50 AMC Lectures Problems Book 1 (10) Area And Area … · 50 AMC Lectures Problems Book 1 (10) Area And Area Method 98 Problem 15: As shown in the figure, D, E, and F are points on

50 AMC Lectures Problems Book 1 (10) Area And Area Method

95

PROBLEMS

Problem 1: ABCD is a square with the side length of a. Half circles are drawn inside the

square with BC and CD as diameters, respectively. Find the shaded area.

Problem 2: As shown in the figure, circle O has the radius of 1. The inscribed angle

ABC = 30. Find the shaded area.

Problem 3: As shown in the figure, isosceles right triangle ABC with BC = 8 cm. BDC is

a half circle with the diameter BC. Find the sum of two shaded areas I and II.

Problem 4: Two diameters of circle O are perpendicular. Using AO, BO, CO, and DO as

diameters to draw circles as shown in the figure. Show that the areas

of four regions inside the big circle but outside the small circles equal

the areas of four regions shared by every two small circles.


96

Problem 5: Two squares with the side lengths of 10 cm and 12 cm, respectively. Find the

shaded area.

Problem 6: (1983 AMC #28) Triangle ABC in the figure has area 10. Points D, E, and F,

all distinct from A, B, and C, are on sides AB, BC and CA respectively, and

AD = 2, DB = 3. If triangle ABE and quadrilateral DBEF have equal areas,

find that area.

Problem 7: D, E, and F are three vertices of rhombus BFDE which are on the sides AC,

AB, and BC of triangle ABC, respectively, as shown in the figure. If AB

= a and BC = b, find the perimeter of the rhombus BFDE.

Problem 8: As shown in the figure, trapezoid ABCD has the area of S. AB = b, CD = a

with a < b and AB//CD. Two diagonals meet at O. The area of BOC

is S9

2. Find

b

a.

Problem 9: In ABC, D is the midpoint of side BC, E is the midpoint of AD, F is the

midpoint of BE, and G is the midpoint of FC. What part of the area of ABC

is the area of EFG?


97

Problem 10: As shown in the figure, ABC is divided into six smaller triangles by lines

drawn from the vertices through a common interior point. The areas of

four of these triangles are as indicated. Find the area ABC.

Problem 11. (1988 AIME) Let P be an interior point of ABC and extend lines from the

vertices through P to the opposite sides. Let a, b, c, and d

denote the lengths of the segments indicated in the figure. Find

the product abc if a + b + c = 43 and d = 3.

Problem 12: (1997 China Middle School Math Competition) As shown in the figure, P is

a point inside the equilateral triangle ABC. The distances from P o each

side are PD = 1, PE = 3, and PF = 5. Find the area of the equilateral

triangle ABC.

Problem 13: The area of ABC is 60. As shown in the figure, E and F trisect BC. D is

the midpoint of CA. BD intersects AE at G, AF at H. Find the area of

quadrilateral AGH.

Problem 14: (2001 AMC 12) In rectangle ABCD, points F and G lie on AB so that AF =

FG = GB and E is the midpoint of DC. Also, AC intersects EF at H

and EG at J. The area of rectangle ABCD is 70. Find the area of

triangle EHJ.


98

Problem 15: As shown in the figure, D, E, and F are points on BC, CA, and AB of an

acute triangle ABC. AD, BE, and CF meet at P. AP = BP = CP = 6.

If PD = x, PE = y, PF = z, and xy + yz + zx = 28. Find the value of

xyz.

Problem 16: As shown in the figure, ABCD is a rectangle. BD is the diagonal. AE and

CF are perpendicular to BD. Find the area of ABCD if BE = 1, and EF = 2.

34 , (B) 53 (C) 6 (D) Undetermined.

Problem 17: As shown in the figure, ABCD is a square. Find the shaded area.

(A)17 (B) 17

290 (C)18 (D)10 3

Problem 18: As shown in the figure, the area of parallelogram ABCD is 1. E and F are

the midpoints of AB and BC, respectively. AF meets CE at G and DE

at H. Find the area of EGH.

