UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS FIFTY-FIRST ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 12, 2008
1) At the beginning of the semester, Mr. Kost had 25 students in his class. At the end of the semester, he had 20 students in his class. What was the percent decrease in the number of students in Mr. Kost's class over the semester?
20 = 25(x)
x = 20
25 = 4
5 Decrease = 20%
2) On a map, a line segment of length 11
4 inches represents 40 miles. What distance is represented by a line segment of
length 2 inches?
5
4inch = 40 miles ï 40
5 4
miles
inch ï 40 · 4 5 = 32
miles
inch ï 2 inches = 64 miles
3) Simplify 5– 2– 5– 1 6– 2– 6– 1. Express your answer as a rational number in lowest terms.
1
52 –
1 5
1 62 −
1 6 =
1 25 –
1 5
1 36 –
1 6 = −
4 25 · −
5 36 =
1
45
4) What rational number between 1
2 and 2
3 is three times as far from 1
2 as it is from 2 3?
q – 1
2 = 3 2
3 −q ï q – 1
2 = 2 – 3q ï 4q = 2 + 1 2 =
5
2 ï q =
5
8
5) Each small square in the given 10 x 10 square has an area of one square unit.
Find the area of the shaded region.
A = 1
2(1)(10) + 1
2(9)(10) + 10 = 5 + 45 + 10(1) = 60
6) If a, b and c are integers such that a + b + c = 91 and abc = 729, determine the value of a2 + b2 + c2.
abc = 729 = 36 ï a = 34 b = 32 c = 1 (81 + 9 + 1) = 91
(a + b + c 2 = a2 + b2 + c2 + 2(ab + ac + bc)
912 = a2 + b2 + c2 + 2(ab + ac + bc) = a2 + b2 + c2 + 2(81 · 9 + 81 · 1 + 9 · 1)
912 = a2 + b2 + c2 + 2(819) ï a2 + b2 + c2 = 912 – 2(819) = 6643
7) The ratio of boys to girls in Mrs. Brown's class is 3 to 2. In Mrs. Smith's class, the ratio of boys to girls is 4 to 3. If there are 30 students in Mrs. Brown's class and 28 students in Mrs. Smith's class, what is the total number of boys in the two classes?
B
G = 3 2
3
2 G + G = 30 ï 5
2 G = 30 ï G = 12 ï B = 3
2(12) = 18
In Mrs. Smith's class
B
G = 4 3
4
3 G + G = 28 ï 7
3 G = 28 ï G = 12 ï B = 4
3(12) = 16
Boys = 18 + 16 = 34
8) The sixth grade class at Williston Central School is raffling off a turkey as a fundraising project. If the turkey costs $22 and the tickets are sold at 75 cents each, how many tickets will have to be sold for the class to make a profit of $20 ?
3
4 x = 42
x = 4
3 42 = 4(14) = 56
9) A sweater, originally priced at $100, is successively discounted by 10%, 15% and 20%. What is the price of the sweater after the third discount?
P = 100 9
10 85 100
80 100 =
9⋅17⋅8 20 = 61.2
10) An integer n > 1 is abundant if the sum of its proper divisors (positive integer divisors smaller than n) is greater than n. Find the smallest abundant integer.
n 2 3 4 5 6 7 8 9 10 11 12
Divisors 1 1 1,2 1 1,2,3 1 1,2,4 1,3 1,2,5 1 1,2,3,4,6
Sum 1 1 3 1 6 1 7 4 8 1 16
n = 12
11) A binary operation is defined by a b=a2b−b2a. Find 2 (3 4).
2 (3 4) = 2 (9 · 4 – 16 · 3) = 2 (–12) = 4(–12) – 144(2) = – 48 – 288 = – 336
12) Larry's retirement portfolio consists of stocks and bonds. In 2000, 30% of the value of his portfolio was in stocks.
By 2008, the value of his stocks has doubled, while the value of his bonds has decreased by 1
3. What fraction of the
value of Larry's 2008 retirement portfolio is in bonds? Express your answer as a rational number in lowest terms.
P =
2 3 70
60+2
3 70
= 140
180+140 = 14 18+14 =
14 32 =
7
16
13) A day of skiing at Misty Valley costs $40. However, if you purchase a Misty Pass for $100, you receive a 30% discount on each day that you ski. Suppose you purchase a Misty Pass at the beginning of the ski season. What is the fewest number of days that you must ski so that your total cost is less than it would have been if you had not purchased a Misty Pass?
Let n = number of days.
No pass Cn = 40n With pass Cw = 100 + .7(40)n = 100 + 28n
12n > 100 ï n > 25
3 ï n = 9
14) Find all real numbers x such that log3 x – 2 = log3 x3 – 8 .
