UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS FIFTY-FIFTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 14, 2012
1) What is the smallest positive integer n such that 2n + 1 and 3n + 1 are both primes?
n 2n+1 3n+1
1 3 4
2 5 7
n = 2
2) Every student in Ms. Math’s class took the AP Calculus Exam and received a score of either a 4 or a 5. If 32 of the students earned a 5 and that was 80% of the class, how many students earned a 4 on the AP Exam?
Let n = number of students.
0.8n = 32 ï n = 32
0.8 = 320
8 = 40
Number scoring 4 = n – 32 40 – 32 = 8
3) Express 1
2+ 3
4+5
6
as a rational number in lowest terms.
1
2+ 3
4+5
6
= 1
2+ 3
29 6
= 1
2+ 18
29
= 761 29
= 2976
4 M,A,BandNare collinear withMA=AB=BN. ArcsMA, AN, NBand MB
are all semicircles. IfMN=9, what is the perimeter of the shaded region ?
P = 2p 3
2 + 2p(3) = 3p + 6p = 9
5) If p and q are the roots of the equation x2– 6x+ 2 = 0, find the value of 1
p +
1
q .
1
p +
1
q = p+q
pq = 6 2 = 3
6) Among the 41 students in a class, 12 have a cat,
5 have a dog, 8 have a robot, 2 have a cat and a dog, 6 have a cat and a robot, 3 have a dog and a robot, and 1 has a cat, a dog and a robot.
How many students have neither a cat nor a dog nor a robot?
Dog Cat
Robot 1
1
1 5
2 5
0
None = 41 – (1 + 1 + 1 + 2 + 5 + 5) = 41 – 15 = 26
7) The total cost of tickets to the school play for one adult and 3 children is $27. If the cost of a ticket for an adult is $1 more than the cost of a ticket for a child, what is the cost of an adult ticket?
A = cost of adult ticket and C = cost of child ticket.
A + 3C = 27 ï A + 3(A – 1) = 27 ï 4A = 30 ï A = 30
4 = 7.50
8) A line passing through the points (3 , – 6) and (9 , 3) intersects the x-axis at the point (a , 0). What is the value of a ?
3,−6
9,3
a,0
3
9 –a = 6
a– 3 ï 3a – 9 = 54 – 6a ï 9a = 63 ï a = 7
9) Let f(n) = 3 n!
3n !. Find f 4
f 4
f 5 =
3ÿ4!
12!
3ÿ5!
15!
= 3ÿ4! 12! ·
15!
3ÿ5! =
15ÿ14ÿ13
5 = 3ÿ14ÿ13 = 546
10) For non-zero real numbers a and b, define a*b=a b – 1
a –
1
b. What is the value of (1 * 2) * 3 ?
1 * 2 = 1(2) – 1
1– 1 2 = 2 –
3 2 =
1 2
1
2 * 3 = 1 2(3) –
1
1
2 – 1
3 = 3 2 – 2 –
1 3 =
3 2 –
7 3 =
9 – 14 6 = –
5 6
11 Two circles, one of radius 3 and the other of radius 5, are externally tangent and are internally tangent to a larger circle. What is the area of the shaded region inside the larger circle that is not contained in either of the two smaller circles?
R = 5+5+3+3
2 = 8 = radius of the large circle
Area = p82 –p52 –p32 = p(64 – 25 – 9) = 30
12) Find all values of k so that the curves y=x2– 3k x–1 and y= 9 k x – 17 intersect in exactly one point.
x2– 3k x–1 = 9 k x – 17
x2– 3k x 9k x– 1 +17 = 0
x2– 12k x + 16 =0
x = 12k≤ 12k2–4 16
2
For a single solution 12k2–4 16 = 0 ï 12k2 = 4(16) ï 12k = ± 8 ï k = ± 2 3
13) Find all real numbers x such that 4 –x + 4 + x =2x .
