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UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FORTY-SEVENTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 4, 2004
1) Express
+
as a rational number in lowest terms.
Solution 1
2) Solve for n:
=
. Express your answer as a rational number in lowest terms.
Solution 2
3) Express
as a rational number in lowest terms.
Solution 3
4) If
= 125, what is the value of
? Express your answer as a rational number in lowest terms .
Solution 4
Solution 5
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Solution 6
7) A triangle has angles α, β and γ. Angle γ is 10° larger than α and the difference between α and β is 40°. If β is the smallest
angle, find the degree measure of angle β.
Solution 7
8) Tank A contains a mixture of lime juice and water, where 80% is water and the rest is lime juice. Tank B contains pure
lime juice. If the contents of Tank A and Tank B are combined to fill a 200 liter tank with a mixture which is half lime
juice and half water, how many liters were in Tank A ?
Solution 8
9) Find the distance from the point (– 4 , 1) to the center of the circle whose equation is
+
+ 2x + 6y = 15.
Solution 9
10) Find all values of x that satisfy
+
–
=5.
Solution 10
11) The lengths of two opposite sides of a rectangle are each increased by 25% while the lengths of the other two sides are each
decreased by 40% . Find the percent decrease in the area of the rectangle.
Solution 11
12) When the two wheels shown in the figure are spun,
two numbers are selected. If the wheels are spun,
what is the probability that the sum of the two
numbers is even ?
Solution 12
13) Find the smallest positive integer divisible by each of 2, 6, 10, 15, 35 and 45.
Solution 13
14) Find the largest integer n such that
+
>
+
.
Solution 14
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15) As shown in the sketch, a line is tangent to a circle
centered at the origin. The point of tangency is
(4 , 3). The line intersects the x-axis at x = a.
Find a.
Solution 15
16) Find all positive real numbers a such that the points ( a , 12 ) and ( 5 , a ) lie on a straight line of slope a.
Solution 16
17) If a, b, c and d are real numbers such that
=
,
=
and
=
, determine the value of
. Express your
answer as a rational number in lowest terms.
Solution 17
18) As shown in the sketch, a square is cut into three
congruent rectangles by two lines parallel to a side.
If each of the three rectangles has a perimeter of
40 cm, determine the area of the square.
Solution 18
19) Express
+
+ 2004 as a rational number in lowest terms.
Solution 19
Solution 20
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21) As shown in the figure, six circles of radius 1 are
arranged so that each circle is tangent to two
others, while a seventh circle of radius 1 is tangent
to each of the other six. What is the area of the
shaded region ?
Solution 21
22) Let θ be an acute angle such that tan(θ) = 3. Find the value of cos(2θ). Express your answer as a rational number in lowest terms.
Solution 22
23) Let x and y be positive real numbers such that
=–
. Find the value of
–
. Express your
answer
as a rational number in lowest terms.
Solution 23
24) Find all real numbers x that satisfy
–
= 16x –
.
Solution 24
25) If x is a number such that 3x +
= 4 , what is the numerical value of
+
?
Solution 25
26) A function f satisfies f (x + y) = 4 f (x) f (y) for all real numbers x and y. If f (3) = 32, find f (1).
Solution 26
27) Express the product
as an integer
.
Solution 27
28) Suppose that x and y are real numbers such that x y = 8 and
Determine the value of
+
y
+ x + y = 144.
.
Solution 28
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29) Eve has a collection of 120 different sweaters. Each sweater is made of either cotton or wool, comes in one of three
styles (full length sleeve, three quarter length sleeve or sleeveless), displays one of four geometric patterns (squares,
triangles, stars, hexagons) and is one of five colors (red, green, blue, yellow, gray). How many of these sweaters
differ from the red, sleeveless wool sweater with the triangle pattern in exactly two ways ?
Solution 29
30) Find positive integers x and y such that
–
=
with x as small as possible.
Solution 30
31) Let f (x) =
. Find the largest value of n such that f (n) = f (2n).
Solution 31
32) If it takes 2004 digits to number the pages of a book, how many pages does the book contain ?