Let y = log3(x)
y – 2 = 3 y – 8
y2 – 4y + 4 = 3y – 8
y2 – 7y + 12 = 0 ï (y – 4)(y – 3) = 0 ï y = 3 , 4 ï x = 33 , 34 ï x = 27 , 81
15) If x4 + x2 + 1 = 0, what is the value of x2+ 1 x2
3
?
x2+ 1 x2
3
= x4+1
x2
3
x4 + x2 + 1 = 0 ï x4 + 1 = – x2
x2+ 1 x2
3
= –x2
x2
3
= – 1
16) Let f(x) be a function such that f(x) + 2 f(3 – x) = 4x + 5 for every real number x. What is f(1)?
f 1 + 2 f 2 = 4+5 = 9
f 2 +2 f 1 = 4 + 5 = 13
f 1 + 2 f 2 =9
– 4 f 1 –2 f 2 = 13
– 3f(1) = – 17 ï f(1) = 17 3
17) Let p(x) be a cubic polynomial such that p(n) = n for n = 0, 1, – 1 and p(2) = 100. What is p(3)?
Let p(x) = ax3 + bx2 +cx +d
p(0) = 0 ï d = 0
p 1 = 1 a+b+c = 1
p −1 = −1 −a+b−c= −1 ï 2b = 0 ï b = 0
p 2 = 100 100 = 8 a+2 c
p 1 = 1 1 = a + c ï
100 = 8 a+2 c
−2 = −2 a −2 c ï 98 = 6a ï a = 49
3
1 = a + c ï c = 1 – 49
3 ï c = – 46
3
p(x) = 49
3 x 3 – 46
3 ï p(3) = 49
3 3 3 – 46
3 (3) = 9(49) – 46 = 395
18) Ship A leaves port at 6:00 a.m. and travels due north at a steady speed of 40 mph. Sometime later, Ship B leaves the same port and travels due east at a steady speed of 36 mph. At 10:00 a.m., the two ships are 200 miles apart.
How many minutes after Ship A left port did Ship B leave port?
Let t be the time n hours ship B has been traveling at 10:00.
362t2 = 4 100 2 – 16(16)(100) = 4(100)(100 – 64)
19) What is the largest positive integer n such that 2008! is an integer multiple of 15n?
If 15n | 2008! then 3n | 2008! and 5n | 2008! Need to find the number times 5 divides the integers § 2008.
22) A cube is inscribed in a sphere of radius 1. Find the surface area of the cube.
a + b – 2 a a+b + a = 1
5 as a rational number in lowest terms.
tan(a+b) = tan α +tan β
25) Find the number of positive integers n<2008 such that 2008
n is divisible by
26) In the next football season, the Flounders will play 16 games and the outcomes of the games will be recorded as a 16-letter sequence of W's and L's (a W for each win and an L for each loss; there will be no ties). How many such sequences correspond to winning exactly six games with no consecutive wins?
f(–4) = a –4 4 + b –4 2 – 12 + 7 = 2008 ï 44a + 42b = 2008 +12 – 7 = 2013
f(4) = a 4 4 + b 4 2 + 12 + 7 = 2013 + 19 = 2032
28) Find the area of the region in the xy plane consisting of all points (x,y) whose coordinates satisfy the inequalities | y | – | x | § 2 and | x | § 3.
x y x + y =2
+ + y−x=2 + − −y−x=2 − + y+x=2 − − −y+x=2
−3 −2 −1 1 2 3
−4
−2
2 4
A = 2 4+10
2 (3) = 42 Area of two trapezoids.
29) Suppose that 6
7 = d2 2! +
d3 3! +
d4 4! +
d5 5! +
d6 6! +
d7
7! , where each dj is an integer such that 0 § dj < j .
Find d2 + d3 + d4 + d5 + d6 + d7 .
6
7 = d2 2! +
d3 3! +
d4 4! +
d5 5! +
d6 6! +
d7
7! Multiply by 7! and simplify:
4320 = 2520 d2 + 840 d3 + 210 d4 + 42 d5 + 7 d6 + d7
At each step pick the largest possible value for each di
d2 = 1
1800 = 840 d3 + 210 d4 + 42 d5 + 7 d6 + d7
d3 = 2
120 = 210 d4 + 42 d5 + 7 d6 + d7
d4 = 0
120 = 42 d5 + 7 d6 + d7
d5 = 2
36 = 7 d6 + d7
d6 = 5
1 = d7
d2 + d3 + d4 + d5 + d6 + d7 = 1 + 2 + 0 + 2 + 5 + 1 = 11
−4 −2 2 4
−4
−2
2 4
−4 −2 2 4
−4
−2 2 4
A = p 2 2 + 2(3)(10) = 4 + 60
33) A standard die with sides labeled 1 through 6 is rolled three times. Find the probability that the product of the three numbers rolled is divisible by 8. Express your answer as a rational number in lowest terms.
