UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS FIFTIETH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 13, 2007
1) Given real numbers x, y and z, define x, y, z = x y y z z x
x2 y2 z2 . Evaluate 3, 2, – 4 .
3, 2, – 4 = 3 2 2 –4 –4 3
32 22 –4 2 =
6 – 8 – 12 9 4 16 = –
14 29
2) Simplify 32 – 2
32 2 .
32 – 2
32 2
=
32 – 2
32 2
32 – 2
32 – 2
=
32 2– 2 2 32 2 2
32 2– 2 2
= 32 – 16 2
32 – 2 = 18 30 =
3 5
3) The midpoints of the longer sides of a 2 by 4 rectangle are joined to the opposite vertices as indicated in the figure. Find the area of the shaded quadrilateral.
Area = 2(4) – 2 1
2 2 2 – 2 1
2 1 2 = 8 – 4 – 2 = 2
4) Given three consecutive integers, the difference between the cubes of the two largest integers is 666 more than the difference between the cubes of the two smallest integers. What is the largest integer ?
Let the integers be n – 1, n and n + 1
n 13 – n3 = 666 + n3 – (n – 13
n3 + 3n2 + 3n + 1 – n3 = 666 + n3 – n3 + 3n2 – 3n + 1
6n = 666 n = 111 n + 1 = 112
5) If w is 10% larger than x, x is 20% larger than y and y is 25% smaller than z, by what percent is w smaller than z ?
w = 1.1x, x = 1.2y and y = .75z w = 1.1(1.2)(.75)z = .99z
Thus w is 1% smaller that z.
6) A ladder leans against a wall. The top of the ladder is 8 feet above the ground. If the bottom of the ladder is then moved 2 feet farther from the wall, the top of the ladder will rest against the foot of the wall. How long is the ladder ?
L
8
x
L = x + 2 x = L – 2 and L2 x2 82
L2 = (L – 22 + 64 L2 = L2 – 4L + 4 + 64 4L = 68 L = 17
7) ABCD is a square, ABE is an equilateral triangle and F is the point of intersection of AC and BE.
Find the degree measure of EAF.
A B
C
D E
F
ABE equilateral EAB = 60°. AC a diagonal of the square FAB = 45°
EAF = EAB – FAB = 60° – 45° = 15°
8) Joe has a bag of marbles. Joe gives Larry half of his marbles and two more. Joe then gives Doug half of the marbles he has left and two more. Finally, Joe gives Jack half of the marbles he has left and two more. Joe has one marble remaining in his bag. With how many marbles did Joe start ?
Let J be the number of marbles Joe has to start.
After first gift: J - 1
2J – 2 = 1 2J – 2
After second gift: 1
2J – 2 – 1 2
1
2 J–2 – 2 = 1 4J – 3
After third gift: 1
4J – 3 – 1 2
1
4J–3 – 2 = 1 8J –
7
2 = 1 1 8J =
9
2 J = 8 9 2 = 36
9) The square AFGH is cut out of the rectangle
ABCD, leaving an area of 92 in2. If FB = 4 inches and DH = 8 inches, find the original area of ABCD.
A
B C
D H
F G
x
y
y–4
x−8
y2 + 4y – y2 + 8y – 16 = 92 12y = 108 y = 9 x = 13 Area = 13(9) = 117
10) Suppose that 60 percent of the population has a particular virus. A medical test accurately detects the virus in 90 percent of the cases in which the patient has the virus, but falsely detects the virus in 20 percent of the cases in which the patient does not have the virus. If the same patient is tested twice, the results are said to be inconsistent if the results of the two tests do not agree. If the test is administered twice to all patients, how many patients out of 250 would expect to get inconsistent results ?
Let H = number of the 250 who have the virus. H = (.6)(250) = 150
Let D = number of the 250 who don’t have the virus. D = (.4)(250) = 100
For those who have the virus the number with different results on the two tests is 2(.9)(.1)(150) = 27
For those who don’t have the virus the number with different results on the two tests is 2(.8)(.2)(100) = 32
Note that the factor of two accounts for the order of the results.
