132 Contest 2015 Solutions

THE UNIVERSITY OF VERMONT

DEPARTMENT OF MATHEMATICS AND STATISTICS FIFTY-EIGHTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION

MARCH 11, 2015

as a rational number in lowest terms.

2) Teresa bought a toy marked 1₄ off the original price. If Teresa paid $60, what was the original price of the toy, in dollars? point of intersection of linesACandDB.If the degree measures of anglesCABandCEBare 12 and 36 respectively, what is the

degree measure of angleDBA? 12

36

From right triangles ABD and BDC, y

2 _h2 _x2

h2 152 92 y

2 ₁₅2 ₉2 _x2

From right triangle ADC, 9 x2 152 y2

Combining: 92 18x x2 152 152 92 x2 18x 2152 92

x 212 2_

18 x 16 AC 9 16 25

6) Express 2 5 4 2 – 5 4 as an integer.

2 5 4 2 – 5 4 2 5 2 2 2 – 5 2 2

2 5 4 2 – 5 4 4 4 5 5 2 4 –4 5 5 2

2 5 4 2 – 5 4 9 4 5 2 9 –4 5 2

2 5 4 2 – 5 4 81 72 5 80 81 72 5 80 322

7) Adam can dig a hole in 2 hours. Ben can dig the same hole in 3 hours. How many hours would it take them to dig the hole if they work together ? Express your answer as a rational number in lowest terms.

Rate for Adam 1₂, rate for Ben 1₃. Hours to together h.

1₂h 1₃h 1 ₆ 5h 1 h 6

5hours.

8) Express 3 5 17 257 1 1 4 as an integer.

3 5 17 257 1 1 4 22 1 22 1 24 1 28 1 11 4 3 5 17 257 1 1 4 _24 _{1 2}4 _{1 2}8 _{1 1}1 4

3 5 17 257 1 1 4 28 1 28 1 11 4 3 5 17 257 1 1 4 216 1 11 4 3 5 17 257 1 1 4 _216_1 4 ₂4 ₁₆

9) Find the sum of the prime factors of 2015.

2015 5 13 31

5 13 31 49

10) Find the degree measure of the angle whose complement is 2₇ of its supplement.

90 Θ 2₇ 180 Θ 630 7Θ 360 2Θ 270 5Θ Θ 54

11) In a group of 4 people, what is the probability that at least 2 of them were born on the same day of the week? Express your answer as a rational number in lowest terms.

number of 4-tuples with no repeats 7 6 5 4

12) Three numbers form a geometric progression. Their sum is 19₂ and the sum of their

reciprocals is 19₁₈. Find these three numbers.

13 Two circles are concentric, as shown. Chord ABof the larger circle is trisected by the smaller circle so thatAC CD DB 1 . The sum of the radii of the larger and smaller circles equals the length

of the chordAB. Find the radius of the larger circle.

A C D 1 B

For their regular walk d r t

The sister travels 15₆₀r t2₂r t3r d

The time the sister travels is ₆₀15 t2 t3 t ₆₀6 t ₆₀9 t2 t3

Using d r t 15₆₀r t2₂r t3r r₆₀9 t2 t3 1₂t2 ₁₀1 t2 1₅hour 12 minutes.

15) Let y be the real number such that 2015y _y20153 .

Find the value of log₂₀₁₅ log₂₀₁₅ y – log₂₀₁₅ y. Express your answer in simplest form.

2015y _y20153

y log₂₀₁₅y20153 20153log₂₀₁₅y log₂₀₁₅y y

20153

log₂₀₁₅ log₂₀₁₅y – log₂₀₁₅ y log₂₀₁₅ y

20153 log2015y

log₂₀₁₅ log₂₀₁₅y – log₂₀₁₅ y log₂₀₁₅ y log₂₀₁₅20153_{ log}

2015y 3

16 The radius of the smaller of the two concentric circles is one meter. The line segments joining the circles consist of the portions of radii of the larger circle that lie outside the smaller circle. These line segments and the smaller circle partition the larger circle into nine sections, each of which has the same area. Find the length of one of the line segments.

