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The MATHEMATICAL ASSOCIATION OF AMERICA

American Mathematics Competitions
Presented by The Akamai Foundation

3rd Annual American Mathematics Contest 10

AMC 10 - Contest A
Solutions Pamphlet
Tuesday, FEBRUARY 12, 2002

This Pamphlet gives at least one solution for each problem on this year’s contest and
shows that all problems can be solved without the use of a calculator. When more than
one solution is provided, this is done to illustrate a significant contrast in methods,
e.g., algebraic vs geometric, computational vs conceptual, elementary vs advanced.
These solutions are by no means the only ones possible, nor are they superior to others
the reader may devise.
We hope that teachers will inform their students about these solutions, both as
illustrations of the kinds of ingenuity needed to solve nonroutine problems and as
examples of good mathematical exposition. However, the publication, reproduction,
or communication of the problems or solutions of the AMC 10 during the period when
students are eligible to participate seriously jeopardizes the integrity of the results.
Duplication at any time via copier, phone, email, the Web or media of any type is a
violation of the copyright law.

Correspondence about the problems and solutions should be addressed to:

Prof. Douglas Faires
Department of Mathematics
Youngstown State University
Youngstown, OH 44555-0001
Orders for prior year Exam questions and Solutions Pamphlets should be addressed to:
Titu Andreescu, AMC Director
American Mathematics Competitions
University of Nebraska-Lincoln, P.O. Box 81606
Lincoln, NE 68501-1606

Copyright © 2002, Committee on the American Mathematics Competitions

The Mathematical Association of America
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 Solutions 2002 3rd AMC 10 A 2

1. (D) We have

102000 + 102002 102000 (1 + 100) 101

2001 2001
= 2000 = ≈ 5.
10 + 10 10 (10 + 10) 20

2. (C) We have
2 12 9 1 4 9 1 + 8 + 27 36
(2, 12, 9) = + + = + + = = = 6.
12 9 2 6 3 2 6 6
2
3. (B) No matter how the exponentiations are performed, 22 always gives 16.
Depending on which exponentiation is done last, we have
 2 2  22 
2
(22 )
22 = 256, 2 = 65, 536, or 22 = 256,

so there is one other possible value.

4. (E) When n = 1, the inequality becomes m ≤ 1 + m, which is satisfied by all
integers m. Thus, there are infinitely many of the desired values of m.
2
5. (C) The large circle has radius

3, so its area is π·3 = 9π. The seven small circles
2
have a total area of 7 π · 1 = 7π. So the shaded region has area 9π − 7π = 2π.
6. (A) Let x be the number she was given. Her calculations produce
x−9
= 43,
3
so
x − 9 = 129 and x = 138.
The correct answer is
138 − 3 135
= = 15.
9 9
 Solutions 2002 3rd AMC 10 A 3

7. (A) Let CA = 2πRA be the circumference of circle A, let CB = 2πRB be the

circumference of circle B, and let L the common length of the two arcs. Then
45 30
CA = L = CB .
360 360
Therefore
CA 2 2 2πRA RA
= so = = .
CB 3 3 2πRB RB
Thus, the ratio of the areas is
2
 2
Area of Circle (A) πRA RA 4
= 2 = = .
Area of Circle (B) πRB RB 9

8. (A) Draw additional lines to cover the entire figure with congruent triangles.
There are 24 triangles in the blue region, 24 in the white region, and 16 in the
red region. Thus, B = W .

9. (B) Adding 1001C − 2002A = 4004 and 1001B + 3003A = 5005 yields 1001A +
1001B + 1001C = 9009. So A + B + C = 9, and the average is
A+B+C
= 3.
3

10. (A) Factor to get (2x + 3)(2x − 10) = 0, so the two roots are −3/2 and 5, which
sum to 7/2.
 Solutions 2002 3rd AMC 10 A 4

11. (B) First note that the amount of memory needed to store the 30 files is

3(0.8) + 12(0.7) + 15(0.4) = 16.8 mb,

so the number of disks is at least

16.8 2
= 11 + .
1.44 3
However, a disk that contains a 0.8-mb file can, in addition, hold only one 0.4-
mb file, so on each of these disks at least 0.24 mb must remain unused. Hence,
there is at least 3(0.24) = 0.72 mb of unused memory, which is equivalent to
half a disk. Since  
2 1
11 + + > 12,
3 2
at least 13 disks are needed.
To see that 13 disks suffice, note that:
Six disks could be used to store the 12 files containing 0.7 mb;
Three disks could be used to store the three 0.8-mb files together with three of
the 0.4-mb files;
Four disks could be used to store the remaining twelve 0.4-mb files.
12. (B) Let t be the number of hours Mr. Bird must travel to arrive on time. Since
three minutes is the same as 0.05 hours, 40(t + 0.05) = 60(t − 0.05). Thus,

40t + 2 = 60t − 3, so t = 0.25.