Problem 19: M is the centroid of ABC. AM = 3, MB = 4, MC = 5. Find the area of

ABC.


99

Problem 20: (1984 AIME) A point P is chosen in the interior of ABC so that when

lines are drawn through P parallel to the sides of ABC , the resulting smaller triangles,

t1, t2 and t3 in the figure, have areas 4, 9 and 49, respectively. Find the area of ABC.

Problem 21: As shown in the figure, in ABC, points D, E, and F are on BC, AC, and

AB, respectively. AD, BE, and CF meet at G. BD = 2CD. The area

S1 = 3, and S2 = 4. Find SABC, the area of ABC.

Problem 22: As shown in the figure, ABCS 1. AD = 3

1AB, BE =

3

1BC and CF =

3

1CA.

CD, AE, BF meet pair wisely at X, Y, and Z. Find the area of XYZ.

Problem 23: For triangle ABC, extend AB to F such that AB = BF, extend BC to D such

that BC = CD, and extend CA to E such that CA = AE. The ratio of the area of triangle

ABC to the area of triangle DEF is 1/7.


100

SOLUTIONS TO PROBLEMS

Problem 1: Solution:

Draw OM CD, ON BC through O, the center of the square.

The unshaded area consists of one smaller square and two quarter circles.

The shaded area is S, and S = 8

)6(])

2(

4

12)

2

1[(

2222 aa

aa


Connect CO to meet the circle at D. Connect AD and AO.

We see that AOC = 60.

The shaded are is the same as the area of the sector OAC – the area of

OAC = 4

3

6

1


Connect CD. Triangle CDB is also an isosceles right triangle with CD

= BD. So I and III have the same areas.

We also see that D is the midpoint of AB. Therefore the sum of the

shaded areas is 1682

1

2

1

2

1 2 ABCS


We label each region as shown in the figure.

We have:

(2) )2(444

(1) 2

2

2

azyx

ayx

4 × (1) – (2): 4x – 4z = 0 4x = 4z

Problem 5: Solution: 50.


101

We connect BF and we know that AC//BF. So the area of triangle ABH is the same as the

area of triangle CHF (Theorem 4(a)).

The shaded area is the area of triangle ACF, which is the same as the

area of triangle ABC, which is 10 10 2 = 50.


Since triangle ABE and quadrilateral DBEF have equal areas, we know that triangle ADG

has the same area as triangle EFG (Theorem 4(b)).

Therefore, AF//DE and ABC is similar to DBE.

BE

DB

CE

AD

BECE

32

BE

CE

3

2

ABE

ACE

S

S

3

2

61023

3

ABES .


Let the side of the rhombus be x.

We write the equation of the areas:

ABC = AED + DFC + SBFDE.

Since AED = DFC = B, we have:

Bacsin2

1 = Bxcx sin)(

2

1 + Bxax sin)(

2

1 + x

2sinB.

ac = x(c – x) + x(a – x) + 2x2 = (a + c)x.

Solving for x: x = ca

ac

.

Therefore the perimeter is ca

ac

4.


Let the areas of DOC= S1 and AOB = S2.


102

.)9

2(

,9

5

2

21

21

SSS

SSS

Solving we get

.9

1

,9

4

2

1

SS

SS

or

.9

4

,9

1

2

1

SS

SS

Since, a < b, 21 SS . Therefore 2

1

2

1 S

S

b

a.

Problem 9: Solution: 1/8.

Draw EC. Since the altitude of BEC is 2

1 the altitude of BAC, and both

triangles share the same base, the area of BEC = 2

1 area of

BAC. Area of EFC = 2

1 area of BEC, and area of EGF =

2

1area of

EFC; therefore area of EGF = 4

1 area of BEC. Thus, since area of

BEC = 2

1 area of ABC, area of EGF =

8

1area of ABC.


Let x and y be the areas for the small triangles as shown in the figure.

