Problem 1 Problem 2 - Purple Comet

PURPLE COMET! MATH MEET April 2018

MIDDLE SCHOOL - SOLUTIONS

Copyright c Titu Andreescu and Jonathan Kane

Problem 1

Find n such that the mean of 74 , 65 , and

1

n

is 1.

Answer: 20

For the mean of three numbers to be 1, their sum must be 3. So,

7

4

+

6

5

+

1

n

= 3. Multiplying by 20n yields

35n + 24n + 20 = 60n, and, therefore, n = 20.

Problem 2

The following figure is made up of many 2 �� 4 tiles such that adjacent tiles always share an edge of length

2. Find the perimeter of this figure.

Answer: 148

The figure is made up of 18 tiles that share 17 edges. Each tile has perimeter 2 �� 2 + 2 �� 4 = 12, and each

shared edge has length 2. It follows that the perimeter of the figure is 18 �� 12 ? 17 �� 2 �� 2 = 148.

Problem

3



1 +1

1+ 1

3

3

3 + 3

The fraction 

1

3+1

3

+



 can be written as

m

n,

where m and n are relatively prime positive integers.

1

1+3

Find m + n.

1

Answer: 31

1



+1

1+ 1

3

3

3 + 3





=

3

1 + 1

3+1

1+3

4

3

3

+

4

3

3

3

1+1

4

4

=

8

9

3

1

2

=

8

9

6

=

4

27 . The requested sum is 4 + 27 = 31.

Problem 4

The diagram below shows a large square with each of its sides divided into four equal segments. The

shaded square whose sides are diagonals drawn to these division points has area 13. Find the area of the

large square.

Answer: 104

There are several ways to see that the shaded square has

1

8

the area of the large square. One way is that

the large square is made up of 5 small squares 4 large triangles, and 4 small triangles. The area of any 2 of

the large triangles and the area of the 4 small triangles is equal to the area of a small square. Thus, the

area of the large square is made up of the area of 5 +

4

2

+

4

4

= 8 small squares. Hence, the requested area of

the large square is 8 �� 13 = 104.

Problem 5

The positive integer m is a multiple of 101, and the positive integer n is a multiple of 63. Their sum is

2018. Find m ? n.

Answer: 2

Note that 2020 = 20 �� 101 and 2016 = 32 �� 63. This shows that 20 �� 101 + 32 �� 63 = 2 �� 2018, so

m = 10 �� 101 = 1010 and n = 16 �� 63 = 1008. The requested difference is 1010 ? 1008 = 2.

Problem 6

5

Find the greatest integer n such that 10n divides

10

210 52

.

1052

Answer: 999

The fraction is

the fraction is

210,000 51024

,

1025

101024

1025

so the largest power of 10 in the numerator is 101024 and the largest power in

= 101024?25 = 10999 . The greatest n is 999.

2

Problem 7

Bradley is driving at a constant speed. When he passes his school, he notices that in 20 minutes he will be

exactly

1

4

of the way to his destination, and in 45 minutes he will be exactly

1

3

of the way to his

destination. Find the number of minutes it takes Bradley to reach his destination from the point where he

passes his school.

Answer: 245

Because

1

3

?

1

4

=

1

12

and 45 ? 20 = 25, it will take Bradley 25 minutes to drive

1

12

of the way to his

destination; that is, it will take him a total of 12 �� 25 = 300 minutes to complete the entire drive to his

destination. In 20 minutes he will have

20 +

3

4

3

4

of the drive remaining, so the number of minutes remaining is

�� 300 = 245.

Problem 8

On side AE of regular pentagon ABCDE there is an equilateral triangle AEF , and on side AB of the

pentagon there is a square ABHG as shown. Find the degree measure of angle AF G.

Answer: 39

The internal angles of a regular pentagon are

3��180?

5

= 108? . Thus,

��GAF = 360? ? (��GAB + ��BAE + ��EAF ) = 360? ? (90? + 108? + 60? ) = 102? . Triangle AF G is

isosceles with AF = AG, so ��AF G = 12 (180? ? ��GAF ) = 39? .

Problem 9

For some k > 0 the lines 50x + ky = 1240 and ky = 8x + 544 intersect at right angles at the point (m, n).

Find m + n.

3

Answer: 44

The slope of the first line is ? 50

k , and the slope of the second line is

the product of their slopes must be ?1, so

50

k

��

8

k

8

k.

