DISINI test_03_Solutions
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 – 1
=
6–1 + 2–2
6–1 + 2–2
85
3
– 93
2
1
=
6
+
32
1
4
27
1
=
6
+
5
1
4
·
12
12
=
2+3
60
=
1
12
3)
a Ñ b = a 2 + ab – b 2 . 7 Ñ x = 59 .
72 + 7x – x 2 = 59
x 2 – 7x + 10 = 0
(x – 2)(x – 5) = 0
x=2,5
4)
1
1 1
= –
2x (x + 2)
x+2
x 2
2x = 2(x + 2) – x(x + 2)
2x = 2x + 4 – x 2 – 2x
x 2 + 2x – 4 = 0
x=
–2≤
4 + 16
2
x=–1±
5
x=–1+
5
=
–2≤
20
2
=
–2≤2
2
5
(positive solution)
5)
Number rebroadcast by C = 460, number rebroadcast by B and not by C = 20.
Total number rebroadcast = 460 + 20 = 480.
%=
6)
480
·
2000
100 = 24%
3G = 6B ï G = 2B
5
2Y = 5B ï Y = 2 B
3
4W = 6B ï W= 2 B
4G + 2Y + 2W = 4(2B) + 2
5
2
B +2
3
2
B = 8B + 5B + 3B = 16B
7)
E
A
B
x
r
C
D
O
x2 = r2 –
r2
4
=
3
4
r2 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
Thus x = 27
10)
500 2
=
127 2 – 123 2
11)
500 ÿ 500
127 + 123 127 – 123
=
500 ÿ 500
250 ÿ 4
= 2 · 125 = 250
100 = 90%
c,d
2,4
−2,2
a,b
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)
B
α α
12
7
β
A
x
π–β
16 –x
D
By the law of sines
sin a
x
=
sin b
7
(1)
C
sin a
16 – x
sin p – b
=
12
=
sin b
Solving (1) and (2) for
sin a
sin b
=
x
7
=
(2)
12
sin a
sin b
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 m y = number of 100,000 miles driven on city roads. The the given information is:
x
mx
y
my
= 3.2
(1)
= 1.8
(2)
x+ y
mx + m y
= 2.4
(3)
Solve (1) and (2) for mx and m y and substitute into (3)
x+ y
x
3.2
+
= 2.4
y
1.8
x+y=
2.4
x
3.2
+
3
2.4
y
1.8
4
x + y = 4x + 3y
1
x
4
1
= 3y ï
x
y
=
4
3
16)
a2 + b 2
= 121
a 2 + b2
Factor as
a2 1 + a–2 b
2
b a
Thus
2
a
b
b
–2
2
+1
=
a2
b2
= 121
= 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)
y=3x+20 10
8
y−5 = −
1
x+1
3
6
−1,5
4
2
-6
-4
-2
2
The closest point is the intersection of the circle and the straight line through (– 1 , 5) which is perpendicular to y = 3x + 20 .
1
The perpendicular straight line has equation y – 5 = – 3 (x + 1).
Thus, we solve simultaneously:
y–5=–
2
x + 1
1
3
x+1
+ y – 5
1
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.
We can find this count by subtracting from the total the number of choices which contain 3 or 4 balls of the same color.
4 of the same color: Red in
5
4
3 of the same color: Red in
5
3
= 5 ways, blue in 1 way and green in 1 way for a total of 7 ways.
10
1
= 100 ways, blue in
4
3
11
1
= 44 ways and green in
188 ways.
Then the desired count is 1365 – 7 – 188 = 1170
Thus the required probability is
1170
1365
6
= 7.
19)
a 0 = 2 , a 1 = 8 and a n =
a2 =
a3 =
a4 =
a1
a0
a2
a1
a3
a2
=
8
2
=4
=
4
8
=
1
2
=
1
8
=
1
4
an–1
an–2
1
=
2
4
1
a5 =
a4
a3
=
8
1
2
1
a6 =
a5
a4
=
4
1
=2
8
a7 =
a6
a5
=
2
1
=8
4
Thus the sequence is periodic with period 6. 2003 ª 5 mod 6 . Thus a 2003 = a 5 =
20
Draw perpendiculars from vertices B and C to side AD.