Square both sides:
4 – x + 4 + x + 2 16 –x2 = 4x2
2 16 –x2 = 4x2– 8 ï 16 –x2 =2x2– 4
16 – x2 = 4x4 – 16 x2 + 16
4x4 – 15x2 = 0
x2 4x2– 15 = 0 ï x = 0 or ± 15
Checking yields the only solution is x = 15 2
14) Let x and y be two positive real numbers such that logx y +logy x = 3. Find the value of logx y 2 + logy x 2.
logx y +logy x 2 = 32
logx y 2 + logy x 2 + 2 logx y logy x = 9 Using logx y = 1
logyx
logx y 2 + logy x 2 + 2 = 9 ï logx y 2 + logy x 2 = 7
15) Let g be a function such that 3g(x) + 2g(1 – x) = 9 + 2x. Find the value of g(2).
Substituting x = 2 and x = – 1
3g(2) + 2g(–1) = 9 + 4 3g(–1) + 2g(2) = 9 – 2
3g(2) + 2g(–1) = 13 2g(2) + 3g(–1) = 7
Multiplying by 3 and 2
9g(2) + 6g(–1) = 39 4g(2) + 6g(–1) = 14
Subtract
5g(2) = 25 ï g(2) = 5
16) Find the exact value of cos3(15°) sin(15°) – cos(15°) sin3 15 ° .
Factoring cos3(15°) sin(15°) – cos(15°) sin3 15 ° = cos(15°) sin(15°)[cos2(15°) – sin2 15 ° ]
Using the identities sin(2a) = 2cos(a)sin(a) and cos(2a) = cos2 a – sin2 a
cos3(15°) sin(15°) – cos(15°) sin3 15 ° = 1
2 sin(30°) cos(30°) = 1 2
1 2
3 2 =
3 8
17) What integer n satisfies n < 5+2 2 + 5 – 2 2 < n+1?
Squaring
n2 < 5+2 2 + 5–2 2 + 2 17 < n+1 2
n2 < 10 + 2 17 < n+12
Since 4 < 17 < 5 , n2 < 18 < 20 < n+12 ï n = 4
18) In a random arrangement of the letters of GREENMOUNTAINS, what is the probability that the vowels are in alphabetical order within the arrangement? Express your answer as a rational number in lowest terms.
There are 14 letters including 2 E’s and 3 N’s. 6 of the 14 letters are vowels.
The vowels must appear in the order AEEIOU and can be placed in 14
6 different positions.
Arrange the remaining 8 letters in 8!3! ways in the remaining positions.
p =
14 6
8!
3!
14!
2!×3!
= 6!14!8! 8!3! 2!14!3! = 2!6! = 2
720 =
1 360
19) If q is an acute angle such that tan(q) = 2, find the value of cos(4q). Express your answer as a rational number in lowest terms.
θ
1
2
sin(q) = 2
5 and cos(q) = 1
5 sin(2q) = 2sin(q)cos(q) = 2 2
5 1
5 = 4 5
cos(4q) = cos(2(2q)) = 1 – 2 sin2 2q = 1 – 2 4 5 2
= 1 – 32
25 = –
7 25
20) Find the area of the region in the plane consisting of all points (x , y) that satisfy | 4x – 8 | §y§ 12 .
Find points of intersection of y = | 4x – 8 | and y = 12
x§ 2 ï 8 – 4x = 12 ï 4x = – 4 ï x = – 1
x¥ 2 ï 4x – 8 = 12 ï 4x = 20 ï x = 5
–1,12 5,12
−2 2 4 6
5 10 15
Area = 1
2(base)(height) = 1
2(5 + 1)(12) = 36
21) Find all positive real numbers x such that log2 x +log2 x– 12 = 6 .
log2 x + log2 x– 12 =6
log2 x x–12 = 6
x(x – 12) = 26 = 64
x2– 12x– 64 = 0
22) Let f be a function such that f x 5 =x
2 +x +2 . Find the sum of all values of w such that f(5w) = 2012.
f(5w) = f(25w
5 = 25w
2 +25w + 2 = 2012
625w2 +25w – 2010 = 0 ï w2 + 1 25w–
2012 625 = 0
Suppose the roots are a and b, then (w – a)(w – b) = 0 ï w2– a + b w +ab = 0
Comparing coefficients, a + b = – 1 25
23) Suppose that 9a =12, 12b =15, 15c = 18, 18d = 21, 21e =24 and 24f =27.