For the lines to be perpendicular,

= 1, and k = 20. The two equations are 50x + 20y = 1240

and 20y ? 8x = 544. Subtracting the second equation from the first yields 58x = 696, so m = 12. Then

20n = 8m + 544 = 8 �� 12 + 544 = 640, so n = 32. The requested sum is 12 + 32 = 44.

Problem 10

The triangle below is divided into nine stripes of equal width each parallel to the base of the triangle. The

darkened stripes have a total area of 135. Find the total area of the light colored stripes.

Answer: 108

The diagram shows nine overlapping triangles that share the same top vertex with all of their bases

parallel. These nine triangles are similar to each other, so their areas are proportional to the squares of

their heights. Suppose the smallest triangle has area A. Then the other triangles have area 22 A, 32 A, 42 A,

52 A, 62 A, 72 A, 82 A, and 92 A. Thus, the area of the dark stripes is




A 1 + 32 ? 22 + 52 ? 42 + 72 ? 62 + 92 ? 82 = 45A. Similarly, the area of the light colored stripes is




A 22 ? 1 + 42 ? 32 + 62 ? 52 + 82 ? 72 = 36A. Thus, the ratio of the area of the dark stripes to the area

of the light stripes is

45

36

= 54 , and the total area of the light colored stripes is 135 ��

4

5

= 108.

Problem 11

Find the number of positive integers less than 2018 that are divisible by 6 but are not divisible by at least

one of the numbers 4 or 9.

Answer: 112

There are b 2018

6 c = 336 positive integers less than 2018 that are divisible by 6. Let A be the set of positive

integers less than 2018 that are divisible by 6 �� 2 = 12 and B be the set of positive integers less than 2018

336

336

that are divisible by 6 �� 3 = 18. Then |A �� B| = |A| + |B| ? |A �� B| = b 336

2 c+b 3 c?b 6 c=

168 + 112 ? 56 = 224. Thus, the number of positive integers less than 2018 divisible by 6 but not divisible

by 4 or 9 is 336 ? 224 = 112.

4

Problem 12

Line segment AB has perpendicular bisector CD, where C is the midpoint of AB. The segments have

lengths AB = 72 and CD = 60. Let R be the set of points P that are midpoints of line segments XY ,

where X lies on AB and Y lies on CD. Find the area of the region R.

Answer: 1080

Let X be a point on AB. Then as Y ranges over CD the midpoints of XY form a line segment

perpendicular to AB half way between X and CD and half its length. It follows that R is a rectangle with

sides parallel to AB and CD. The rectangle has length 36 and width 30, so its area is 36 �� 30 = 1080.

Problem 13

Suppose x and y are nonzero real numbers simultaneously satisfying the equations

x+

2018

= 1000 and

y

9

+ y = 1.

x

Find the maximum possible value of x + 1000y.

Answer: 1991

Multiply the first equation by y and the second equation by x to obtain xy + 2018 = 1000y and 9 + xy = x.

Subtraction yields 2009 = 1000y ? x. Solving this for y and substituting into 9 + xy = x yields

x2 + 1009x + 9000 = 0, which factors as (x + 9)(x + 1000) = 0. This gives the two possible solutions




(x, y) = (?9, 2) and (x, y) = ?1000, 1009

1000 . The requested maximum is ?9 + 1000 �� 2 = 1991.

Problem 14

Find the number of ordered quadruples of positive integers (a, b, c, d) such that ab + cd = 10.

Answer: 58

There are 9 ordered pairs of positive integers (m, n) such that m + n = 10. For each such ordered pair,

there are quadruples of positive integers (a, b, c, d) where ab = m and cd = n. Clearly, there are as many

pairs of (a, b) with ab = m as there are positive integer divisors of m, and, thus, for each pair (m, n) with

m + n = 10, the number of quadruples (a, b, c, d) with ab + cd = 10 and ab = m is equal to the product of

the number of positive integer divisors of m and the number of positive integer divisors of n. The number

of positive integer divisors of the positive integers form 1 to 9 are given in the following table.

m

1

2

3

4

5

6

7

8

9

number of divisors

1

2

2

3

2

4

2

4

3

The number of quadruples is, therefore, 1 �� 3 + 2 �� 4 + 2 �� 2 + 3 �� 4 + 2 �� 2 + 4 �� 3 + 2 �� 2 + 4 �� 2 + 3 �� 1 = 58.

Problem 15

There are integers a1 , a2 , a3 , . . . , a240 such that x(x + 1)(x + 2)(x + 3) �� �� �� (x + 239) =

240

P

n=1

number of integers k with 1 �� k �� 240 such that ak is a multiple of 3.

5

an xn . Find the

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