10
B
C
25
x
h
5
A
E
20
D
F
From triangle ACF h 2 = 25 2 – 15 2 = 400 ï h = 20
From triangle CFD x 2 = 20 2 + 5 2 = 425 ï x = 5
21)
17 . Thus CD = 5
17
1
4
4
3
11
1
= 44 ways for a total of
B
x
x
C
x
y
y
x
x
y
A
y
y
y
D
3
Area of each small rectangle is xy. The total area is 7xy = 756. By construction, 3x = 4y ï y = 4 x.
Thus 756 = 7x
756 =
x2 =
21
4
3
4
x
x2
4 756
21
3
= 144 ï x = 12 ï y = 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)
14
14
r
R
If the radii of the circles C 1 and C 2 are R and r, the area between the circles if p ( R 2 – r 2 ). From the figure R 2 – r 2 = 14 2 = 196.
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. Thus x5 = 206 – 127 = 79
25)
z 1 = 0 and z n + 1 = z n
2
+i
z2 = 02 + i = i
z3 = i2 + i = – 1 + i
z 4 = –1 + i
z5 = i
2
2
+ i = 1 – 2i – 1 + i = – i
+i=–1+i
z 4 = –1 + i
2
+ i = 1 – 2i – 1 + i = – i
Thus for n > 2 if n ª 0 mod, z n = – i and if n mod 2 = 1, z n = – 1 + i. Then z 2003 = – 1 + i .
26)
x log 3 2 = 81
x log 3 2
ï
2
= 34
x log 3 2 = 3 4
log 3 2
and
x log 3 2
4
= 3 4 log 3 2 = 3 log 3 2 = 16
27)
x
8
4 + x =6
4
Let t = 4 x . 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
4 x = 2 or 4 x = 4 ï x =
1
2
or x = 1
Thus the sum of the real solutions is
28)
1
2
3
+ 1 = 2.
2
=
x log 3 2
log 3 2
ï
y
x
A 12,32
r
r
x
r
B 24,32
r
r
y
y
C 24,16
x
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 =
12 2 + 16 2 = 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
From the given information 75P + 35F = 60(F + P)
P
F+P
3
75P + 35F = 60F + 60P ï 15P = 25F ï F = 5 P
P
F+P
=
P
3
5
1
=
P+P
8
5
8
=
5
31)
1
2
2 cos(2x) + 1 = 0 ï cos(2x) = –
Thus 2x =
x=
p
3
2p
3
2p
3
+ kp or
For each k,
p
3
4p
3
+ 2kp or
+ kp
+ kp +
for k = 0, 1, 2, · · · 99
+ 2kp
for k = 0, 1, 2, · · · 99
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 · 100 2 = 10000
32)
C
α
D
E
α
F
G
α
A
B
1
Area triangle CDE = 2 CD cos(a) DE = b
(1)
1
Area triangle CFG = 2 CF cos(a) DE = 2b
(2)
1
Area triangle CAB = 2 CA cos(a) AB = 3b
(3)
1
2
=
CD ÿ DE
CF ÿ FG
=
1
2
By similar triangles
DE
FG
=
CD
CF
ï
CD 2
CF
=
1
2
ï
CD
CF
=
1
1
3
=
CD ÿ DE
CA ÿ AB
=
1
3
By similar triangles
DE
AB
=
CD
CA
ï
CD 2
CA
=
1
3
ï
CD
CA
=
1
CD
FA
=
CD
CA – CF
CD
FA
=
3 –
CD
=
3 CD –
2 CD
=
1
3 –
2
Rationalizing the denominator
2
33)
Let r 1 , r 2 and r 3 be the zeros of the x 3 + a x 2 + b x + c. Then
x 3 + a x 2 + b x + c = (x – r 1 )(x – r 2 )(x – r 3 )
2
3
ï CF =
2 CD
ï CA =
3 CD
= x3 – r1 + r2 + r3 x2 + r1 r2 + r1 r3 + r2 r3 x + r1 r2 r3
Thus a = – r 1 + r 2 + r 3
b = r1 r2 + r1 r3 + r2 r3
Given r 1 + r 2 + r 3 = 2r 1 r 2 r 3