Thus, the number with different results on the two tests is 27 + 32 = 59
11) Find all real solutions of 9x + 2 3x 2 = 243.
9x + 2 3x 2 = 243 32x + 2 32 · 3x = 243
Let y = 3x y2 + 18 y – 243 = 0 (y + 27)(y – 9) = 0 y = – 27 , 9
y = 3x 9 = 3x x = 2
12) Suppose that sin(2x) = 1
7 . Express sin
4 x + cos4 x as a rational number in lowest terms.
sin(2x) = 2sinx cosx = 1
7 sin(x)cos(x) = 1 2 7
1 = sin2 x cos2 x 2 = sin4 x 2 sin2 x cox2 x cos4 x
sin4 x cos4 x = 1 – 2 sin x cosx 2
sin4 x cos4 x = 1 – 2 1 2 7
2
= 1 – 1
14 = 13 14
13) For each positive integer k, let ak be the sum of the first k positive integers. If exactly x of the ak consist of one digit, exactly y of the ak consist of two digits and exactly z of the ak consist of three digits, what is the product x · y · z ?
ak = 1 + 2 + 3 + · · · + k = k k 1
2
k k 1
2 < 10 k(k+1) < 20 k
2 k– 20 0 (k + 5)(k – 4) < 20 k < 4 k = 3 x = 3
k k 1
2 < 100 k(k+1) < 200 k
2 k– 200 0 k < – 1 1 800
2 k 13 y = 10
k k 1
2 < 1000 k(k+1) < 2000 k
2 k– 2000 0 k < – 1 1 8000
2 k 44 z = 31
14) A rectangular solid has a top face with
surface area of 28 square feet, a front face with surface area of 20 square feet and a side face with surface area of 70 square feet. What is the volume of this solid ?
xy = 28, yz = 20 and xz = 70 Multiplying these three equations:
x2y2z2 = 28(20)(70) xyz2 = 22 7 22 5 2 5 7 = 25 52 72 xyz = 4 2 5 7 = 140 2
15) John rolls two standard six-sided dice. Given that the sum of the numbers on the top faces is not 6, what is the probability that the sum is 7 ?
Express your answer as a rational number in lowest terms.
There are 6(6) = 36 possible pairs of numbers (x,y).
There are 5 pairs adding to 6 are (1,5), (2,4), (3,3), (4,2) and (5,1).
So total minus those adding to 6 is 36 – 5 = 31
The are 6 pairs adding to 7 are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1).
Thus the probability of a 7 given not 6 is 6 31.
16) A hardware store sells three different kinds of widgets. Green widgets cost 25¢ each, blue widgets cost $1 each and red widgets are $3 each. Traci bought 75 widgets for a total cost of exactly $75. What is the greatest number of green widgets she could have bought?
Let G, B and R be the numbers of green, blue and red marbles purchased.
G + B + R = 75 B + R = 75 – G (1)
1
4G + B + 3R = 75 B + 3R = 75 – 1
4G (2)
(2) – 3(1) B – 3B = 75 – 1
4G – 3(75) + 3G
– 2B = –2(75) 11
4G B = 75 – 11
8G 0 G = 48
17) Find the area of the convex polygon in the plane with vertices at the points whose coordinates are 2, 3 , 1, 10 , 5, 10 , 8, 7 and 4, 0 .
7 3
19) A box contains 6 red balls, 4 blue balls and 2 green balls. Three balls are randomly drawn (without replacement) from the box. What is the probability that the three balls selected are all the same color? Express your answer as a rational number in lowest terms.
