UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS FORTY-SIXTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 12, 2003
Solution 1
1) 1 2–
1 5 2 ·
10 10 =
5 – 2 20 =
3 20
2)
85 3– 93 2 6– 1+2– 2
– 1
= 6– 1+2– 2 85 3– 93 2 =
1 6+
1 4 32 27 =
1 6+
1 4 5 ·
12 12 =
2+3 60 =
1 12
3)
a Ñ b = a2 + ab – b2. 7 Ñ x = 59 . 72 + 7x – x2 = 59
x2 – 7x + 10 = 0 (x – 2)(x – 5) = 0 x = 2 , 5
4) 1
x+2 = 1
x –
1
2 2x (x + 2) 2x = 2(x + 2) – x(x + 2) 2x = 2x + 4 – x2 – 2x x2 + 2x – 4 = 0
x = – 2≤ 4+16 2 =
– 2≤ 20 2 =
– 2≤2 5 2
x = – 1 ± 5
x = – 1 + 5 (positive solution) 5)
Number rebroadcast by C = 460, number rebroadcast by B and not by C = 20. Total number rebroadcast = 460 + 20 = 480.
% = 480
3G = 6B ï G = 2B
2Y = 5B ï Y = 5 2B 4W = 6B ï W= 3
2B 4G + 2Y + 2W = 4(2B) + 2 5
2 B + 2 3
2B = 8B + 5B + 3B = 16B 7)
A
B
C
D
O
E
x
r
x2 = r2 – r 2
4 = 3 4 r
2 ï x = 3 2 r =
3
2 3 = 3 2
8)
Let x be the number of pieces of candy in each jar. Bert gets 25 + x – 17 = x + 8 pieces
Karla gets x – 25 + 17 = x – 8 pieces
Thus Bert has x + 8 – (x – 8) = 16 more pieces. 9)
Quiz 1 13
20 = 65% – 14% , below the 79% average Quiz 2 41
50 = 82% + 3% , above the 79% average Quiz 3 x
30 must be + 11% , above the 79% average or x
30· 100 = 90% Thus x = 27
10) 5002 1272– 1232 =
500ÿ500 127+123 127 – 123 =
500ÿ500
2,4
−
2,2
a,b
c,d
Equating the changes in the x and y coordinates of opposite sides of the square; a– 2 = – 2 –c
b– 4 = 2 –d a + 2 = 2 –c
b– 2 = 4 –d
Adding these four equations;
2a + 2b – 6 = – 2c – 2d + 6 ï 2a + 2b + 2c + 2d = 12 Thus a + b + c + d = 6
12)
4x – 12 = 2(x – 12) · 4 4x – 12 = 8x – 96 4x = 84 ï x = 21 Question 12
Solution 13
Solution 13 13)
A
B
C
D
α α
β
π
–
β
x
16 –x
7
12
By the law of sines sina
x = sinb
sina
16 –x = sinp–b
12 = sinb
12 (2) Solving (1) and (2) for sina
sinb
sina
sinb =
x 7 =
16 –x 12
12x = 112 – 7x ï 19x = 112 ï x = 112 19
14)
Number the positions 1 through 8 from left to right. First place the two C ' s.
Place the two C ' s in two of positions 1, 2 and 3. This can be done in 3 ways. This forces the three B ' s to be placed in positions 7, 8 and whichever of 1, 2 and 3 not occupied by a C. For example, CCBAAABB.
Place the two C ' s in two of positions 4, 5 and 6. This can be done in 3 ways. This forces the three A ' s to be placed in positions 7, 8 and whichever of 4, 5 and 6 not occupied by a C. For example, BBBCACAA.
Place one C in one of positions 1, 2 and 3 and the other C in one of positions 4, 5 and 6. This can be done in 3 · 3 = 9 ways. This forces two of the B ' s into the two of 1, 2 and 3 not occupied by a C and two of the A ' s into the two of 4, 5 and 6 not occupied by a C. The remaining A and B can be placed in positions 7 and 8 in two ways for a total or 2 · 9 = 18 ways. For example, CBBCAABA. Thus the total is 3 + 3 + 18 = 24 ways.
