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2006AMC10-Bsolutions

2006AMC10-Bsolutions
2006AMC10-Bsolutions

The MATHEMATICAL ASSOCIATION OF AMERICA

american Mathematics Competitions

7th Annual American Mathematics Contest 10 aMC 10 – Contest b

Solutions Pamphlet

Wednesday, February 15, 2006

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 at any time via copier, telephone, email, the World Wide Web or media of any type is a violation of the competition rules.

Correspondence about the problems and solutions for this

AMC 10 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/a217949475.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:

Prof. Douglas Faires, Department of Mathematics

Youngstown State University, Youngstown, OH 44555-0001

Copyright ? 2006, The Mathematical Association of America

1.(C)Because

(?1)k=

1,if k is even,?1,if k is odd,

the sum can be written as

(?1+1)+(?1+1)+···+(?1+1)=0+0+···+0=0.

2.(A)Because4?5=(4+5)(4?5)=?9,it follows that

3?(4?5)=3?(?9)=(3+(?9))(3?(?9))=(?6)(12)=?72.

3.(A)Let c and p represent the number of points scored by the Cougars and the

Panthers,respectively.The two teams scored a total of34points,so c+p=34.

The Cougars won by14points,so c?p=14.The solution is c=24and p=10, so the Panthers scored10points.

4.(D)The circle with diameter3has areaπ

3

2

2

.The circle with diameter1has

areaπ

1

2

2

.Therefore the ratio of the blue-painted area to the red-painted area

is

π

3

2

2

1

2

2

π

1

2

2=8.

5.(B)The side length of the square is at least equal to the sum of the smaller

dimensions of the rectangles,which is2+3=5.

4

3

5

2

3

If the rectangles are placed as shown,it is in fact possible to contain them within

a square of side length5.Thus the smallest possible area is52=25.

6.(D)Since the square has side length2/π,the diameter of each circular section is

2/π.The boundary of the region consists of4semicircles,whose total perimeter is twice the circumference of a circle having diameter2/π.Hence the perimeter

of the region is

π·

2

π

=4.

7.(A)We have

x

1?x?1

x =

x

x?x+1

x

=

x

1

x

=

x2=|x|.

When x<0,the given expression is equivalent to?x.

8.(B)The square has side length

40.

Let r be the radius of the semicircle.Then

r2= √

40

2

+

40

2

2

=40+10=50,

so the area of the semicircle is1

2

πr2=25π.

9.(B)Francesca’s600grams of lemonade contains25+386=411calories,so200

grams of her lemonade contains411/3=137calories.

10.(A)Let the sides of the triangle have lengths x,3x,and15.The Triangle

Inequality implies that3x

11.(C)Since n!contains the product2·5·10=100whenever n≥10,it su?ces

to determine the tens digit of

7!+8!+9!=7!(1+8+8·9)=5040(1+8+72)=5040·81.

This is the same as the units digit of4·1,which is4.

12.(E)Substituting x=1and y=2into the equations gives

1=2

4

+a and2=

1

4

+b.

It follows that

a+b=

1?

2

4

+

2?

1

4

=3?

3

4

=

9

4

.

OR

Because

a=x?y

4

and b=y?

x

4

we have a+b=

3

4

(x+y).

Since x=1when y=2,this implies that a+b=3

4(1+2)=9

4

.

13.(E)Joe has 2ounces of cream in his cup.JoAnn has drunk 2ounces of the 14

ounces of co?ee-cream mixture in her cup,so she has only 12/14=6/7of her 2ounces of cream in her cup.Therefore the ratio of the amount of cream in Joe’s co?ee to that in JoAnn’s co?ee is 267

·2=76.14.(D)Since a and b are roots of x 2?mx +2=0,we have

x 2?mx +2=(x ?a )(x ?b )and ab =2.

In a similar manner,the constant term of x 2?px +q is the product of a +(1/b )and b +(1/a ),so

q = a +1b b +1a =ab +1+1+1ab =92

.15.(C)Since ∠BAD =60?,isosceles BAD is also equilateral.As a consequence,

AEB , AED , BED , BF D , BF C ,and CF D are congruent.These six triangles have equal areas and their union forms rhombus ABCD ,so each has area 24/6=4.Rhombus BF DE is the union of BED and BF D ,so its area is 8.A

F

E

D

C B

OR Let the diagonals of rhombus ABCD intersect at O .Since the diagonals of a

rhombus intersect at right angles, ABO is a 30–60–90?triangle.Therefore

AO =√3·BO .Because AO and BO are half the length of the longer diagonals of rhombi ABCD and BF DE ,respectively,it follows that

Area(BF DE )Area(ABCD )= BO AO 2=13

.Thus the area of rhombus BF DE is (1/3)(24)=8.

A

F

O

E

D C

B

16.(E)In the years from2004through2020,Each Leap Day occurs3·365+

366=1461days after the preceding Leap Day.When1461is divided by7the remainder is5.So the day of the week advances5days for each4-year cycle.In the four cycles from2004to2020,the Leap Day will advance20days.So Leap Day in2020will occur one day of the week earlier than in2004,that is,on a Saturday.

17.(D)After Alice puts the ball into Bob’s bag,his bag will contain six balls:two

of one color and one of each of the other colors.After Bob selects a ball and places it into Alice’s bag,the two bags will have the same contents if and only if Bob picked one of the two balls in his bag that are the same color.Because there are six balls in the bag when Bob makes his selection,the probability of selecting one of the same colored pair is2/6=1/3.