COD

BOD

ACO

ABO

S

S

S

S

30

40

70

84

y

x

Similarly, we have CEO

AEO

BCO

ABO

S

S

S

S

y

x 70

3040

84

Or y

y 70

4

3

70

70

.

x = 56 and y = 35. The total area is 84 + 70 + 40 + 30 + 35 + 56 = 315.


103

Problem 11. Solution:

By Theorem 9, we have

BAC

BPC

S

S

= ad

d

(1)

CBA

CPA

S

S

= bd

d

(2)

ACB

APB

S

S

= cd

d

(3)

We also know that BPCS + CPAS + APBS = ABCS

Adding (1), (2), and (3), we get: ad

d

+

bd

d

+

cd

d

= 1.

Simplifying into: 2d 3 + (a + b + c)d

2 – abc = 0.

Therefore abc = 2d 3 + (a + b + c)d

2 = 2 × 3

2 + 43 × 3

2 = 441.


Connect PD, PE, and PF.

We have ABCS = 2

1 (PD + PE + PF) BC =

2

9 BC.

We know that BPCS = 2

1 PD BC =

2

1 BC.

Let the height on BC be h. We have h

PD =

ABC

PBC

S

S

= 9

1.

So we get h = 9.

Then the side of triangle ABC is 363

329 .

Therefore ABCS = 327)36(4

3 2 .


Connect FG and FD. Since AD = DC, EF = FC, we know that DF

is parallel to AE. HFGADH SS .

We also know that 302

1 ABCBDC SS and 20

3

2 BDCBFD SS .

Since BE = EF and DF // GE, BG = GD, 54

1 BFDBEG SS , and


104

5 BEGBEFG SS , and 1552053

1 ABCBEGABEBGA SSSS .

Let aS AHG , bS Hdf , and cS ADH .

So c + b = 20 – 10 = 10, and a + c = 30 – 15 = 15.

From Theorem 8, we have 2

3

10

15

HD

GH

cb

ca.

So by Theorem 6 we have 2

3

c

a 2a = 3c

We already know that a + c = 15. Therefore a = 9 and b = 6.

Problem 14: Solution: (C).

The area of triangle EFG is (1/6)(70) = 35/3.

Triangles AFH and CEH are similar, so 3/2 = EC/AF = EH/HF and

EH/EF = 3/5.

Triangles AGJ and CEJ are similar, so 3/4 = EC/AG = EJ/JG and

EJ/EG = 3/7.

Since the areas of the triangles that have a common altitude are proportional to their

bases, the ratio of the area of EHJ to the area of EHG is 3/7, and the ratio of the area

of EHG to that of EFG is 3/5.

Therefore, the ratio of the area of EHJ to the area of EFG is (3/5)(3/7) = 9/35.

Thus, the area of EHJ is (9/35)(35/3) = 3.


Draw PMBC at M, ANBC at N.

We know that PBCS = 2

1 PM BC and ABCS =

2

1 AN BC.

Therefore 6

x

x

AD

PD

AN

PM

S

S

ABC

PBC .

Similarly 6

y

y

S

S

ABC

PAC and 6

z

z

S

S

ABC

PAB .

Adding them together we get:


105

6x

x +

6y

y +

6z

z =

ABC

PABPACPBC

S

SSS

= 1

1 – 6

6

x + 1 –

6

6

y + 1 –

6

6

z = 1

6

3

x +

6

3

y +

6

3

z = 1.

Simplifying:

3(yz + zx + xy) + 36(x + y + z) + 324 = xyz + 6(xy + yz + zx) + 36(x + y + z) + 216.

We are given that xy + yz + zx = 28.

Hence xyz = 108 – 3( xy + yz + zx) = 24.

Problem 16: Solution: (A).

Since AB = CD, ABD = CDB, RtABE RtCDF.

BE = DF.

We know that BE = 1, EF = 2.

FD = 1, ED = EF + FD = 2 + 1 = 3

BD = BE + ED = 1 + 3 = 4.

