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THE UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-EIGHTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 11, 2015
3
1) Express
4
5
6
3
4
5
6
–
–
–
–
4
3
6
as a rational number in lowest terms.
5
4
60
60
3
6
35
45 80
50 72
22
5
2) Teresa bought a toy marked
1
4
off the original price. If Teresa paid $60, what was the original price of the toy,
in dollars?
Let x be the original price.
3
4
x 60
x
$80
x
4
3
60 80
3) If 45% of x is 300, what is 75% of x?
.45 x 300
.45
.75
.45
x
.75
.45
300
.75 x
15
9
300 500
4 In the figure A, B, C and D are points on the circle and E is the
point of intersection of lines AC and DB. If the degree measures
of angles CAB and CEB are 12 and 36 respectively, what is the
degree measure of angle DBA ?
180
36
12
DBA 180
D
E
C
36
12
A
B
DBA 24
5 In right triangle ACD , AD is perpendicular to DC, AD
perpendicular to AC. If AB 9, find AC.
15 and DB is
D
h
x
15
9
A
C
y
B
y2 h2 x2
h2 152 92
From right triangles ABD and BDC,
From right triangle ADC, 9 x
Combining: 92
2 122 
18
x
18 x x2
x 16
2
152
5
2
5
2
5
2
5
92
x2
92
18 x 2 152
x2
92 
AC 9 16 25
4
5
2–
4
2
152
y2
152
4
6) Express 2
152
y2
5
as an integer.
2
5
2 2
4
2–
5
2–
5
2–
5
2–
5
4
4
4
5
2
4 4
5
9 4
5
2
5
4
4
2 2
2–
4 –4
5
2
5
2
9 –4
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.
1
,
2
Rate for Adam
1
2
h
1
3
1
h
5
6
h 1
3 5 17 257 1
14
3 5 17 257 1
14
3 5 17 257 1
14
3 5 17 257 1
14
3 5 17 257 1
14
6
h
3 5 17 257 1
8) Express
1
.
3
rate for Ben
22
24
28
216
216 
5
14
h.
hours.
as an integer.
1 22
1 24
1 24
1 28
1 28
1 1
1 1
14
Hours to together
24
1 28
1 1
1 1
14
14
14
14
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
90
630
270
Θ
2
7
180
7 Θ 360
5Θ
Θ
2
7
of its supplement.
Θ
2Θ
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.
Total number of 4-tuples of days
74 .
number of 4-tuples with no repeats
7 6 5 4
number of 4-tuples with at least one match = total nimus no match
74 7 6 5 4
74
p
73 6 5 4
73
223
343 120
343
343
19
2
12) Three numbers form a geometric progression. Their sum is
19
.
18
reciprocals is
and the sum of their
Find these three numbers.
Let a and r be the first term and common ratio.
1
a
19
2
19
18
ar 2
a ar
1
ar
1
ar 2
Multiplying the second equation by a2 r 2 :
19
18
ar 2
r
ar
3
a2
13 a 18 0
1
r
19
2
a2 r 2
3
a
3
a aa
19
18
ar 2
a2 r 2
9
a a2
2a 9 a 2
a2 r 2
19
2
r
1
9
2 a2
0
19
18
a2 r 2
ar
3
13 a 18 0
9
2
a
or a 2
2
3
r
or r
3
2
In both cases the set of three numbers is  2 , 2, 3
9
3 leads to the equation 2 a2
Using ar
25 a 18 0 whose solutions are not integers.
9
Thus the solution is 2, 3, 
2
13 Two circles are concentric, as shown. Chord AB of the larger circle
is trisected by the smaller circle so that AC CD DB 1 . The
sum of the radii of the larger and smaller circles equals the length
of the chord AB . Find the radius of the larger circle.
r
R 3
r
2
R2
h2
2
r2
R2
9
4
R2
9 6 R R2  2
1 2
r2
h2
h2
1
4
C
D
h r
1
B
R
3 R
h2
3 2
A
R2
9
4
R2
1
4
r2
r2
2
6 R 11
R2
3 R
R
2
2
11
6
14) A brother and sister walk home from school every day at the same constant speed. One day,
15 minutes after leaving school, the boy realized that he had forgotten his lunch bag at school
and ran back to get it. In the meantime, the girl continued to walk home at half her usual speed.
When the boy caught up with her, they resumed walking at their usual speed and arrived home
6 minutes later than usual. How many minutes did the girl walk alone?
For their regular walk d
The sister travels
15
60
r
rt
t2
r
2
t3 r
15
60
The time the sister travels is
Using d
15
60
rt
r
t2
r
2
d
t2
9
t3 r
r 60
t3
t2
6
60
t
t3 
15) Let y be the real number such that 2015 y
9
60
t
1
t
2 2
t2
1
10
t3
t2
1
5
hour
12 minutes.
3
y 2015 .
Find the value of log 2015 log 2015 y – log 2015 y . Express your answer in simplest form.
2015 y
y 2015
y log 2015 y 2015  20153 log 2015 y
3
3
log 2015 log 2015 y – log 2015 y
log 2015 
y