Average 5 w1 w2 w3 w4 w5
5 =
w1 w2 w3 w4
4 + 2 = w1 + 6 + 2 = w1 + 8 w1 w2 w3 w4 + w5 = 5w1 + 40
w1 w2 w3 w4 + w5 = 5w1 + 40 4w1 + 24 + w5 = 5w1 + 40
w5 = w1 + 16
21) The lines x – 2y = 2, – 3x + y = 4 and 2x + y = 4 intersect in pairs to determine the vertices of a triangle. Find the area of this triangle.
–2,–2
2,0 0,4
−3 −2 −1 1 2 3
−4 −2
2 4 6
Find the points of intersection:
x–2 y=2 2 x+y=4
x–2 y=2
4 x+2 y=8 adding 5x = 10 x = 2 y = 0
x–2 y=2 –3 x+y=4
x–2 y=2
–6 x+2 y=8 adding –5x = 10 x = – 2 y = – 2
2 x+y=4
–3 x+y=4 subtracting 5x = 0 x = 0 y = 4
Area of the triangle = area of the rectangle minus the area of the three triangles.
A = 4(6) – 1
2(4)(2) – 1
2(4)(2) – 1
2(2)(6) = 24 – 4 – 4 – 6 = 10
22) Rectangle A has vertices at (0 , 0), (0 , 6), ( – 4 , 6) and ( – 4 , 0). Rectangle B has vertices at (2 , 0), (20 , 0), (20, 8) and (2, 8). A line divides each rectangle into two regions of equal area. What is the slope of this line ?
−2,3
11,4
5 10 15 20
2 4 6 8
Fact: Any straight line which divides a rectangle into two equal areas must pass through the centroid (center) of the rectangle.
Given this fact, the specified line must pass through the points (–2,3) and (11,4) with slope m = 4 – 3
11 – –2 = 1
13
1
2(4)[a + b] = 1
2(4)[6 – a + 6 – b]
1
2(18)[c + d] = 1
2(4)[8 – c + 8 – d]
b–a
4 =
d–c
18
d–a
24 =
c–b
2
Solving gives: a = 37
13, b = 41 13, c =
43
13 and d = 61 13 m =
b–a
4 =
41 13–
37 13
4 = 1 13
−4,a 0,b
2,c
20,d 20,8
−4,6
a 6−a
b 6−b
c 8−c
d 8−d
23) Suppose that P is a point that lies outside of two concentric circles C1 and C2, with C1 inside C2. Suppose that T is a point on
C1 such that PT is tangent to C1 and PT = 17; suppose that S is a point on C2 so that PS is tangent to C2 and PS = 15. If Q is the
point on C2 that is on PT (with Q between P and T), find TQ.
P
Q
S T
C
r1 r2
r2
Let r1 be the radius of C1 and r2 be the radius of C2.
PC2 = r12 + 172 (1)
PC2 = r22 + 152 (2)
r22 = TQ2 + r12 (3)
(3) TQ2 = r22 – r12
Thus TQ2 = 64 TQ = 8
24) Find the sum of all real numbers x such that | x – 2006 | + | x – 2007 | = 3.
x–2007 2007–x 2007–x x–2007
x–2006 2006–x x–2006 x–2006
x–2006 + x–2007 =3 4013–2x=3 1=3 2x–4013=3
2x=4010 2x=4016
x=2005 2008
2006 2007
Sum = 2005 + 2008 = 2013
25) Suppose that a, b and c are positive integers such that a log144(3) + b log144(2) = c . What is the value of a b
c ?
a log144(3) + b log144(2) = c
log144 3a + log
1442b = c
log144 3a 2b = c
3a 2b = 144c 3a 2b = 24 32 c = 24c 32c
a = 2c and b = 4c a b
c = 2c 4c
c =
6c c = 6
26) Let f (x) = a x + b. Find all real values of a and b such that f (f (f (1))) = 29 and f (f (f (0))) = 2.