What is the value of a · b · c · d · e · f ?
Using the given equations from last to first,
27 = 24f = 21ef= 18def = 15cdef = 12bcdef = 9abcdef
33 =32abcdef ï 2abcdef = 3 ï abcdef = 3 2
24) If x, y and z are positive real numbers such that x y= 40, x z= 60 and y z= 96, what is the value of x + y + z ?
(xz)(yz) = 60(96) ï xy z2 = 60 96 ï 40z2 = 60 96 ï z2 = 60 96
40 = 144 ï z = 12
12x= 60 ï x = 5 and 12y = 96 ï y = 8
x + y+ z =5 +8 +12 = 25
25) Let S= 5, 52, 53, ÿ ÿ ÿ, 510 . Suppose that a and b are distinct integers chosen from S.
For how many ordered pairs a, b is logab an integer?
Let a = 5k and b = 5n with k and n in { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
loga b œZ+ ï log5k 5n œZ+ ï k divides n
k n number
1 2, 3, 4, 5, 6, 7, 8, 9, 10 9
2 2, 4, 6, 8 4
3 6, 9 2
4 8 1
5 10 1
17
A B
C D
E
26 SquareABCDhas side length 6. A semicircle whose diameter isABis drawn internally. The line that is tangent to the semicircle atTand passes through vertexC intersects sideADatE. FindET.
A B
C D
E
T
Since CB and CT are tangents to the circle from the same point, CB = CT = 6. Similarly, ET = EA = x.
From triangle EDC, CD2 + DE2 = CE2
Substituting, 62 + 6 –x2 = 6+x2
36 + 36 – 12x + x2 = 36 + 12x + x2 ï 36 = 24x ï x 3
2
27) Let S be the set of all three-digit positive integers, each of whose digits is 1 , 3 , 5 , 7 or 9. For example, 197 and 331 are two elements in S. How many integers in S are divisible by 3?
An integer is divisible by 3 in and only if the sum of its digits is divisible by 3. So considering the possible sets of three allowed digits and number of different arrangements:
Digits Arrangements
1, 3, 5 6
1, 5, 9 6
3, 5, 7 6
5, 7, 9 6
1, 1, 7 3
3, 3, 9 3
7, 7, 1 3
9, 9, 3 3
1, 1, 1 1
3, 3, 3 1
5, 5, 5 1
7, 7, 7 1
9, 9, 9 1
Total = 41
Number = 41
28) Consider the binary sequence 01001000100001000001··· , where each block of 0’s contains one more 0 that the preceding block of 0’s. Note that the first 1 appears in position 2. In what position does the 200th 1 appear?
Position 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ⋅ ⋅ ⋅
Digit 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1
Let pk = position of kth 1
p1 = 2
p3 = 2 + 3 + 4 What is the probability that the elements of T can be arranged to form an arithmetic progression with positive common difference? Express your answer as a rational number in lowest terms.
p = 16 chosen on sidesBCandACrespectively so that triangleADEis similar to triangleABCandAE>AD>DE. Find the area of triangleADE.
From DAFD, tan(b) = h
z = 4
3 ï h = 4 3 z
From DFBD, tan(b) = h
24 –z = 4
3 ï h = 4 3 24 –z
4
3z = 4
3 24 –z ï z = 24 – z ï 2z = 24 ï z = 12
h = 4
3 z ï h = 4
3 12 = 16
From DAFD, b2 = 122 +162 = 400 ï b = 20
From DABC, tan(a) = 24
32 = 3 4
From DADE, tan(a) = a
b = 3
4 ï a = 3 4 b =
3
4 20 = 15
Area = 1
2ab = 1
2 15 20 = 150
34) In how many ways can the squares in a 2 by 5 array be colored red (R), green (G) or blue (B) so that no two squares with a common edge are the same color? One such example is
R
B
G
B
R
B
G
B
R
B
.Let Nk be the number of colorings of a 2 by k grid.