c = – r1 r2 r3
ï a = 2c
r 1 2 + r 2 2 + r 3 2 = 3r 1 r 2 r 3 ï r 1 2 + r 2 2 + r 3 2 = – 3c
p(1) = 1 ï a + b + c = 0 ï b = – a – c
r1 + r2 + r3
–2c
2
2
= r 12 + r 22 + r 3 2 + 2 r 1 r 2 + r 1 r 3 + r 2 r 3
Using the above:
= – 3c + 2(b) = – 3c + 2(– 2c – c)
9
4
4 c 2 = – 9c ï c(4c + 9) = 0 ï c = 0 or c = –
Since c cannot be zero (c = 0 would imply a = b = 0) c = –
34)
xy
·
2
x + y2
1
and
1
3
§y§
Then the minimum value of f is the maximum value of
x
y
+
Let f =
xy
=
1
1
x
y
xy
2
5
if
y
+
§x§
1
2
and
1
3
§y§
3
8
ï
1
a
is increasing on
Thus the maximum of a +
3
8
1
a
16
15
=
3
2
y
x
=a+
2
1
ï
16
15
§a§
3
3
3
§ a § 2 , the maximum occurs when a = 2 .
+
2
3
=
13
6
and the minimum value of
B
3
5
F
D
t
1
β
A
1
E
x
where a = y .
3
2
35)
r
1
a
1
§a§
5
3
8
Since a +
1
2
x
2
2
5
§x§
2
C
xy
x
2
y2
is
6
.
13
9
4
1
1
1
1
The area of the triangle is 2 · 4 · 3 = 2 · 5 · r + 2 · 3 · r + 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
Then t = 4 · sin
2
5
t2 = 8 ·
=
b
2
ï t 2 = 16 · sin 2
b
2
= 16 ·
1 – cos b
2
3
= 8(1 – 5 )
since cos(b) =
3
5
b
2
t
=
2
2
from the given right triangle.
16
5
36)
B
a
C
A
r
r
D
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) =
sin(B) =
b
2r
a
.
2r
and sin(C) =
c
2r
sin(A) + sin(B) + sin(C) =
37)
Similarly,
a
b c
2r
=
31
20
C m,n
B 15,20
20
15
10
5
5
A
10
The length of AB is
15
152 + 20 2 = 25 and an equation of the line containing AB is
4
y – 20 = 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 ¥1 ï h ¥
If m and n are integers and C is not on the line,
Thus the minimum occurs when h =
1
5
1
and the minimum area is 2 · 25 ·
1
5
4m–3n
5
1
5
=
5
2
38)
D
3α
A
x
B
α
C
β
O
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 quadrilateral OBDC, 3a + p + b = 2p. Also by comparing arcs, b = 2p – 2x. These together give 2x – 3a = p.
From triangle BAC, x + 2a = p. Solve
ï 7x = 5p ï x =
2x–3α = π
ï
x + 2α = π
4x–6α = 2π
3x + 6α = 3π
5
7
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)
r1
r2
Let 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
3
r1
This pattern continues. Hence r3 =
1
3
r2 =
Then the sum of the areas is A = p r1 2 + p
A = p r1 2 [ 1 +
1
9
+
1
92
+
1
93
+ · · · ] = p r1 2
1
32
r1 · · · etc.
r1 2
3
1
1–
+p
9
1
9
r1 2
32
= 8 p r1 2
+···
Again from the sketch, tan
p
6
=
r1
1
ï r1 =
2
9
Thus A = 8 p
3
6
2
=
9
8
·
3
36
·p=
3
32
1
2
tan
p
6
=
1
2
1
3
=
3
6
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 – 1
=
6–1 + 2–2
6–1 + 2–2
85
3
– 93
2
1
=
6
+
32
1
4
27
1
=
6
+
5
1
4
·
12
12
=
2+3
60
=
1
12
3)
a Ñ b = a 2 + ab – b 2 . 7 Ñ x = 59 .
