T he MATheMATICAL ASSOCIATION Of AMeRICA
American Mathematics Competitions
9th Annual American Mathematics Contest 10
AMC 10
Contest B
Solutions Pamphlet
Wednesday, FEBRUARY 27, 2008
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 u 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. The solu-tions are by no means the only ones possible, nor are they superior to others the reader may devi.
We hope that teachers will inform their students about the 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 riously jeopardizes the integrity of the results. Dismination via copier, telephone, e-mail, World
Wide Web or media of any type during this period is a violation of the competition rules.
After the contest period, permission to make copies of problems in paper or electronic form including posting on web-pages for educational u is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear the copyright notice.
Correspondence about the problems/solutions for this AMC 10 and orders for any publications should be addresd to:
American Mathematics Competitions
University of Nebraska, P.O. Box 81606, Lincoln, NE 68501-1606
Phone: 402-472-2257; Fax: 402-472-6087; email: The problems and solutions for this AMC 10 were prepared by the MAA’s Committee on the AMC 10 and AMC 12 under the direction of AMC 10 Subcommittee Chair:
Chair: LeRoy Wenstrom, Columbus, MS
Copyright © 2008, The Mathematical Association of America
1.Answer (E):The number of points could be any integer between 5·2=10and 5·3=15,inclusive.The number of possibilities is 15−10+1=6.
12348910111516171822
2324252.Answer (B):The two sums are 1+10+17+22=50and
4+9+16+25=54,so the positive difference between the sums
is 54−50=4.
Query:If a different 4×4block of dates had been chon,the
answer would be unchanged.Why?
3.Answer (D):The properties of exponents imply that
3
x √x = x ·x 12 13= x 32 13=x 12.4.Answer (C):A single player can receive the largest possible salary only when the other 20players on the team are each receiving the minimum salary of $15,000.Thus the maximum salary for any player is $700,000−20·$15,000=$400,000.
5.Answer (A):Note that (y −x )2=(x −y )2,so
(x −y )2$(y −x )2=(x −y )2$(x −y )2= (x −y )2−(x −y )2 2=02=0.
6.Answer (C):Becau AB +BD =AD and AB =4BD ,it follows that
BD =1
5·AD .By similar reasoning,CD =110·AD .Thus
BC =BD −CD =15·AD −110·AD =110
·AD.7.Answer (C):The side length of the large triangle is 10times the side length of each small triangle,so the area of the large triangle is 102=100times the area of each small triangle.
8.Answer (C):The total cost of the carnations must be an even number of dollars.The total number of dollars spent is the even number 50,so the number of ros purchad must also be even.In addition,the number of ros purchad cannot exceed 50
3.Therefore the number of ros purchad must be one of the
even integers between 0and 16,inclusive.This gives 9possibilities for the number of ros purchad,and conquently 9possibilities for the number of bouquets.
9.Answer (A):The quadratic formula implies that the two solutions are
x 1=2a +√4a 2−4ab 2a and x 2=2a −√4a 2−4ab 2a
,so the average is 12(x 1+x 2)=12
2a 2a +2a 2a =1.OR The sum of the solutions of a quadratic equation is the negative of the coefficient of the linear term divided by the coefficient of the quadratic term.In this ca the sum of the solution is −(−2a )a
=2.Hence the average of the solutions is
1.10.Answer (A):Let O be the center of the circle,
and let D be the interction of OC and AB .Be-cau OC bicts minor arc AB ,OD is a perpen-
dicular bictor of chord AB .Hence AD =3,and
applying the Pythagorean Theorem to ADO yields
OD =√52−33=4.Therefore DC =1,and apply-
ing the Pythagorean Theorem to ADC yields AC =√32+12=√10.11.Answer (B):Note that u 5=2u 4+9and 128=u 6=2u 5+u 4=5u 4+18.
Thus u 4=22,and it follows that u 5=2·22+9=53.
12.Answer (A):During the year Pete takes
44×105+5×104=44.5×105
steps.At 1800steps per mile,the number of miles Pete walks is
44.5×10518×10=44.518
×103≈2.5×103=2500.13.Answer (B):Becau the mean of the first n terms is n ,their sum is n 2.