The central angle between two of the radii of the larger circle is 2₈Π Π₄

The area of the smaller circle equals the area of one of the sectors. The area of the smaller circle is Π 12 Π

If the length of one of the segments is r, then Π 1₂ Π₄ 1 r 2 1₂ Π₄ 1

Π 1₂ Π₄ 1 r 2 1₂ Π₄ 1 r2 2r 8 0 r 2 r 4 0 r 2 orr 4. Since the length must be positive, r 2 meters

17 Four circles of radius 2 are pairwise tangent as shown in the figure. A fifth circle of radius 2 is drawn so as to pass through the common points of tangency. Find the total area of the region

B A A

Area of triangle 1₂ 2 2 B A A B 2 B 2 A

Area of sector 1₂Π₂ 22 B A A 2A B Π Combining: Π 2A 2 A A Π 2

Area 8A 8 Π 2

18) Solve the equation log

x 5 logx4 16 1

2 for x. Express your answer as a rational number in

lowest terms.

logxay z xa z y xaz y az log_xy log_xay 1

alogxy

Using the above:

log

x 5 logx4 16 1

2 2 logx5 1

4logx 16 1 2

2 logx 5 1₄logx 16 1₂ logx 25 logx161 4 1₂

log_x 50 1₂ x1 2 50 x 2500

19) Find the sum of all of the real solutions to the equation x x– 6 7.

x 6 6 x x 6

x x 6 7 x 6 x 7 x x 6 7

6x x2 _{7 x}2 _{6x 7}

x2 _{6x 7 0 x}2 _{6x 7 0}

x 6 36 28 2 x 7 x 1 0

x 3 2 x 7

6

Sum = 3 2 3 2 7 13

20) Find the area of the region in the plane that simultaneously satisfies the inequalities x2 _y2 _{6x– 10y 30 andy 5 – x 3 .}

x2 y2 6x– 10y 30 x 3 2 y 52 64

x 3 x 3 x 3

y 5 x 3 y x 8 y 2 x

10 5 5

5 10

A 1₂3₂Π 82_{ 48Π}

21 A gameboard in the shape of an equilateral triangle is partitioned into 25 congruent equilateral triangular regions; these are numbered 1 to 25, as shown in the figure. A blue checker is placed on one of the regions and a red checker is placed on a different region. How many ways can this be done so that the two checkers are not in adjacent regions ? Two regions are adjacent if they share a common edge.

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21

22 23 24 25

Place the blue checker on any of the 25 triangular regions and count the number of non-adjacent regions on which the red checker can be placed. These counts are undicated below.

The total number of possible ways is the sum of these 25 numbers.

23 21 22 21 22 21 22 21 23

22 21 21 21 21 21 22

22 21 21 21 22

22 21 22 23

Total 3 23 9 22 13 21 540

22) If k is a non-negative integer, define k as the sum of the cubes of the digits of k. For example, 889 83 ₈3 ₉3 _{512 512 729 1753. The sequence a}

n is defined by

a1 43 333 43 433 172

a2 172 13 73 23 352

a3 352 33 53 23 160

a4 160 13 63 03 217

a5 217 23 13 73 352

a6 352 33 53 23 160

Thus a3k 160, a3k 1 217 and a3k 2 352

a2015 a3 671 2 352

23 How many different paths are there from the point labeled START at the bottom of the figure to the point labeled END at the top of the figure, travelling diagonally upward to the left, diagonally upward to the right or straight up along lines in the figure ?