72 + 7x – x 2 = 59
x 2 – 7x + 10 = 0
(x – 2)(x – 5) = 0
x=2,5
4)
1
1 1
= –
2x (x + 2)
x+2
x 2
2x = 2(x + 2) – x(x + 2)
2x = 2x + 4 – x 2 – 2x
x 2 + 2x – 4 = 0
x=
–2≤
4 + 16
2
x=–1±
5
x=–1+
5
=
–2≤
20
2
=
–2≤2
2
5
(positive solution)
5)
Number rebroadcast by C = 460, number rebroadcast by B and not by C = 20.
Total number rebroadcast = 460 + 20 = 480.
%=
6)
480
·
2000
100 = 24%
3G = 6B ï G = 2B
5
2Y = 5B ï Y = 2 B
3
4W = 6B ï W= 2 B
4G + 2Y + 2W = 4(2B) + 2
5
2
B +2
3
2
B = 8B + 5B + 3B = 16B
7)
E
A
B
x
r
C
D
O
x2 = r2 –
r2
4
=
3
4
r2 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
Thus x = 27
10)
500 2
=
127 2 – 123 2
11)
500 ÿ 500
127 + 123 127 – 123
=
500 ÿ 500
250 ÿ 4
= 2 · 125 = 250
100 = 90%
c,d
2,4
−2,2
a,b
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)
B
α α
12
7
β
A
x
π–β
16 –x
D
By the law of sines
sin a
x
=
sin b
7
(1)
C
sin a
16 – x
sin p – b
=
12
=
sin b
Solving (1) and (2) for
sin a
sin b
=
x
7
=
(2)
12
sin a
sin b
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 m y = number of 100,000 miles driven on city roads. The the given information is:
x
mx
y
my
= 3.2
(1)
= 1.8
(2)
x+ y
mx + m y
= 2.4
(3)
Solve (1) and (2) for mx and m y and substitute into (3)
x+ y
x
3.2
+
= 2.4
y
1.8
x+y=
2.4
x
3.2
+
3
2.4
y
1.8
4
x + y = 4x + 3y
1
x
4
1
= 3y ï
x
y
=
4
3
16)
a2 + b 2
= 121
a 2 + b2
Factor as
a2 1 + a–2 b
2
b a
Thus
2
a
b
b
–2
2
+1
=
a2
b2
= 121
= 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)
y=3x+20 10
8
y−5 = −
1
x+1
3
6
−1,5
4
2
-6
-4
-2
2
The closest point is the intersection of the circle and the straight line through (– 1 , 5) which is perpendicular to y = 3x + 20 .
1
The perpendicular straight line has equation y – 5 = – 3 (x + 1).
Thus, we solve simultaneously:
y–5=–
2
x + 1
1
3
x+1
+ y – 5
1
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.
We can find this count by subtracting from the total the number of choices which contain 3 or 4 balls of the same color.
4 of the same color: Red in
5
4
3 of the same color: Red in
5
3
= 5 ways, blue in 1 way and green in 1 way for a total of 7 ways.
10
1
= 100 ways, blue in
4
3
11
1
= 44 ways and green in
188 ways.
Then the desired count is 1365 – 7 – 188 = 1170
Thus the required probability is
1170
1365
6
= 7.
19)
a 0 = 2 , a 1 = 8 and a n =
a2 =
a3 =
a4 =
a1
a0
a2
a1
a3
a2
=
8
2
=4
=
4
8
=
1
2
=
1
8
=
1
4
an–1
an–2
1
=
2
4
1
a5 =
a4
a3
=
8
1
2
1
a6 =
a5
a4
=
4
1
=2
8
a7 =
a6
a5
=
2
1
=8
4
Thus the sequence is periodic with period 6. 2003 ª 5 mod 6 . Thus a 2003 = a 5 =
20
Draw perpendiculars from vertices B and C to side AD.