AE2 = EB DE = 1 × 3 = 3 AE = 3 .

ABCDS = 2 ABDS = 2 2

1 BD AE = 4 3

Problem 17: Solution: (B).

Both BGDE and CHAF are parallelograms. PQRS is also a parallelogram.

Since ADE BAF, DE = AF, ADE = BAF.

Therefore QPS = PAD + ADP = PAD + BAF = 90.

So PQRS is a rectangle.

In RtABF, AF = 3422 BFAB .

In parallelogram AFCH, AFCHS = FC × AB = AF × PS.

So 34

10PS .

Similarly, 34

10PQ .

So PQRS is a square. 17

502 PSSPQRS .


106

Therefore the shaded area is: AFCHS + PQRSBGDE SS = 2 × 5 + 2 × 5 17

290

17

50 .


Connect BG and DG.

Since EBCABCABF SSS 2

1, CFGAEG SS .

Therefore AEGBEG SS BFGCFG SS = ABFS3

1 = ABCS

6

1 =

ABCDS12

1 =

12

1.

We also know that ADGS + BCGS = 2

1ABCDS . So ADGS =

3

1.

So DH

HE =

ADH

AEH

S

S

= GDH

GEH

S

S

= GDHADH

GEHAEH

SS

SS

=

ADG

AEG

S

S

= 4

1

3

112

1

.

AEHS = 5

1DAES =

5

1 × ( ABCDS

4

1) =

20

1.

Therefore 30

1

20

1

12

1 AEHAEGEGH SSS .


Extend AD to G such that MD = DG. Connect CG.

So we have MG = 2MD = AM = 3, GC = BM = 4, and MC = 5.

So MGC = 90.

We also have 2

1MGCS MG GC = 6.

32

1 MGCMDC SS

Thus 186 MDCABC SS .


Let MP = P, PN = q, RT = r, AB = c, and the area of ABC be S.


107

Three small triangles are similar to ABC. The ratio of the square roots of their areas is

the same as the ratio of their corresponding sides.

So we have c

q

S

2 (1)

c

p

S

3 (2)

c

r

S

7 (3)

Adding (1), (2), and (3) together we get:

1732

c

c

c

rqp

S

Therefore 12S and S = 144.

Note: 1tS + 2t + 3t


Since BD = 2CD, 22

3 S

S.

So S3 = 8.

Therefore 41

32

S

SS

GE

BG.

46

54

S

SS. (1)

We also know that

16

54

SS

SS2

2

3 S

S, (2)

And

23

16

SS

SS

FB

AF

S

S

4

5 (3)

Solving the system of equations (1), (2), and (3), we have

S4 = 8, S5 = 4, S6 = 3.

So ABCS S1 + S2 + S3 + S4 + S5 + S6 = 3 + 4 + 8 + 8 + 4 + 3 = 30.



108

Connect CY and let SS BEY .

Then we have 3

1

BC

BE

S

S

YBC

YBE .

So SS YBC 3 .

Since FA

CF

S

S

YAF

YCF

and FA

CF

S

S

BAF

BCF

,

BCY

BAY

S

S

= YCFBCF

YAFBAF

SS

SS

= 2

CF

FA.

So SS BAY 6 .

Therefore we have SSSS BEYBAYABE 7 .

Since 3

1

ABC

ABE

S

S, 121 SS ABC .

So we get S = 21

1.

21

6BAYS . Similarly, we have

21

6 BCZCAX SS .

Therefore ABCXYZ SS ( BCZCAXBAY SSS ) = 7

13

21

61 .

Problem 23: Proof:

Connect FC, DA, and EB as shown in the figure. All the seven triangles have the same

areas.

Therefore, the ratio of the area of triangle ABC to the area of triangle DEF

is 1/7.

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50 AMC Lectures Problems Book 1 (10) Area And Area … · 50 AMC Lectures Problems Book 1 (10) Area And Area Method 98 Problem 15: As shown in the figure, D, E, and F are points on