18.(E)Note that the?rst several terms of the sequence are:

2,3,3

2

,

1

2

,

1

3

,

2

3

,2,3,...,

so the sequence consists of a repeating cycle of6terms.Since2006=334·6+2, we have a2006=a2=3.

19.(A)Since OC=1and OE=2,it follows that∠EOC=60?and∠EOA=30?.

The area of the shaded region is the area of the30?sector DOE minus the area of congruent triangles OBD and OBE.First note that

Area(Sector DOE)=1

12

(4π)=

π

3

.

In right triangle OCE,we have CE=√

3,so BE=

3?1.Therefore

Area( OBE)=1

2

(

3?1)(1).

The required area is consequently

π3?2

3?1

2

=

π

3

+1?

3.

OR

Let F be the point where ray OA intersects the circle,and let G be the point where ray OC intersects the

circle.

Let a be the area of the shaded region described in the problem,and b be the area of the region bounded by AD,AF,and the minor arc from D to F.Then b is also the area of the region bounded by CE,CG,and the minor arc from G to E.By the Inclusion-Exclusion Principle,

2b?a=Area(Quartercircle OF G)?Area(Square OABC)=π?1. Since b is the area of a60?sector from which the area of OAD has been deleted,we have

b=2π

3

?

3

2

.

Hence the area of the shaded region described in the problem is

a=2b?π+1=2

3

?

3

2

?π+1=

π

3

+1?

3.

20.(E)The slope of line AB is(178?(?22))/(2006?6)=1/10.Since the line

AD is perpendicular to the line AB,its slope is?10.This implies that

?10=y?(?22)

8?6

,so y=?10(2)?22=?42,and D=(8,?42).

As a consequence,

AB=

20002+2002=200

101and AD=

22+202=2

101.

Thus

Area(ABCD)=AB·AD=400·101=40,400.

21.(C)On each die the probability of rolling k,for1≤k≤6,is

k

1+2+3+4+5+6=

k

21

.

There are six ways of rolling a total of7on the two dice,represented by the ordered pairs(1,6),(2,5),(3,4),(4,3),(5,2),and(6,1).Thus the probability of rolling a total of7is

1·6+2·5+3·4+4·3+5·2+6·1

212=

56

212

=

8

63

.

22.(D)The total cost of the peanut butter and jam is N(4B+5J)=253cents,so

N and4B+5J are factors of253=11·23.Because N>1,the possible values of N are11,23,and253.If N=253,then4B+5J=1,which is impossible since B and J are positive integers.If N=23,then4B+5J=11,which also has no solutions in positive integers.Hence N=11and4B+5J=23,which has the unique positive integer solution B=2and J=3.So the cost of the jam is11(3)(5¢)=$1.65.

23.(D)Partition the quadrilateral into two triangles and let the areas of the trian-

gles be R and S as shown.Then the required area is T=R+S.

3

77

R S

a

b

Let a and b,respectively,be the bases of the triangles with areas R and3,as indicated.If two triangles have the same altitude,then the ratio of their areas is the same as the ratio of their bases.Thus

a b =

R

3

=

R+S+7

3+7

,so

R

3

=

T+7

10

.

Similarly,

S 7=

S+R+3

7+7

,so

S

7

=

T+3

14

.

Thus

T=R+S=3

T+7

10

+7

T+3

14

.

From this we obtain

10T=3(T+7)+5(T+3)=8T+36, and it follows that T=18.

24.(B)Through O draw a line parallel to AD intersecting P D at F .

Then AOF D is a rectangle and OP F is a right triangle.Thus DF =2,F P =2,and OF =4√2.The area of

trapezoid AOP D is 12√2,and the area of hexagon AOBCP D is 2·12√2=24√2.

OR

Lines AD,BC,and OP intersect at a common point H

.Because ∠P DH =∠OAH =90?,triangles P DH and OAH are similar with ratio of similarity 2.Thus 2HO =HP =HO +OP =HO +6,so HO =6

and AH =√HO 2?OA 2=4√2.Hence the area of OAH is (1/2)(2)(4√2)=4√2,and the area of P DH is (22)(4√2)=16√2.The area of the hexagon is twice the area of P DH minus twice the area of OAH ,so it is 24√2.

25.(B)The 4-digit number on the license plate has the form aabb or abab or baab ,

where a and b are distinct integers from 0to 9.Because Mr.Jones has a child of age 9,the number on the license plate is divisible by 9.Hence the sum of the digits,2(a +b ),is also divisible by 9.Because of the restriction on the digits a and b ,this implies that a +b =9.Moreover,since Mr.Jones must have either a 4-year-old or an 8-year-old,the license plate number is divisible by 4.These conditions narrow the possibilities for the number to 1188,2772,3636,5544,6336,7272,and 9900.The last two digits of 9900could not yield Mr.Jones’s age,and none of the others is divisible by 5,so he does not have a 5-year-old.Note that 5544is divisible by each of the other eight non-zero digits.

The

American Mathematics Competitions

are Sponsored by

The Mathematical Association of America

University of Nebraska-Lincoln

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

Canada/USA Mathcamp

Canada/USA Mathpath

Casualty Actuarial Society

Clay Mathematics Institute

Institute for Operations Research and the Management Sciences

L. G. Balfour Company

Mu Alpha Theta

National Council of Teachers of Mathematics

National Assessment & Testing

Pedagoguery Software Inc.

Pi Mu Epsilon

Society of Actuaries

U.S.A. Math Talent Search

W. H. Freeman and Company

Wolfram Research Inc.

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