA s
2
22 2
V r h
SA r rh
Some other formulas:
2
2
1
3
2
V s h
SA b b
3
2
4
3
4
V r
SA r
2
2 2
1
31
2
V r h
SA r r
Right Cone
SphereRight Square Pyramid
Round 2 – Pythagorean Relations in Rectilinear Figures
The Pythagorean Theorem: For right triangles with legs a and b and hypotenuse c, the sum of the squares of the legs is equal to the square of the hypotenuse.
2 2 2a b c a
b
c
In this round, if a right triangle is not given, look to construct right triangles from the figure that is described or drawn.
Pythagorean Triples: One of the main shortcuts in this section is to know some Pythagorean Triples; the most popular are the 3 – 4 – 5 and the 5 – 12 – 13.
3
12
13
4
5 5
Generating more Pythagorean Triples: There are a few ways to generate Pythagorean Triples, here is one:
Consider the sequence: 1 2 3 41 , 2 , 3 , 4 ,3 5 7 9
. . .
Starting with the mixed number 1 1/3 for each iteration, increase the whole number and numerator each by 1 and increase the denominator by 2.
Change each mixed number into an improper fraction…4 12 24 40, , , ,3 5 7 9
. . .
Each fraction generates a triple with the numerator and denominator as the legs and the hypotenuse is one more than the numerator. (This does not generate all Pythagorean triples.)
Another method to use to generate Pythagorean Triples is to pick integers m and n such that m > n. The three sides are then, 2mn, m2 – n2, and m2 + n2. See the table below.
m, n 2mn m2 – n2 m2 + n2 Primitive or Multiple
2, 1 4 3 5 Primitive
3, 2 12 5 13 Primitive
3, 1 6 8 10 2*(3 – 4 – 5)
4, 3 24 7 25 Primitive
4, 2 16 12 20 4*(3 – 4 – 5)
4, 1 8 15 17 Primitive
5, 4 40 9 41 Primitive
5, 3 30 16 34 2*(8 – 15 – 17)
This method will find all the Primitive Pythagorean Triples, but will not find all of the multiples.
Primitive Pythagorean Triples are those that are not multiples of another triple. With the primitive triples, we can easily find the multiples. Here is a list of some Primitive Pythagorean Triples
a b c
3 4 5
5 12 13
8 15 17
7 24 25
20 21 29
12 35 37
9 40 41
28 45 53
11 60 61
33 56 65
16 63 65
48 55 73
13 84 85
Simplifying Radicals: Since not all sides of the right triangle will be whole numbers, simplifying radicals in a necessary skill for this round.
To simplify radicals, factor inside the radical using perfect squares, then pull out their square roots.
Ex) Ex)150
25 6
25 6
5 6
2 11
8 10
8 10
4 5
18
9 2
9 2
3 2
x y
x y y
x y y
x y y
Rationalize Denominators: According to the rules, denominators must be rationalized if a radical appears there.
Ex)
Ex)
2
2
6
6 6 23
2
222
6
6
1
3
1 6 6
3 63
6
186
Round 3 – Linear Equations
This round obviously consists of solving linear equations, but they tend to weave other random topics throughout the round. Be prepared to solve systems of equations, and solve implicitly, work with percents such as mixture problems, compare parallel and perpendicular lines and see some notation that can get confusing. Also, converting English to algebra seems to be prominent in this section.
Solving Systems of Equations:By Addition (or cancelation): Line up like terms and constants from each equation. “Fix” at least one equation so that when the equations are added, one of the variable terms cancels out. Solve the resulting equation for the variable that is left over then back fill to find the value of the other variable.
Ex) 3 2 7
5 8 21
x y
x y
3 2( 4)( ) 7
5 8 21
( 4)x y
x y
12 8 28
5 8 21
x y
x y
7 7 x 1x
3 2 7x y
3( ) 2 7
3 2 7
2 4
2
1 y
y
y
y
Solving Implicitly
Ex) Solve for y:1
3 2 23
x y xy
1(3 23 3) 2
3x y xy
9 6 6x y xy 9 9x xy xy x
6 6 9y xy x
(6 ) 6 9y x x (6 ) (6 )x x
6 9 9 6
6 6
x xy or y
x x
Remove fractions.
Get all terms with a y on one side and everything else on the other.
Factor out a y.
Divide both sides by 6 – x.
Converting a Repeating Decimal to a Fraction(There is only a very slight chance that you would see this in rounds.)
Step 1: Let x equal the repeating decimal you are trying to convert to a fraction
Step 2: Examine the repeating decimal to find the repeating digit(s)
Step 3: Place the repeating digit(s) to the left of the decimal point
Step 4: Place the repeating digit(s) to the right of the decimal point
Step 5: Subtract one equation from the other(As you subtract, just make sure that the difference is positive for both sides.)
Step 6: Solve for x.See the example on the next slide.
5
9x
0.55555555...x
0. 55555 5....
10 5.5555555...x
0.55555551 5...x
9 5x
Step 1: Let x equal the repeating decimal you are trying to convert to a fraction. (We will use this equation to form two others.)
Step 2: Examine the repeating decimal to find the repeating digit(s). (only the 5 repeats)
Step 3: Place the repeating digit(s) to the left of the decimal point. (Multiply both sides of the original equation by 10 in this case.)
