当前位置：搜档网 › 2008-AMC10-Bsolutions

2008-AMC10-Bsolutions

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 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 solu-tions 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. Dissemination 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 use 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 addressed to:

American Mathematics Competitions

University of Nebraska, P.O. Box 81606, Lincoln, NE 68501-1606

Phone: 402-472-2257; Fax: 402-472-6087; email: amcinfo@https://www.sodocs.net/doc/394919954.html, 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

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 di?erence between the sums

is 54?50=4.

Query:If a di?erent 4×4block of dates had been chosen,the

answer would be unchanged.Why?

3.Answer (D):The properties of exponents imply that

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):Because 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 roses purchased must also be even.In addition,the number of roses purchased cannot exceed 50

3.Therefore the number of roses purchased must be one of the

even integers between 0and 16,inclusive.This gives 9possibilities for the number of roses purchased,and consequently 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 coe?cient of the linear term divided by the coe?cient of the quadratic term.In this case 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 intersection of OC and AB .Be-cause OC bisects minor arc AB ,OD is a perpen-

dicular bisector 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):Because the mean of the ?rst 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):Because 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√

.Hence A is in the

second quadrant and

A =

√

3,5

15.Answer(A):By the Pythagorean Theorem we have a2+b2=(b+1)2,so

a2=(b+1)2?b2=2b+1.

Because b is an integer with b<100,a2is an odd perfect square between1and 201,and there are six of these,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 these

cases,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 tosses are tails,which is(1

2)2=1

.Thus the requested probability is

1?1

17.Answer(B):The responses on these three occasions,in order,must be

YNN,NYN,or NNY,where Y indicates approval and N indicates disapproval.

The probability of each of these 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

?10)bricks in an hour,

5 n

?10

bricks in5hours to complete the chimney.Thus

5 n

?10

=n,

and the number of bricks in the chimney is n=900.

Suppose 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 expressed as9x,

10y ,or 5(x +y ?10).The equality of these 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 sector with two right triangles

removed as shown.The hypotenuse of each triangle is 4,

and the vertical leg is 2,so each is a 30–60–90?triangle.Therefore the sector has a central angle of 120?,and the

area of the sector 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 seated ?rst.The ?rst woman may sit in

any of the 10chairs.Because men and women must alternate,the number of choices for the remaining women is 4,3,2,and 1.Thus the number of possible seating arrangements for the women is 10·4!=240.Without loss of generality,suppose that a woman sits in chair 1.Then this woman’s spouse 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 spouses 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 spouses sitting in chairs 2,6,10,and 4,respectively.So for each possible seating arrangement for the women there are two arrangements for the men.Hence,there are 2·240=480possible seating 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

one of four con?gurations:positions 1,3,and 5,positions 2,4,and 6,positions 1,3,and 6,or positions 1,4,and 6.In the ?rst two cases,the blue bead can be placed in any of the three remaining positions.In the last two cases,the blue bead can be placed in either of the two adjacent remaining positions.In each case,the placement of the white beads is then determined.Hence there are 2·3+2·2=10orders that meet the required condition,and the requested

probability is 1060=1

23.Answer (B):Because the area of the border is half the area of the ?oor,the

same is true of the painted rectangle.The painted rectangle measures a ?2by b ?2feet.Hence ab =2(a ?2)(b ?2),from which 0=ab ?4a ?4b +8.Add 8to each side of the equation to produce

8=ab ?4a ?4b +16=(a ?4)(b ?4).

Because the only integer factorizations of 8are

8=1·8=2·4=(?4)·(?2)=(?8)·(?1),

and because b >a >0,the only possible ordered pairs satisfying this equation for (a ?4,b ?4)are (1,8)and (2,4).Hence (a,b )must be one of the two ordered pairs (5,12),or (6,8).

A B C

D M A

B C D O 24.Answer (C):Let M be on the same side of line BC as

A ,such that BMC is equilateral.Then ABM and

MCD are isosceles with ∠ABM =10?and ∠MCD =

110?.Hence ∠AMB =85?and ∠CMD =35?.There-

fore

∠AMD =360??∠AMB ?∠BMC ?∠CMD

=360??85??60??35?=180?.It follows that M lies on AD and ∠BAD =∠BAM =

85?.

Let ABO be equilateral as shown.

Then

∠OBC =∠ABC ?∠ABO =70??60?=10?.

Because ∠BCD =170?and OB =BC =CD ,

the quadrilateral BCDO is a parallelogram.Thus

OD=BC=AO and AOD is isosceles.Letα=∠ODA=∠OAD.The sum of the interior angles of ABCD is360?,so we have

360=(α+60)+70+170+(α+10)andα=25.

Thus∠DAB=60+α=85?.

25.Answer(B):Number the pails consecutively so that Michael is presently at

pail0and the garbage truck is at pail1.Michael takes200/5=40seconds to walk between pails,so for n≥0he passes pail n after40n seconds.The truck takes20seconds to travel between pails and stops for30seconds at each pail.

Thus for n≥1it leaves pail n after50(n?1)seconds,and for n≥2it arrives at pail n after50(n?1)?30seconds.Michael will meet the truck at pail n if and only if

50(n?1)?30≤40n≤50(n?1)or,equivalently,5≤n≤8.

50time (sec)

distance (ft)

200truck

Michael

Hence Michael?rst meets the truck at pail5after200seconds,just as the truck leaves the pail.He passes the truck at pail6after240seconds and at pail7 after280seconds.Finally,Michael meets the truck just as it arrives at pail8 after320seconds.These conditions imply that the truck is ahead of Michael between pails5and6and that Michael is ahead of the truck between pails7 and8.However,the truck must pass Michael at some point between pails6and 7,so they meet a total of?ve times.

The

American Mathematics Competitions

are Sponsored by

The Mathematical Association of America

The Akamai Foundation

Contributors

American Mathematical Association of Two Year Colleges

American Mathematical Society

American Society of Pension Actuaries

American Statistical Association

Art of Problem Solving

Awesome Math

Canada/USA Mathcamp

Canada/USA Mathpath

Casualty Actuarial Society

Clay Mathematics Institute

IDEA Math

Institute for Operations Research and the Management Sciences

L. G. Balfour Company

Mu Alpha Theta

National Assessment & Testing

National Council of Teachers of Mathematics

Pedagoguery Software Inc.

Pi Mu Epsilon

Society of Actuaries

U.S.A. Math Talent Search

W. H. Freeman and Company

Wolfram Research Inc.

2008-AMC10-Bsolutions

相关文档

最新文档