The distance from his home to work is 40(0.25 + 0.05) = 12 miles. Therefore,
his average speed should be 12/0.25 = 48 miles per hour.

OR

Let d be the distance from Mr. Bird’s house to work, and let s be the desired
average speed. Then the desired driving time is d/s. Since d/60 is three minutes
too short and d/40 is three minutes too long, the desired time must be the
average, so  
d 1 d d
= + .
s 2 60 40
This implies that s = 48.
 Solutions 2002 3rd AMC 10 A 5

13. (B) First notice that this is a right triangle, so two of the altitudes are the legs,
whose lengths are 15 and 20. The third altitude, whose length is x, is the one
drawn to the hypotenuse. The area of the triangle is 21 (15)(20) = 150. Using 25
as the base and x as the altitude, we have
1 300
(25)x = 150, so x= = 12.
2 25

25
20
x

15

OR
Since the three right triangles in the figure are similar,
x 20 300
= , so x = = 12.
15 25 25
14. (B) Let p and q be two primes that are roots of x2 − 63x + k = 0. Then
x2 − 63x + k = (x − p)(x − q) = x2 − (p + q)x + p · q,
so p + q = 63 and p · q = k. Since 63 is odd, one of the primes must be
2 and the other 61. Thus, there is exactly one possible value for k, namely
k = p · q = 2 · 61 = 122.
15. (E) The digits 2, 4, 5, and 6 cannot be the units digit of any two-digit prime, so
these four digits must be the tens digits, and 1, 3, 7, and 9 are the units digits.
The sum is thus
10(2 + 4 + 5 + 6) + (1 + 3 + 7 + 9) = 190.
(One set that satisfies the conditions is {23, 47, 59, 61}.)
16. (B) From the given information,
(a + 1) + (b + 2) + (c + 3) + (d + 4) = 4(a + b + c + d + 5),
so
(a + b + c + d) + 10 = 4(a + b + c + d) + 20
10
and a + b + c + d = − .
3
OR
Note that a = d + 3, b = d + 2, and c = d + 1. So,
a + b + c + d = (d + 3) + (d + 2) + (d + 1) + d = 4d + 6.
Thus, d + 4 = (4d + 6) + 5, so d = −7/3, and
 
7 10
a + b + c + d = 4d + 6 = 4 − +6=− .
3 3
 Solutions 2002 3rd AMC 10 A 6

17. (D) After the first transfer, the first cup contains two ounces of coffee, and the
second cup contains two ounces of coffee and four ounces of cream. After the
second transfer, the first cup contains 2 + (1/2)(2) = 3 ounces of coffee and
(1/2)(4) = 2 ounces of cream. Therefore, the fraction of the liquid in the first
cup that is cream is 2/(2 + 3) = 2/5.
18. (D) There are six dice that have a single face on the surface, and these dice can
be oriented so that the face with the 1 is showing. They will contribute 6(1) = 6
to the sum. There are twelve dice that have just two faces on the surface because
they are along an edge but not at a vertex of the large cube. These dice can be
oriented so that the 1 and 2 are showing, and they will contribute 12(1 + 2) = 36
to the sum. There are eight dice that have three faces on the surface because
they are at the vertices of the large cube, and these dice can be oriented so that
the 1, 2, and 3 are showing. They will contribute 8(1 + 2 + 3) = 48 to the sum.
Consequently, the minimum sum of all the numbers showing on the large cube
is 6 + 36 + 48 = 90.
19. (E) Spot can go anywhere in a 240◦ sector of radius two yards and can cover
a 60◦ sector of radius one yard around each of the adjoining corners. The total
area is  
2 240 2 60
π(2) · + 2 π(1) · = 3π.
360 360

60 2
1
1
240

20. (D) Since △AGD is similar to △CHD, we have HC/1 = AG/3. Also, △AGF
is similar to △EJF , so JE/1 = AG/5. Hence,