Step 4: Place the repeating digit(s) to the right of the decimal point. (Multiply both sides of the original equation by 1 in this case; the 5 was already to the right of the decimal.)
Step 5: Subtract one equation from the other
As you subtract, just make sure that the difference is positive for both sides
0.5 55555510 5....x
10 5.5555555...x 1 0.5555555...x
Step 6: Solve for x.
Example: Convert the repeating decimal 0.555555… to a fraction
45 1
990 22x
0.045454545...x
45The repeats
1000 45.45454545...x
990 45x
Step 1: Let x equal the repeating decimal you are trying to convert to a fraction. (We will use this equation to form two others.)
Step 2: Examine the repeating decimal to find the repeating digit(s).
Step 3: Place the repeating digit(s) to the left of the decimal point. (Multiply both sides of the original equation by 1000 in this case.)
Step 4: Place the repeating digit(s) to the right of the decimal point. (Multiply both sides of the original equation by 10 in this case)
Step 5: Subtract one equation from the other
As you subtract, just make sure that the difference is positive for both sides
1000 45.45454545...x 10 0.45454545... x
Step 6: Solve for x.
0.045 4541000 545....x
0.0 45454541 ..0 5..x 10 0 45454545....x
More ChallengingExample: Convert the repeating decimal
0.045454545… to a fraction
Round 4 – Algebra 1: Fractions and Mixed Numbers
In this section, some of the part A problems simply have you add or subtract fractions, mixed numbers and decimals. Of course, you could also see square roots and you are very likely to see some complex fractions (fraction expressions within a fraction)
Simplifying Complex Fractions.One way to simplify this type of expression is to add/subtract to simplify the numerator and denominator separately. Then you will be left with one fraction divided by one fraction. To divide, just multiply by the reciprocal and simplify.
a cb ds da c
Short-cut: If you can make every denominator the same, all the denominators will cancel.Ex) 3 2
4 31 12 4
3 24 31 12 4
3 43 46 36 3
9 812 126 312 12
9 8
6 3
1
9
Round 5 – Inequalities and Absolute Value
Remember that inequalities give you a range of answers. Often times you will be asked for the largest possible answer or the smallest, but sometimes you need to use interval notation to describe your answer. You could also see graphs of inequalities.
(When solving an inequality, don’t forget to flip the inequality symbol when you multiply or divide both sides by a negative.)
2 0 2 4 6
Simple: 2x
Compound:
2 0 2 4 6
0 6x and x or using interval notation: [0,6)
Bracket for closed (including) Parenthesis for open
(not including)
Complex:
2 0 2 4 6
[ 3,0) [3,6]
Absolute Value
Absolute value is, by definition, a piecewise function:
x 0
0
x if x
x if x
Ex) Solve for x: 4 3x x
OR
4 4
1
x
x
4 3
2 4
2
x x
x
x
OR
4 3x x If 4 – 3x is greater than zero.
4 3( )x x If 4 – 3x is less than zero.
Absolute Value
The absolute value of a number is the number’s distance from zero on a number line.
If you see an absolute value inequality, the above definition may help you set up two inequalities.
Ex) 3 2x (x – 3)’s distance from zero on a number line is at least 2.
3 2x 3 2x OR
5x 1x
2 0 2 4 6
OR
Round 6 – Algebra 1: Evaluations
In this round, expect to evaluate expressions with roots, fractions, and exponents. You may also see operations with numbers whose base is not 10. You may be asked to complete a sequence of numbers or find the sum of a sequence of numbers. Also, you may be asked to perform a defined operation that uses a random symbol like .
Ex) Given , Evaluate . 2a b
a bb
3 2
3a 2b
3 2 2( ) ( )
( )
2
2
3
6 2
2
84
2
3 2 4 Therefore,
Defined Operations
Base 10 NumbersWe use base 10 numbers all of the time. We have memorized the places of base 10 numbers since we were young children.
For Example, the number has a 2 in the one’s place, a 9 in the ten’s place and a 6 in the hundred’s place. (The subscript of 10 simply means that it is a base 10 number. When there is no subscript, we always assume that the number is base 10)
10692
To understand other bases, we need to know where the names of the place values come from.
6 9 2010110210Base 10 place values:
Evaluate to get the names: 110100
So the number 69210 is really:
Six 100’s = 600+ Nine 10’s = 90+ Two 1’s = 2
= 692
Base 2 NumbersBase 2 numbers’ place values work the same as base 10:
When we are in base 10, remember that we can only use the numbers from 0 to 9. In base 2, we can only use numbers from 0 to 1. So to evaluate 101102 :
021222
Base 2 place values:
Evaluating this base 2 number: 13242
124816In base 2, each placevalue is worth…
0 1 1 0
16
0212223242
Using these place values:8 4 2 1
Yields: One 16 + Zero 8’s + One 4 + One 2 + Zero 1’s16 + 0 + 4 + 2 + 0Which is: =22
101102 = 2210Therefore:
Base 3 Numbers
If you can figure out the base 10 equivalent to 12013 then you’ve got it.
Solution: 27 + 18 + 0 + 3 = 48