15)
Let x = number of accidents on rural roads, mx = number of 100,000 miles driven on rural roads, y = number of accidents on city roads and my = number of 100,000 miles driven on city roads. The the given information is:
x
mx = 3.2 (1)
y
my = 1.8 (2)
x+y
mx+my = 2.4 (3)
Solve (1) and (2) for mx and my and substitute into (3) x+y
x 3.2+
y 1.8
= 2.4
x + y = 2.4 3.2x +
2.4 1.8y
x + y = 3 4x +
4 3y
1 4x =
1 3y ï
x y =
4 3
16)
a2+b 2
a2 1 +a– 2b 2
b2 a 2b– 2+1 = a2 b2 = 121
Thus a
b = 11 ï a = 11b
a + b = 11b + b = 12 b § 1000 1000
12 = 83 + 1
3 Thus there are 83 choices for the pair (a , b). 17)
-6 -4 -2 2
2 4 6 8 10
−1,5 y=3x+20
y−5 = −1 3 x+1
The closest point is the intersection of the circle and the straight line through (– 1 , 5) which is perpendicular to y = 3x + 20 .
The perpendicular straight line has equation y – 5 = – 1 3(x + 1). Thus, we solve simultaneously:
y– 5=– 1
3 x+1 1
x + 1 2+ y – 5 2 =10 2
Substituting from (1) into (2)
x+1 2 + 1 9 x+1
2 = 10 ï 10 9 x+1
2 = 10 ï x+1 2 = 9
x + 1 = ± 3 ï x = – 4 or 2 The desired value is the smaller solution, i.e. x = – 4. Substituting in (1) gives y = 6. Thus the closest point is (x , y) = (– 4 , 6)
18)
The total number of balls is 15. The total number of ways to pick 4 of the 15 balls is 15
4 = 1365.
The desired count is the number of choices which contain 0, 1 or 2 balls of the same color.
4 of the same color: Red in 5
4 = 5 ways, blue in 1 way and green in 1 way for a total of 7 ways.
3 of the same color: Red in 5
3 10
1 = 100 ways, blue in 4 3
11
1 = 44 ways and green in 4 3
11
1 = 44 ways for a total of
188 ways.
Then the desired count is 1365 – 7 – 188 = 1170
Thus the required probability is 1170 1365 =
6 7. 19)
a0 = 2 , a1 = 8 and an =
an– 1 an– 2
a2 = a1 a0
= 8 2 = 4
a3 = a2 a1 =
4 8 =
1 2
a4 = a3 a2 =
1 2 4 =
1 8
a5 = a4 a3 =
1 8 1 2
= 1 4
a6 = a5 a4 =
1 4 1 8
= 2
a7 = a6 a5 =
2 1 4
= 8
Thus the sequence is periodic with period 6. 2003 ª 5 mod 6 . Thus a2003 = a5 = 1 4
20
Draw perpendiculars from vertices B and C to side AD.
10
20
25
A
B
C
D
x
h
E
F
5
From triangle ACF h2 = 252 – 152 = 400 ï h = 20
A
B
C
D
x
x
x
x
x
y
y
y
y
y
y
Area of each small rectangle is xy. The total area is 7xy = 756. By construction, 3x = 4y ï y = 3 4x. Thus 756 = 7x 3
4x 756 = 21
4 x 2
x2 = 4 756
21 = 144 ï x = 12 ï y = 3 4· 12 = 9 The perimeter of ABCD is 5x + 6y = 5(12) + 6(9) = 114 22)
cos(81°) + cos(39°) = cos(b)
cos(A + B) = cos(A) cos(B) – sin(A) sin(B) cos(A – B) = cos(A) cos(B) + sin(A) sin(B) Solving for cos(A + B) + cos(A – B) cos(A + B) + cos(A – B) = 2cos(A) cos(B) Then A + B = 81° and A – B = 39°.
Solving gives A = 60° and B = 21°. Since 2cos(60°) = 1, B = 21° 23)
r
R
14
14
Thus the required area is 196 . 24)
Let S = { x1 , x2 , x3 , x4 , x5 } with x1 < x2 < x3 < x4 < x5.