HC AG/3 5
= = .
JE AG/5 3

21. (D) The values 6, 6, 6, 8, 8, 8, 8, 14 satisfy the requirements of the problem,

so the answer is at least 14. If the largest number were 15, the collection would
have the ordered form 7, , , 8, 8, , , 15. But 7 + 8 + 8 + 15 = 38,
and a mean of 8 implies that the sum of all values is 64. In this case, the four
missing values would sum to 64 − 38 = 26, and their average value would be 6.5.
This implies that at least one would be less than 7, which is a contradiction.
Therefore, the largest integer that can be in the set is 14.
 Solutions 2002 3rd AMC 10 A 7

22. (C) The first application removes ten tiles, leaving 90. The second and third
applications each remove nine tiles leaving 81 and 72, respectively. Following
this pattern, we consecutively remove 10, 9, 9, 8, 8, . . . , 2, 2, 1 tiles before we are
left with only one. This requires 1 + 2(8) + 1 = 18 applications.

OR

Starting with n2 tiles, the first application leaves n2 − n tiles. The second
application reduces the number to n2 − n − (n − 1) = (n − 1)2 tiles. Since two
applications reduce the number from n2 to (n − 1)2 , it follows that 2(n − 1)
2
applications reduce the number from n2 to (n − (n − 1)) = 1, and 2(10 − 1) =
18.
23. (D) Let H be the midpoint of BC. Then EH is the perpendicular bisector of
AD, and △AED is isosceles. Segment EH is the common altitude of the two
isosceles triangles △AED and △BEC, and
p
EH = 102 − 62 = 8.

Let AB = CD = x and AE = ED = y. Then 2x+2y+12 = 2(32), so y = 26−x.

Thus,
82 + (x + 6)2 = y 2 = (26 − x)2 and x = 9.

E
y y
10 10
A x B 6 H 6 C x D

24. (A) There are ten ways for Tina to select a pair of numbers. The sums 9, 8,
4, and 3 can be obtained in just one way, and the sums 7, 6, and 5 can each
be obtained in two ways. The probability for each of Sergio’s choices is 1/10.
Considering his selections in decreasing order, the total probability of Sergio’s
choice being greater is
  
1 9 8 6 4 2 1 2
1+ + + + + + +0+0+0 = .
10 10 10 10 10 10 10 5
 Solutions 2002 3rd AMC 10 A 8

25. (C) First drop perpendiculars from D and C to AB. Let E and F be the feet
of the perpendiculars to AB from D and C, respectively, and let

h = DE = CF, x = AE, and y = F B.

D 39 C
5 12
h h
A x E 39 F y B

Then
25 = h2 + x2 , 144 = h2 + y 2 , and 13 = x + y.
So

144 = h2 + y 2 = h2 + (13 − x)2 = h2 + x2 + 169 − 26x = 25 + 169 − 26x,

which gives x = 50/26 = 25/13, and

s  2 r r
2
25 25 144 60
h= 5 − =5 1− =5 = .
13 169 169 13

Hence
1 60
Area (ABCD) = (39 + 52) · = 210.
2 13

OR

Extend AD and BC to intersect at P . Since △P DC and △P AB are similar,

we have
PD 39 PC
= = .
PD + 5 52 P C + 12
So P D = 15 and P C = 36. Note that 15, 36, and 39 are three times 5, 12,
and 13, respectively, so 6 AP B is a right angle. The area of the trapezoid is the
difference of the areas of △P AB and △P DC, so
1 1
Area(ABCD) = (20)(48) − (15)(36) = 210.
2 2

D C
A B

OR
 Solutions 2002 3rd AMC 10 A 9

Draw the line through D parallel to BC, intersecting AB at E. Then BCDE

is a parallelogram, so DE = 12, EB = 39, and AE = 52 − 39 = 13. Thus
DE 2 + AD2 = AE 2 , and △ADE is a right triangle. Let h be the altitude from
D to AE, and note that
1 1
Area(ADE) = (5)(12) = (13)(h),
2 2
so h = 60/13. Thus
60 1
Area(ABCD) = · (39 + 52) = 210.
13 2

D C
A E B
 The
American Mathematics Contest 10 (AMC 10)
Sponsored by
Mathematical Association of America
The Akamai Foundation
University of Nebraska – Lincoln
Contributors
American Mathematical Association of Two Year Colleges
American Mathematical Society
American Society of Pension Actuaries
American Statistical Association
Casualty Actuarial Society
Clay Mathematics Institute
Consortium for Mathematics and its Applications
Institute for Operations Research and the Management Sciences
Kappa Mu Epsilon
Mu Alpha Theta
National Association of Mathematicians
National Council of Teachers of Mathematics
Pi Mu Epsilon
School Science and Mathematics Association
Society of Actuaries