10
B
C
25
x
h
5
A
E
20
D
F
From triangle ACF h 2 = 25 2 – 15 2 = 400 ï h = 20
From triangle CFD x 2 = 20 2 + 5 2 = 425 ï x = 5
21)
17 . Thus CD = 5
17
1
4
4
3
11
1
= 44 ways for a total of
B
x
x
C
x
y
y
x
x
y
A
y
y
y
D
3
Area of each small rectangle is xy. The total area is 7xy = 756. By construction, 3x = 4y ï y = 4 x.
Thus 756 = 7x
756 =
x2 =
21
4
3
4
x
x2
4 756
21
3
= 144 ï x = 12 ï y = 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)
14
14
r
R
If the radii of the circles C 1 and C 2 are R and r, the area between the circles if p ( R 2 – r 2 ). From the figure R 2 – r 2 = 14 2 = 196.
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. Thus x5 = 206 – 127 = 79
25)
z 1 = 0 and z n + 1 = z n
2
+i
z2 = 02 + i = i
z3 = i2 + i = – 1 + i
z 4 = –1 + i
z5 = i
2
2
+ i = 1 – 2i – 1 + i = – i
+i=–1+i
z 4 = –1 + i
2
+ i = 1 – 2i – 1 + i = – i
Thus for n > 2 if n ª 0 mod, z n = – i and if n mod 2 = 1, z n = – 1 + i. Then z 2003 = – 1 + i .
26)
x log 3 2 = 81
x log 3 2
ï
2
= 34
x log 3 2 = 3 4
log 3 2
and
x log 3 2
4
= 3 4 log 3 2 = 3 log 3 2 = 16
27)
x
8
4 + x =6
4
Let t = 4 x . 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
4 x = 2 or 4 x = 4 ï x =
1
2
or x = 1
Thus the sum of the real solutions is
28)
1
2
3
+ 1 = 2.
2
=
x log 3 2
log 3 2
ï
y
x
A 12,32
r
r
x
r
B 24,32
r
r
y
y
C 24,16
x
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 =
12 2 + 16 2 = 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
From the given information 75P + 35F = 60(F + P)
P
F+P
3
75P + 35F = 60F + 60P ï 15P = 25F ï F = 5 P
P
F+P
=
P
3
5
1
=
P+P
8
5
8
=
5
31)
1
2
2 cos(2x) + 1 = 0 ï cos(2x) = –
Thus 2x =
x=
p
3
2p
3
2p
3
+ kp or
For each k,
p
3
4p
3
+ 2kp or
+ kp
+ kp +
for k = 0, 1, 2, · · · 99
+ 2kp
for k = 0, 1, 2, · · · 99
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 · 100 2 = 10000
32)
C
α
D
E
α
F
G
α
A
B
1
Area triangle CDE = 2 CD cos(a) DE = b
(1)
1
Area triangle CFG = 2 CF cos(a) DE = 2b
(2)
1
Area triangle CAB = 2 CA cos(a) AB = 3b
(3)
1
2
=
CD ÿ DE
CF ÿ FG
=
1
2
By similar triangles
DE
FG
=
CD
CF
ï
CD 2
CF
=
1
2
ï
CD
CF
=
1
1
3
=
CD ÿ DE
CA ÿ AB
=
1
3
By similar triangles
DE
AB
=
CD
CA
ï
CD 2
CA
=
1
3
ï
CD
CA
=
1
CD
FA
=
CD
CA – CF
CD
FA
=
3 –
CD
=
3 CD –
2 CD
=
1
3 –
2
Rationalizing the denominator
2
33)
Let r 1 , r 2 and r 3 be the zeros of the x 3 + a x 2 + b x + c. Then
x 3 + a x 2 + b x + c = (x – r 1 )(x – r 2 )(x – r 3 )
2
3
ï CF =
2 CD
ï CA =
3 CD
= x3 – r1 + r2 + r3 x2 + r1 r2 + r1 r3 + r2 r3 x + r1 r2 r3