f (f (f (1))) = 29 f (f (a+b)) = 29
f(a(a+b)+b) = 29 f a2 ab b = 29
a a2 ab b + b = 29 a3 a2b ab b = 29
f (f (f (0))) = 2 f (f (b)) = 2 f(ab+b) = 2
a ab b + b = 29 a2b ab b = 2
a3 + 2 = 29 a3 = 27 a = 3
9a + 3b + b = 2 13b = 2 b = 2
13
a = 3 b = 2
29) Let C1 be a circle of radius 1, let C2 be a pair of congruent
31) Alice, Bob, Carl and Dave decide to play a game in which they take turns spinning a wheel on which the numbers 1, 2, 3 and 4 are equally likely to be selected. The winner is the first person who spins a 3. The order of play is Alice, Bob, Carl and Dave, repeated until a winner is decided. Find the probability that Alice wins. Express your answer as a rational number in lowest terms.
The probability Alice wins on her first turn is 1
32) In triangle ABC, AC 12. If one of the trisectors of angle B is the median to AC and the other trisector of angle B is the altitude
to AC, find the area of triangle ABC.
A
B C
D E
α α α
6
3 3
EDB DBA ED = DA = 6
2 = 3.
From EDB tan( ) = 3
DB
From CDB tan(2 ) = 9
DB
tan(2 ) = 2 tan
1 – tan2 =
2 3
DB
1 –DB9
= 9
DB
6
1 – 9
DB2
= 9 1 – 9
DB2 =
6 9 =
2 3
9 DB2 =
1
3 DB
2 = 27 DB = 3 3
Area = 1
2 · 12 · 3 3 = 18 3
33) Given a finite sequence of n real numbers, let Sk be the sum of the first k terms of the sequence. The Cesaro sum of this sequence is defined to be S1 S2 Sn
n . Suppose that the Cesaro sum of s = { a1 , a2 , · · · a49} is 100.
Find the value of a0 so that the Cesaro sum of s = { a0 , a1 , a2 , · · · a49} is 101.
For s = { a1 , a2 , · · · a49} S1 S2n Sn = a1 a1 a2 a1 a492 a3 a1 a49 = 100
a1 a1 a2 a1 a2 a3 a1 a49 = 100(49) = 4900
For s = a0, a1 , a2 , · · · a49} S1 S2 Sn
n =
a0 a0 a1 a0 a1 a2 a0 a49
50 = 101
50a0 + a1 a1 a2 a1 a2 a3 a1 a49 = 101(50) = 5050
50a0 = 5050 – 4900 = 150 a0 = 3
x
x
x
x
y
y
y
y
z
z
z
z
Edge length = 4(x + y + z) = 124 x + y + z = 31
Surface area 2(xy + xz + yz) = 622
x y z2 = x2 y2 z2 2 xy xz yz = 312
312 = d2 + 622 d2 = 961 – 622 = 339
35) Find the largest positive integer p such that 57 can be expressed as the sum of p consecutive positive integers.
Let n be the first integer:
57 =
k 0
p–1
(n + k) =
k 0
p–1 n +
k 0
p–1
k = np + p–1 p
2
2 · 57 = 2np + p2 – p = p(2n – 1 + p) = p
Since p is the smaller of the two factors, p | 2 · 57 and p 2 5 53 p = 2 · 53 = 250
36) Find the number of ordered pairs (a , b) of positive integers such that loga(b) + 10 logb(a) = 7 and 2 a b 2007.
loga(b) + 10 logb(a) = 7 loga(b) + 10
logab = 7
If x = loga b x + 10
x = 7 x
2 – 7x + 10 = 0 (x – 2)(x – 5) = 0 x = 2 , 5
loga b = 2 a2 = b a2 2007 a = 2, 3, 4, ··· 44 43 values
loga b = 5 a5 = b a5 2007 a = 2, 3, 4 3 values
Number of pairs 43 + 3 = 46
37) If f (x) = x3 – 4x2 + 18x – 9, f a = 0 , f b = 0 , f (c) = 0 and a, b and c are all distinct, find f 1
a
1
b
1
c .
f 1
a
1
b
1
c = f(
ab ac bc
(x–a)(x–b)(x–c) = x3– a b c x2 ab ac bc x–abc = x3 – 4x2 + 18x – 9
ab + ac + bc = 18 abc = 9
f 1
a
1
b
1
c = f(
ab ac bc
abc ) = f
18
9 f 2 = 2
3– 4 22 18 2 – 9 = 8 – 16 + 36 – 9 = 19
38) Find bases a and b such that 386a = 272b and 146a = 102b .