There are 5 4–element subsets of S. Each element of S appears in 4 of these subsets. Thus the total of the 5 subsets is: 4( x1 + x2 + x3 + x4 + x5) = 169 + 153 + 182 + 193 + 127 = 824 ï x1 + x2 + x3 + x4 + x5 = 206
By construction, the smallest sum is x1 + x2 + x3 + x4 = 127. Thusx5 = 206 – 127 = 79 25)
z1 = 0 and zn+1 = zn 2 + i
z2 = 02 + i = i z3 = i2 + i = – 1 + i
z4 = –1+i 2 + i = 1 – 2i – 1 + i = – i z5 = i 2 + i = – 1 + i
z4 = –1+i 2 + i = 1 – 2i – 1 + i = – i
Thus for n > 2 if n ª 0 mod, zn = – i and if n mod 2 = 1, zn = – 1 + i. Then z2003 = – 1 + i. 26)
x
log32 = 81 ïx
log32 = 34 andx
log32 2=
x
log32 log32ï
x
log32 2= 34 log32 = 34 log32 = 3log324 = 16
27)
4
x
+ 8 4
x = 6
Let t = 4x. Then the equation is
t + 8
t = 6 ï t
2 – 6t + 8 = 0
(t – 2)(t – 4) = 0 ï t = 2 or t = 4
4x = 2 or 4x = 4 ï x = 1
2 or x = 1 Thus the sum of the real solutions is 1
x
y
r
r r x
x
r
r
y y
B 24,32 A 12,32
C 24,16
Construct perpendiculars from the center of the circle to the sides of the circumscribed triangle. Label as indicated above.
x + r = 12 and y + r = 16 ï x + y = 122 + 162 = 20
Then the perimeter of triangle ABC is 2x + 2y + 2r = 12 + 16 + 20 = 48 ï x + y + r = 24 Now x + y + r = 24 and x + r = 12 ï y = 12
Now x + y + r = 24 and y + r = 16 ï x = 8
If the center of the circle is (h , k), then (h , k) = (12 + 8 , 16 + 12) = (20 , 28) 29)
2003! = 2003 · 2002 · 2001 · · · 3 · 2 · 1
This product will end in one zero for each pair of factors 2 and 5 appearing in the product.
Since there are more occurrences of 2 than 5, we need only count the number of 5's in the product. Let x = integer part of x.
Then the number of 5's in the product is: 2003
5 + 2003
25 + 2003
125 + 2003
625 = 400 + 80 + 16 + 3 = 499 30)
Let P = number who passed and F = number who failed. Find P F+P
75P + 35F = 60F + 60P ï 15P = 25F ï F = 3 5P P
F+P = P 3 5P+P
= 18 5
= 58
31)
2 cos(2x) + 1 = 0 ï cos(2x) = – 1 2
Thus 2x = 2p
3 + 2kp or 4p
3 + 2kp for k = 0, 1, 2, · · · 99
x = p
3 + kp or 2p
3 + kp for k = 0, 1, 2, · · · 99 For each k, p
3 + kp + 2p
3 + kp = (2k + 1) p Thus the sum of all the real zeros in [0 , 100p] is:
p + 3p + 5p + · · · + 199p = p(1 + 3 + 5 + · · · + 199) = p · 1002 = 10000 32)
C
A
B
F
G
D
E
α α αArea triangle CDE = 1
2CD cos(a) DE = b (1) Area triangle CFG = 1
2CF cos(a) DE = 2b (2) Area triangle CAB = 1
2CA cos(a) AB = 3b (3) 1
2 = CDÿDE CFÿFG =
1
2 By similar triangles DE FG =
CD CF ï
CD CF 2
= 1 2 ï
CD CF =
1
2 ï CF = 2 CD 1
3 = CDÿDE CAÿAB =
1
3 By similar triangles DE AB =
CD CA ï
CD CA 2
= 1 3 ï
CD CA =
1
3 ï CA = 3 CD CD
FA = CD CA – CF =
CD
3 CD – 2 CD = 1
3 – 2 Rationalizing the denominator CD
FA = 3 – 2 33)
Let r1, r2 and r3 be the zeros of the x3 + ax2 + bx + c. Then
= x3 – r1 +r2 +r3 x2 + r1r2 +r1r3 +r2r3 x + r1r2r3 Thus a = – r1 +r2 +r3 b = r1r2 + r1r3 + r2r3 c = – r1r2r3
Given r1 +r2 +r3 = 2r1r2r3 ï a = 2c
r12 + r
22 + r32 = 3r1r2r3 ï r12 + r22 + r32 = – 3c p(1) = 1 ï a + b + c = 0 ï b = – a – c
r1 +r2 +r3 2 = r12 + r22 + r32 + 2 r1r2 +r1r3 +r2r3 Using the above: –2c 2 = – 3c + 2(b) = – 3c + 2(– 2c – c)
4c2 = – 9c ï c(4c + 9) = 0 ï c = 0 or c = – 9
4 Since c cannot be zero (c = 0 would imply a = b = 0) c = – 9 4
34)
Let f = x y
x2+y2 · 1 xy 1 xy
= x1 y+ y x
if 2 5 § x §
1 2 and
1 3 § y §
3 8
Then the minimum value of f is the maximum value of x y +
y x = a +
1
a where a = x y.