Thus a = – r 1 + r 2 + r 3
b = r1 r2 + r1 r3 + r2 r3
Given r 1 + r 2 + r 3 = 2r 1 r 2 r 3
c = – r1 r2 r3
ï a = 2c
r 1 2 + r 2 2 + r 3 2 = 3r 1 r 2 r 3 ï r 1 2 + r 2 2 + r 3 2 = – 3c
p(1) = 1 ï a + b + c = 0 ï b = – a – c
r1 + r2 + r3
–2c
2
2
= r 12 + r 22 + r 3 2 + 2 r 1 r 2 + r 1 r 3 + r 2 r 3
Using the above:
= – 3c + 2(b) = – 3c + 2(– 2c – c)
9
4
4 c 2 = – 9c ï c(4c + 9) = 0 ï c = 0 or c = –
Since c cannot be zero (c = 0 would imply a = b = 0) c = –
34)
xy
·
2
x + y2
1
and
1
3
§y§
Then the minimum value of f is the maximum value of
x
y
+
Let f =
xy
=
1
1
x
y
xy
2
5
if
y
+
§x§
1
2
and
1
3
§y§
3
8
ï
1
a
is increasing on
Thus the maximum of a +
3
8
1
a
16
15
=
3
2
y
x
=a+
2
1
ï
16
15
§a§
3
3
3
§ a § 2 , the maximum occurs when a = 2 .
+
2
3
=
13
6
and the minimum value of
B
3
5
F
D
t
1
β
A
1
E
x
where a = y .
3
2
35)
r
1
a
1
§a§
5
3
8
Since a +
1
2
x
2
2
5
§x§
2
C
xy
x
2
y2
is
6
.
13
9
4
1
1
1
1
The area of the triangle is 2 · 4 · 3 = 2 · 5 · r + 2 · 3 · r + 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
Then t = 4 · sin
2
5
t2 = 8 ·
=
b
2
ï t 2 = 16 · sin 2
b
2
= 16 ·
1 – cos b
2
3
= 8(1 – 5 )
since cos(b) =
3
5
b
2
t
=
2
2
from the given right triangle.
16
5
36)
B
a
C
A
r
r
D
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) =
sin(B) =
b
2r
a
.
2r
and sin(C) =
c
2r
sin(A) + sin(B) + sin(C) =
37)
Similarly,
a
b c
2r
=
31
20
C m,n
B 15,20
20
15
10
5
5
A
10
The length of AB is
15
152 + 20 2 = 25 and an equation of the line containing AB is
4
y – 20 = 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 ¥1 ï h ¥
If m and n are integers and C is not on the line,
Thus the minimum occurs when h =
1
5
1
and the minimum area is 2 · 25 ·
1
5
4m–3n
5
1
5
=
5
2
38)
D
3α
A
x
B
α
C
β
O
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 quadrilateral OBDC, 3a + p + b = 2p. Also by comparing arcs, b = 2p – 2x. These together give 2x – 3a = p.
From triangle BAC, x + 2a = p. Solve
ï 7x = 5p ï x =
2x–3α = π
ï
x + 2α = π
4x–6α = 2π
3x + 6α = 3π
5
7
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)
r1
r2
Let 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
3
r1
This pattern continues. Hence r3 =
1
3
r2 =
Then the sum of the areas is A = p r1 2 + p
A = p r1 2 [ 1 +
1
9
+
1
92
+
1
93
+ · · · ] = p r1 2
1
32
r1 · · · etc.
r1 2
3
1
1–
+p
9
1
9
r1 2
32
= 8 p r1 2
+···
Again from the sketch, tan
p
6
=
r1
1
ï r1 =
2
9
Thus A = 8 p
3
6
2
=
9
8
·
3
36
·p=
3
32
1
2
tan
p
6
=
1
2
1
3
=
3
6