386a = 272b 3a2 8a 6 = 2b2 7b 2 (1)
146a = 102b a2 4a 6 = b2 2 (2)
(1) – (2) 2a2 4a b2 7b (3)
Rewriting (1) 2a2 4a a2 4a 6 2b2 7b 2 (4)
Substitute (3) into (4) b2 7b a2 4a 6 2b2 7b 2
Or a2 4a 4 b2 a 22 b2 a + 2 = b (5)
Substituting (5) into (3) 2a2 4a a 2 2 7 a 2 2a2 4a a2 4a 4 71 14
a2– 7a– 18 0 (a – 9)(a + 2) = 0 a = 9
a = 9 b = 11
39) Let p(x) = x3 + x12 + x21 + x30 + · · · x2010 = k2230 x3 9k and q(x) = x6 – x3.
Find the remainder when p(x) is divided by q(x).
p x
q x = f(x) + r x
q x p(x) = f(x)q(x) + r(x)
p(x) = x3 + x12 + x21 + x30 + · · · x2010 = x31 x9 x18 x27 x2007 x3P x
q(x) = x6 – x3 = x3 x3– 1 = x3Q(x)
Consider dividing P(x) by Q(x). P x
Q x = F(x) + R x
Q x P(x) = F(x)Q(x) + R(x)
Let y x3. Then P(y) = 1 y3 y6 y9 y669 and Q(y) = y– 1
Then P(y) = F(y)(y–1) + R degree R < degree Q(y) R = constant.
P(1) = F(1)(1–1) + R R = P(1) = 1 + 1 + 1 + ··· 1 = 224
r(x) = x3R = 224x3
A B
C D
40) Let ABCD be a square with A= (0 , 0) , B = (5 , 0) , C = (5 , 5) and D = (0 , 5). Let P be a point on the line y = 2x. Find the value of x such that the sum of the squares of the distances from P to the vertices of the square is a minimum.
A 0,0 B 5,0
C 5,5 D 0,5
P
x,2x
S2 = PA2 PB2 PC2 PD2
S2 = x2 4x2 x–5 2 4x2 x–52 2x–5 2 x2 2x–52
S2 = x2 4x2 x2–10x 25 4x2 x2–10x 25 4x2–20x 25 x2 4x2–20x 25
S2 = 20x2– 60x 100 = 20 x2–3x 5 = 20 x2–3x 9 4–
9
4 5 = 20 x– 3 2
2 11 4
Minimum occurs when x = 3
2
41) In ABC, AB = 7, BC = 5 and CA = 6. Locate points P1, P2, P3 and P4 on BC so that
this side is partitioned into five congruent segments, each of length 1. For k = 1, 2, 3 and 4 , let qk = APk.
Find q12 + q22 + q32 + q42.
P1
P2
P3
P4
q1 q2
q3
q4
α
A B
C
6
7
1 1 1 1 1
Applying the law of cosines to ABC: 62 72 52– 2 7 5 cos cos( ) = 19 35
Applying the law of cosines to ABPk for k = 1, 2, 3 and 4:
q12 72 12– 2 7 1 19
35
q22 72 22– 2 7 2 19 35
q32 72 32– 2 7 3 1935
q42 72 42– 2 7 4 19 35
Adding: q12 q22 q32 q42 4 72 1 4 9 16 2 7 19
35 1 2 3 4