Therefore the n th term is n 2−(n −1)2=2n −1,and the 2008th term is 2·2008−1=4015.
14.Answer (B):Becau OAB is a 30–60–90◦triangle,we have BA =
5√33.
Let A and B be the images of A and B ,respectively,under the rotation.Then
B =(0,5),B A is horizontal,and B A =BA=5√
3
3
.Hence A is in the
cond quadrant and
A =
−
5
3
√
3,5
.
15.Answer(A):By the Pythagorean Theorem we have a2+b2=(b+1)2,so
a2=(b+1)2−b2=2b+1.
Becau b is an integer with b<100,a2is an odd perfect square between1and 201,and there are six of the,namely,9,25,49,81,121,and169.Hence a must be3,5,7,9,11,or13,and there are6triangles that satisfy the given conditions.
16.Answer(A):If one die is rolled,3of the6possible numbers are odd.If two
dice are rolled,18of the36possible outcomes have odd sums.In each of the
cas,the probability of an odd sum is1
2.If no die is rolled,the sum is0,which
is not odd.The probability that no die is rolled is equal to the probability that
both coin toss are tails,which is(1
2)2=1
4
.Thus the requested probability is
1−1
4
·
1
2
=
3
8
.
17.Answer(B):The respons on the three occasions,in order,must be
YNN,NYN,or NNY,where Y indicates approval and N indicates disapproval.
The probability of each of the is(0.7)(0.3)(0.3)=0.063,so the requested probability is3(0.063)=0.189.
18.Answer(B):Let n be the number of bricks in the chimney.Then the
number of bricks per hour Brenda and Brandon can lay working alone is n
9and
n 10,respectively.Working together they can lay(n
9
+n
10
−10)bricks in an hour,
or
5 n
9
+
n
10
−10
bricks in5hours to complete the chimney.Thus
5 n
9
+
n
10
−10
=n,
and the number of bricks in the chimney is n=900.
OR
Suppo that Brenda can lay x bricks in an hour and Brandon can lay y bricks in an hour.Then the number of bricks in the chimney can be expresd as9x,
10y ,or 5(x +y −10).The equality of the expressions leads to the system of equations
4x −5y =−50
−5x +5y =−50.
It follows that x =100,so the number of bricks in the chimney is 9x =
900.
19.Answer (E):The portion of each end of the tank that
is under water is a circular ctor with two right triangles
removed as shown.The hypotenu of each triangle is 4,
and the vertical leg is 2,so each is a 30–60–90◦triangle.Therefore the ctor has a central angle of 120◦,and the
area of the ctor is 120360·π(4)2=163π.The area of each triangle is 12(2) 2√3 ,so the portion of each end that is underwater has area 163π−4√3.The length of the cylinder is 9,so the volume of the water is 9 163
π−4√3 =48π−36√3.20.Answer (B):Of the 36possible outcomes,the four pairs (1,4),(2,3),(2,3),
and (4,1)yield a sum of 5.The six pairs (1,6),(2,5),(2,5),(3,4),(3,4),and (4,3)yield a sum of 7.The four pairs (1,8),(3,6),(3,6),and (4,5)yield a sum of 9.Thus the probability of getting a sum of 5,7,or 9is (4+6+4)/36=7/18.Note:The dice described here are known as Sicherman dice.The probability of obtaining each sum between 2and 12is the same as that on a pair of standard dice.
21.Answer (C):Let the women be ated first.The first woman may sit in
any of the 10chairs.Becau men and women must alternate,the number of choices for the remaining women is 4,3,2,and 1.Thus the number of possible ating arrangements for the women is 10·4!=240.Without loss of generality,suppo that a woman sits in chair 1.Then this woman’s spou must sit in chair 4or chair 8.If he sits in chair 4then the women sitting in chairs 7,3,9,and 5must have their spous sitting in chairs 10,6,2,and 8,respectively.If he sits in chair 8then the women sitting in chairs 5,9,3,and 7must have their spous sitting in chairs 2,6,10,and 4,respectively.So for each possible ating arrangement for the women there are two arrangements for the men.Hence,there are 2·240=480possible ating arrangements.
22.Answer (C):There are 6!/(3!2!1!)=60distinguishable orders of the beads
on the line.To meet the required condition,the red beads must be placed in