20153
log 2015 log 2015 y – log 2015 y
log 2015 y
log 2015 y
y
20153
log 2015 y
log 2015 20153  log 2015 y
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
2Π
8
Π
4
The area of the smaller circle equals the area of one of the sectors.
The area of the smaller circle is Π 1 2 Π
If the length of one of the segments is r, then Π
1
2
Π
4
Π
r
2 r
1
2
Π
4
1 r
2
1
2
Π
4
1
r2
2r
8 0
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
shaded in the figure.
1 r
2
4
1
2
0
Π
4
1
r
2 or r
4.
B
A
A
1
2
Area of triangle
Combining: Π
B A
Π
18) Solve the equation log
x
B 2 A
2A B Π
A Π 2
2A 2 A
8A 8 Π 2
Area
A B 2
 2  22  B A A
1
2
Area of sector
2 2
5
1
2
log x4 16
for x. Express your answer as a rational number in
lowest terms.
xa
logxa y z
z
xaz y
y
az logx y
logxa y
1
a
logx y
Using the above:
log
5
x
1
2
2 logx 5
1
2
logx 25
log x4 16
1
4
2 logx 5
logx 16
1
2
logx 50
x1 2
50
x
1
4
1
2
logx 16
logx 161 4 
1
2
2500
19) Find the sum of all of the real solutions to the equation x
x 6
x x 6
6 x
7
7
x x 6
7
6x x2 7
x2 6x 7
x2 6x 7 0
x2 6x 7 0
36 28 2
x 3
7.
x 6
x 6 x
x 6
x–6
x 7
2
x 1
0
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
6 x – 10 y 30 and y 5 – x 3 .
x2
y2
6 x – 10 y 30
x 3
y 5
x 3
x 3
y x 8
10
5
x 3
2
x 3
y 2 x
y 5
2
64
10
5
10
1
2
A
5

5
3Π
 82 
2
48 Π
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.
25
22 23 24
17 18 19 20 21
10 11 12 13 14 15 16
1
2
3
4
5
6
7
8
9
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
23
Total 3 23
22
21
22
22
21
21
21
22
22
21
21
21
21
21
22
21
22
21
22
21
22
21
9 22
13 21
540
22) If k is a non-negative integer, define
889
a0
3
8
8
3
43 333 and a n
9
3
23
k
as the sum of the cubes of the digits of k. For example,
512 512 729 1753. The sequence a n is defined by
an–1
for n 1. Find the value of a 2015 .
a1
43 333
a2
a3
a4
a5
a6
Thus a3 k
a2015
13
33
13
23
33
172
352
160
217
352
a3
73
53
63
13
53
160, a3 k
671 2
4 33  172
43
23
23
03
73
23
352
160
217
352
160
217 and a3 k
1
352
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 ?
END
START
END
226
92
29
8
1
92
42
21
6
1
29
21
10
5
1
8
6
5
3
1
1
1
1
1
START
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?
Let a number of accumulated puzzles and d
the number of days to catch up at 4 per day.
a d
4
a d 10
3
d
d
10
a d 4d
10 3 d