START END

START END

1 1

1 3 1

1 5 5 1

1 6 10 6 1

8 21 21 8

29 42 29

92 92

226

Number of paths = 226

24) Every day a crossword puzzle is placed on Jenny s desk. While Jenny is on vacation, the puzzles accumulate on her desk. When she returns from vacation, Jenny begins to solve the accumulated puzzles as well as the new puzzles that appear each day. Jenny determines that if she solves exactly four puzzles a day, she will completely catch up with her puzzle solving ten days sooner than she would if she solves exactly three puzzles each day. How long was Jenny on vacation?

28 A circle of radius 25 passes through two adjacent vertices of a square and is tangent to the side opposite the side joining the adjacent vertices. What is the length of one side of the square ?

s 8₅ 25 40

30) Find the smallest positive integer n such that n is divisible by 101000.

Each factor of 5 in n will produce a 10 in n . Hence n must be a multiple of 5

31 Find the length of either tangent line from the origin to the circle

x2 _y2_{– 6x– 8y 21 0 .}

33) Suppose that x, yandz are positive real numbers such that x y z 85, y z x 120 andz x y 105.

cos2k _{Φ 7, determine the value of cos 2Φ. Express your answer as a rational number in lowest terms.}

!

35 Circles with centersCandDhave respective radii 3 and 10. A common tangent intersects the circles atEandF,

C P _D

36) Find the smallest positive integer n such that ₆n– 17

n 11 is a positive rational number that is not in lowest terms.

Reducible there exist integer p 1 such that p n 17 andp 6n 11

p n 17 andp 6n 11 p 6n 11 6 n 17

p 11 6 17 p 113 Prime

n 17 113 6n 11 791 113₇₉₁ 1₇ n 130

Let y k 2013 y2 y 1 100 y3 y2 100 0 y 5 y2 4y 20 0 y 5 and 2 4i

y 5 k 2013 5 k 2018

39) Let S be the set of all 8-digit positive integers obtained by rearranging the digits of 12345678. For example, 13578642 and 78651234 are elements of S. How many elements of S are divisible by 11 ?

The integer n is divisible by 11 ' the sum of the even position digits = sum of the odd position digits.

1 2 3 4 5 6 7 8 36 so the (even)(odd) sums must be 18.

1,2,7,8 3,4,5,6

1,3,6,8 2,4,5,7

1,4,6,7 2,3,5,8

1,4,5,8 2,3,6,7

Number = 8 ways to pick (evens) * 4! * 4! ways to order = 8(24)(24) = 4608

40) Find the smallest positive real number x such that x2 – x,x- = 6, where ,t- is the largest integer less than or equal to t. For example, ,3.25- 3 and ,13- 13.

Let x n swhere nisapositive integer and 0 s 1.

x2–x,x- 6  n s2–n s n 6 n2 2ns s2–n2 ns 6 Since n2 _{is an integer the equation reduces to 2ns s}2_{–ns 6}

This implies that ns is an integer, hence 2ns is an integer and the equation reduces to ns 6 s 6_n. SInce s is less than one the smallest value for n is 7.

x 7 6₇ 55 7

41) A fair coin is flipped 10 times. What is the probability that, for every time the coin lands heads, either the flip immediately before it was heads or the flip immediately after it was heads (or both)? For example, the sequence of flips HHTHHHHTTT has the desired property, but HTHHTTTHHH does not. Express your answer as a rational number in lowest terms.

Let an be the number of sequences of length n that satisfy the given condition.

The any sequence of length n can end in a

T appended to any sequence counted by an 1 HH appended to any sequence counted by an 2 THHH appended to any sequence counted by an 4

Thus an an 1 an 2 an 4

a1 1 T

a2 2 TT, HH

a3 4 HHH, HHT, THH, TTT

a4 7 HHHH, HHHT, THHH, HHTT, THHT, TTHH, TTTT

a5 a4 a3 a1 7 4 1 12

6 5 4 2

a7 a6 a5 a3 21 12 4 37

a8 a7 a6 a4 37 21 7 65

a9 a8 a7 a5 65 37 12 114

a10 a9 a8 a6 114 65 21 200

Probability = 200₂10

200 1024