Three 2 ' s 2×2×2 1
Two 2 ' s 2×2×4 3
Two 2 ' s 2×2×6 3
One 2 2×4 odd 3×3! = 18
One 2 2×4×4 3
One 2 2×4×6 6
One 2 2×6×6 3
No 2 4×4 odd 3 3 = 9
No 2 4×4×4 1
No 2 4×4×6 3
No 2 4×6 odd 3×3! = 18
No 2 4×6×6 3
No 2 6×6×6 1
Total = 1 + 3 + 3 + 18 + 3 + 6 + 3 + 9 + 1 + 3 + 18 + 3 + 1 = 72
p = 72
63 =
1 3
34) Suppose that a, b, c are real numbers such that a+b+c=10
a2+b2+c2 =50 . Find the maximum value of a.
c = 10 – a – b
b2 + c2 = 50 – a2
b2 + (10 – a – b 2 = 50 – a2
b2 + 100 + a2 + b2 – 20a – 20b + 2ab = 50 – a2
2 b2 + b(2a – 20) + 50 – 20a + 2 a2 = 0 If b is to be real then
4 a2 – 80a + 400 – 400 + 160a – 16 a2 ¥ 0
By the Law of Sines sin γ
AN = sin 3π
4
1 ï AN = 2sin(g)
By the Law of Sines sin γ
BN = sin 3π
4
2
ï BN = 2 sin(g)
Area of ABC = Area of ANB + Area of ANC + Area of CNB
1
2 = 1
2· AN sin(g) · 2 + 1
2 · CN sin(g) · 1 + 1
2 · BN sin(g) · 1
1 = 2 sin(g) sin(g) · 2 + sin(g) sin(g) · 1 + 2 sin(g) sin(g) · 1
1 = 2 sin2 γ + sin2 γ + 2 sin2 γ ï sin2 γ = 1
5
Area of ABC = 1
2 · CN · BN = 1
2sin γ · 2 sin(γ = sin 2 γ
= 1 5
39) ABCD is a rectangle with AB = 4 and AD = 3. Let P
be any point on side AB. Let E be the point on diagonal DB such that PE DB. Let F be the point
on diagonal AC such that PF AC. Find PE + PF.
A B
C D
P
E
F
x 4 – x
α
From D ABC sin(a) = 3
5
From D APF sin(a) = PF
x = 3
5 ï PF = 3 5x
From D ABC sin(a) = PE
4 –x = 3
5 ï PE = 3 5(4 – x)
PE + PF = 3
5(4 – x + x) =
12 5
40) If a + b + c = 3 , ab + ac + bc = 4 and abc = 5 , find 1
a2 + 1 b2 +
1 c2.
1
a2
+
1 b2+
1 c2 =
b2c2+a2c2+a2b2
abc2
(ab + ac + bc2 = 42 = a2b2 +a2c2 +b2c2 + 2 a2bc + ab2c +abc2
16 = a2b2 +a2c2 + b2c2 + 2 5 3
a2b2 + a2c2 +b2c2 = 16 – 2 5 3 = – 14
1
a2
+
1 b2+
1 c2 =
–14 5 2 = –
14 25
41) Lines l1 and l2 intersect in the plane as shown. Circles
A and B, each of radius 1 cm, are drawn tangent to l1
and l2 as shown. Circle C is then constructed tangent
to l1, l2, and A (externally), and circle D is constructed
tangent to l1,l2, and B (externally), as shown. If the
radius of circle C is 4 cm, find the radius of circle D, in cm.
A B
C
D
l1
l2
A B
C
E X F
Y Z
G
H
Let GX = y, YX = t, ZX = w, EX = x,
—
CXG =—
YXH = a and—
HXZ =—
ZXF = b. Given CG = 4.By similar triangles CGX and AEX, 4
y = 1
x ï y = 4x
A B
E X F
Y
Z
α β
α β
A B
E X F
Y
Z
α β
α β
H
By similar triangles CGX and AEX, t+1
x = 4+2+t
y ï t+1
x = 6+t
4x ï 4t + 4 = 6 + t ï t = 2 3
2a + 2b = p ï a + b = p
2
From triangle AEX, sin(a) = 1
1+23
ï sin(a) = 3
5 ï cos(a) = 4
5. Since a and b are complimentary,
sin(b) = 4
5 and cos(b) = 3 5.
From triangle XBF, sin(b) = 1
1+w = 4
5 ï w = 1 4
So if R is the radius of circle D, sin(b) = R
R+2+1
4
= 4
5 ï 4 5 =
R
R+9
4