a

3d a 0
2d a
20
d
30
d
20 and a 60
20 days to catch up.
25) Let g be a function such that g 1
of log3 g3200 ?
g3
g3 1
1 and g 3 n
1 30
1 g1
g32  g 3 3 3 g 3 3 31
g33  g3 32  32 g32  32 31
g34  g3 33  33 g33  33 31
g3k  g3 3k 1  3k
g3200  3200
1
26) Let Sn
n
k
1
1
2 3
,
2
n
1
1k k 1
199 2
k
,
1
3 4
31 2
31
1
,
1
1

1
31
2 3
2
k 2
31
2 3
k 1
3k k
1 2
log3 g3200  19 900
,
1
k
2
g3k 1  3k
1
319 900
1
1 k
n g n for any positive integer n. What is the value
1
n 1
n
1
2
1
2
. Find the average value of the elements of S2015 .
1
3
1
n
1
n
1
1
1
n
n
1
n
1
n
1
n 1
n 1
An
n
1
A 2015
2016
27) Find the positive integer b such that
3 b2
2b b 5
b2
8b 0
2b
bb 8
20 b 15 b
2 b2
10 b 3 b2
0
b 8
320 b .
2b
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 ?
r
r
r2
r2
5 s2
4
2
 2 2
2 sr
r2
s r
s2
2 rs 0
s
s2
4
5
s 4 s 2 r 0
s 0, s
8
5
r
s r
s2
8
5
s
25
40
1
x
29) Given that x
x
1 2

x
32
x2
x
1 3

x
33
x3
2 x x 
1
x2
1
x2
x3
1 2

x3
182
x6
1
x2
x2
1
x
1
3 x
x6
1
x6
1
1

x
2
1
x3
27 x3
2 x 3  x3 
1
1
x6
x6
x6
1
x6
182
2
.
9 2 7
3 x2  x  3 x x2 
x3
1
x3
1
x2
x2
2
1
x6
3 , find the integer value of x6
3 x
x3
1
x3
1
x3
3 3
1
x6
1