Consider any coloring of a 2 by k grid and look at the last two squares. For example, suppose the are colored R and G.
R
G
For the 2 by k + 1 grid, the added top square could possibly be G or B and bottom added square could possibly be R or B.
R G
G R
R G
G B
R B
G R
R B
G B
One of these leads to matching colors, B and B. Thus for this “ending” choice of the 2 by k grid there are 3 possible new colorings. Hence Nk+1 = 3*Nk .
For k =1 there are 3 · 2 choices. So N1 = 2 · 3 , N2 = 2 · 3 · 3 = 2 · 32 · · · Nk = 2 · 3k
35 LetS1be a square with side length 2
8 . A sequence of smaller squares is
constructed by joining the midpoints of the sides of each previous square to form a new square. That is, the corners of squareS2are the midpoints of the sides of squareS1; the corners of squareS3are the midpoints of squareS2; and so on. The first 5 such squares are shown in the figure. LetPibe the
perimeter of squareSi. Find
k=1 ¶
Pk.
Consider consecutive squares, one with side s and the next with side z.
So must pick 0 §s§ 12, 0 §t§ 5, 0 §u§ 2 and 0 §v§ 1
Number of n = 13(6)(3)(2) = 468
40 LetSbe the square with vertices 0, 0 , 3, 0 , 3, 3 and 0, 3 . Each side of this square is trisected and the points of trisection are joined as indicated in the sketch to form two overlapping squares. Find the area of the region common to these two squares.
Line (0,1) to (2,0) y = 1 – 1
2 x
Line (0,2) to (1,0) y = – 2(x – 1)
Intersect 1 – 1
2 x = – 2(x – 1) ï 1 – 1
2 x = – 2x + 2 ï 3
2x = 1 ïx = 2
3 ï y = 2 3
Line (0,1) to (1,3) y = 2x + 1 Line (0,2) to (1,0) y = – 2(x – 1)
Intersect 2x + 1 = – 2(x – 1) ï 2x + 1 = – 2x + 2 ï 4x = 1 ïx = 1
4 ï y = 3 2
Side if inscribed square = 12+22 = 5 Let A = area of square = 5.
Length (0,1) to (1/4,3/2) = 1
4 2
+ 1 2
2
= 1
16+ 1 4 =
5 4
Length (0,1) to (2/3,2/3) = 2
3 2
+ 1 3
2
= 4
9+ 1 9 =
5 3
Let T = area triangle ((0,1) - (1/4,3/2) - (2/3,2/3) = 1
2 5 4
5 3 =
5 24
Shaded area = area of square – 4T = 5 – 4 5
24 = 5 – 5 6 =
25 6
A
B
C
41 Two circles of radius 16 have their centers on the circumference of each other. ABis the diameter of the right–hand circle that passes through the
centers of the two circles. A smaller circle is constructed tangent toABand the two given circles as in the sketch.
Find the radius of the smaller circle.
A
B
C
D
E
Let the radius of each of the large circles be R and the radius of the small circle be r.
AD = R, AC = R + r DC = R – r
AE2 = (R + r2 – r2 = R2 + 2Rr DE = (R – r2 – r2 = R2 – 2Rr
AD = AE – DE
R = R2+2Rr – R2–2Rr
R2 = R2 + 2Rr + R2 – 2Rr – 2 R4– 4R2r2
R2 = 2 R4– 4R2r2
R4 = 4R4 – 16R2r2
R2 = 4R2 – 16r2
r2 = 3
16R 2
r = 3
4 R
r = 3