2 5 § x §
1 2 and
1 3 § y §
3 8 ï
2 5 3 8
§ a § 1 2 1 3
ï 16 15 § a §
3 2
Since a + 1
a is increasing on 16 15 § a §
3
2, the maximum occurs when a = 3 2. Thus the maximum of a + 1
a = 3 2 +
2 3 =
13
6 and the minimum value of x y
The area of the triangle is 1
2· 4 · 3 = 1
2· 5 · r + 1
2· 3 · r + 1
2· 4 · r where r is the radius of the inscribed circle.
Thus r = 1. The line CD is a perpendicular bisector of line EF. If b is the angle at C, sin b 2 =
t 2 2
Then t = 4 · sin b 2 ï t
2 = 16 · sin2 b 2 = 16 ·
1 – cosb
2 = 8(1 – 3
5) since cos(b) = 3
5 from the given right triangle. t2 = 8 · 2
5 = 16
5
36)
A
B
C
D
a
r
r
Construct line CD through the center of the circle. Add line BD. The angles at A and D are the same since they subtend the same arc. The angle DBC is a right angle since it is inscribed in a semicircle.
Thus sin(A) = sin(D) = a
2r. Similarly, sin(B) = b
2r and sin(C) = c 2r
sin(A) + sin(B) + sin(C) = a b c 2r =
31 20
5 10 15 5
10 15
20
B 15,20
A
C m,n
The length of AB is 152 +202 = 25 and an equation of the line containing AB is
y – 20 = 4
5(x – 15) or 4x – 3y = 0.
The height h of the triangle is distance from C to this line. h = 4m– 3n 42+32
= 4m– 3n 5
If m and n are integers and C is not on the line, 4m– 3n ¥ 1 ï h ¥ 1 5
Thus the minimum occurs when h = 1
5 and the minimum area is 1 2· 25 ·
1 5 =
5 2
38)
A
B
C
D
α
3
α
O
β
x
Draw segments from points B and C to the center O of the circle. These segments are perpendicular to the respective tangents to the circle.
From triangle BAC, x + 2a = p. Solve 2x– 3α = π x + 2α = π ï
4x– 6α = 2π 3x + 6α = 3π
ï 7x = 5p ï x = 57
39)
n(a,b) = aba + bab = 100 a + 10 b + a + 100 b + 10 a + b = (100 + 10 + 1)(a + b) = 111 (a + b) There are 100 pairs (a,b) for a total of 200 digits. Each digit occurring 20 times.
Thus the total is 20(0 + 1 + 2 + · · · + 9)(111) = 20(45)(111) = 99900 40)
Let x = a + h where a is an integer and 0 § h < 1.
If x > 0, the largest value of [ x ] is 9 ï 9 § x < 10 and f(x) = 9x ï 81 § f(x) < 90 ï max(f(x)) = 89
If x < 0, the smallest value of [ x ] is – 10 ï – 10 < x § – 9 and f(x) = – 10x ï 90 § f(x) < 100 ï max(f(x)) = 99 Thus for all x max(f(x)) = 99
41)
r
1r
2Let r1 and r2 be the radii of two circles as indicated. From the right triangle in the sketch, sin p
6 = 1 2 =
r1–r2 r1+r2
ï r2 = 1 3r1
This pattern continues. Hence r3 = 1 3 r2 =
1
32r1 · · · etc.
Then the sum of the areas is A = p r12 + p r1 3
2 + p r1
32 2
+ · · ·
A = p r12[ 1 + 1 9 +
1 92 +
1
93 + · · · ] = p r1 2 1
1 –1 9
= 9 8p r1
Again from the sketch, tan p 6 =
r1 1 2
ï r1 = 1 2 tan
p
6 = 1 2
1
3 = 3 6
Thus A = 9 8p
3 6
2 = 9
8 · 3 36 · p =