x
x3
1
x3
18
2
2 324 2 322
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
no greater than 5000.
n
5
n
25
n
125
n
625
n
3125
1000
n 625 125 25 5 1
n
1000 3125
781
Multiply by 3125
1000 3125
7812 n 1000 3125
4011.25
Hence n 4005
31 Find the length of either tangent line from the origin to the circle
x2 y2 – 6 x – 8 y 21 0.
6
5
r
4
r
3
2
l
d
1
l
1
x2
y2
6 x 8 y 21 0
d2
h2
k2 and d 2
l2
32
42
22
l2
x 3
2
y 4
l2
h2
k2
r2
9 16 4 21
l
32) The string of digits 01 001 000 100 001
2
2
3
4
22
r2
21
consists of blocks of zeros followed by a single one.
Each successive block of zeros contains one more zero than the previous block. Ones
appear in positions 2, 5, 9,
. Find the position of the 100th one.
Position of the nth one pn
p100
100 103
2
50 103
nn 1
2
5150
n
5
n2 n 2 n
2
nn 3
2
33) Suppose that x, y and z are positive real numbers such that x y z
85, y z x
120 and z x y
105.
What is the value of the product x y z?
xy z
yz x
zx y
85
120
105
1
2
3
xy xz 85
yz yx 120
zx zy 105
1
2
3
2
1
yz xz 35
3
4
2 yz 140
In (2) xy 70 120
x2 y2 z2
yz 70
xy 50
In (1) 50 xz 85
xy xz yz
4
xz 35
50 35 70
xyz
2
2 52 5 7 2 5 7
22 54 72
xyz 2 52 7 350
"
34) If
! cos2 k Φ
7, determine the value of cos 2 Φ . Express your answer as a rational number in lowest terms.
k 0
"
! cos2 k Φ
"
! cos2 Φ 
7
k 0
"
k
! cos2 Φ 
k
k 0
1
1 cos2 Φ
7
Using the trig identity cos 2 Φ
cos 2 Φ
7
Using the sum of a geometric series:
k 0
1 cos2 Φ
2 cos2 Φ
1
7
1
cos2 Φ
cos 2 Φ
6
7
2 7 1
6
5
7
35 Circles with centers C and D have respective radii 3 and 10. A
common tangent intersects the circles at E and F,
respectively. The line segments CD and EF intersect at P.
Given that C P 5, find E F. Express your answer as a
rational number in lowest terms.
Α
Α
F
C
P
E
D
F
y
C
P
5
Α
Α
3
10
x
10
D
E
From right triangle PCE, PE
y
2
2
10
cos Α
y
2
x 10
4
5
100
y
x 10
25 2
y
16
5
4
x 10
25
 16
40
3
EF y PE
4
2
1 y
y
2
9
16
100
y2
p
p
6 n 11
11 6 17
n 17 113
y
4
3
10
3
n – 17
6 n 11
is a positive rational number that is not in lowest terms.
there exist integer p 1 such that p
n 17 and p
100
52
4
36) Find the smallest positive integer n such that
Reducible
16
9
y2
100
p
6 n 11
n 17 and p
6 n 11
6 n 17
p 113 Prime
113
791
6 n 11 791
1
7
n 130
37) Given that 2 31 – 1 is prime, find the sum of the reciprocals of all of the positive integer divisors
of 2 30 2 31 – 1.
If 2p
1 is prime, the divisors of 2p
2
3
1, 2, 2 , 2 ,
2
p 1
p
,2
1, 2 2
Sum  1
1
2
1
22
1

2p 1
Sum  1
1
2
1
22
 1
2p 1
1
1
Sum
p
1 are:
1 , 2 2p
1
2p
2p
2
1
1
1
1
2p 1
2p
2  2p   2p 1 
1
1
2p
1
2

1
2p
1,
1
2p
1
2

log
k 2015
k 2013
2
log k 2014
k 2014
k 2013

1
k 2013 k 2014 k 2013
1
2p 1 1

2p 1
38) Find the real number k such that log k – 2015
log k 2015
2p
1

2p 1
1
22
1
1
2
1
100
log k – 2014
2 Log k 2013
k 2015
k 2013
k
k 2014
k 2013
2013 k
2 2 log k – 2013
2
1
100
2014 k
1
k 2013 k 2014 k 2013
2013
100
1
100
.
y2 y 1
Let y k 2013
y 5 and
y 5
100
y3
y2
100 0
y 5 y2
4 y 20 0
2 4i
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 s where n is a positive integer and 0 s 1.
x2  –x ,x-
 n s 2 – n s n
6
6
n2
2 ns s2  – n2
Since n2 is an integer the equation reduces to 2 ns s2  – ns
ns
6
6
This implies that ns is an integer, hence 2 ns is an integer and the equation reduces to
ns 6
x 7
6
.
n
s
6
7
SInce s is less than one the smallest value for n is 7.
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
Thus an
a1
a2
a3
a4
1
2
4
7
an
1
an
2
an
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
4
T
TT, HH
HHH, HHT, THH, TTT
HHHH, HHHT, THHH, HHTT, THHT, TTHH, TTTT
a5 a4 a3 a1 7 4 1 12
a6 a5 a4 a2 12 7 2 21
6
5
4
2
a7
a6
a5
a3
21 12 4 37
a8
a9
a7
a8
a6
a7
a4
a5
37 21 7 65
65 37 12 114
a10
a9
a8
Probability =
a6
200
210
114 65 21 200
200
1024
25
128