Solution 32
33) Express
·
·
···
as a rational number in lowest terms.
Solution 33
34) In a random arrangement of the letters AAAABBBCCD, what is the probability that no two A's are next to each other ?
Solution 34
35) After a difficult mathematics competition between two teams of nine students each, the eighteen competitors are ranked
1 through 18 (there are no ties). The team score is the sum of the ranks of the team members and the team with the
lower score is the winner of the competition. How many different winning scores are possible ?
Solution 35
36) Let
be a square with side length 1. In one
of the triangles formed by the diagonals of
inscribe a square
. In one of the triangles
formed by the diagonals of
inscribe a square
. This process is continued indefinitely.
If
is the area of square
, find
.
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Solution 36
37) Let ABC be a right triangle with
hypotenuse 4. On each side of ∆ ABC an
equilateral triangle is constructed outward
as shown in the figure. Find the sum of
the areas of the equilateral triangles.
Solution 37
38) For how many integers n, 1 ≤ n ≤ 2004, is the rational number
NOT in lowest terms ?
Solution 38
39) From a point P which lies outside a circle of
radius r units, two secants are drawn. The first
secant intersects the circle at points A and B
and the second secant intersects the circle at
points C and D. Given that AB = 14, CD = 2,
PA = 6 and
∠APC = 60°, find
.
Solution 39
40) Find the area of the region in the plane bounded by the straight lines y = x , y = 1 – x , y =
x and y = 1 – 2x .
Solution 40
41) Find all ordered pairs (x , y) of real numbers such that
.
Solution 41
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Solution 1
+
=
+
=
+
=
+
=
+
=
+
=
Question 1
Solution 2
=
=
⇒
=
⇒
⇒ 4n = 3n + 3 ⇒ n = 3
Question 2
Solution 3
=
=2·
=
Question 3
Solution 4
= 125 ⇒
=
⇒
=5 ⇒
= 25 ⇒
= 25 ⇒
=
Question 4
Solution 5
Question 5
Solution 6
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Question 6
Solution 7
Given γ = α +
(A) – (B)
and α – β = 40°
3β = 90° ⇒ β = 30°
Question 7
Solution 8
Let x = number of liters in Tank A and y = number of liters in Tank B.
.8x = 100 ⇒ x =
= 125
Solution 9
+
+ 2x + 6y = 15
+ 2x + 1 +
Complete the square
+ 6y + 9 = 15 + 1 + 9 ⇒
Distance between ( – 4 , 1 ) and ( – 1 , – 3)
+
d=
= 25 ⇒ center = ( – 1 , – 3)
=
=5
Solution 10
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+
–
=5
multiply by x (x + 3)
5 x (x) + (x + 2)(x + 3) – 6 = 5 x(x + 3) ⇒
+
+ 5x + 6 – 6 =
+ 15x
– 10x = 0 ⇒ x(x – 10) = 0 ⇒ x = 0 or x = 10 but x = 0 is extraneous so x = 10
Question 10
Solution 11
Original dimensions x × y new dimensions 1.25x × .6y
=
= .75 ⇒ 25% decrease
Question 11
Solution 12
There are 16 possible pairs. The pairs with even sum are (2,8), (4,8), (3,1), (3,5), (3,9), (5,1), (5,5), (5,9) or eight pairs.
Thus the probability of an even sum =
=
OR
P(E , E) + P(O , O) =
·
+
·
=
=
Question 12
Solution 13
2 = 2, 6 = 2 · 3, 10 = 2 · 5, 15 = 3 · 5, 35 = 5 · 7, 45 =
Thus the least common multiple = 2 ·
·5
· 5 · 7 = 630
Question 13
Solution 14
+
>
+
⇒
>
+
–
⇒
>
⇒ n>
⇒ n>6
Question 14
Solution 15
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Slope of line OP =
. This is perpendicular to the
tangent line. Slope of tangent line = –
–
=
. Thus
⇒ 4a = 25 ⇒ a =
Question 15
Solution 16
= a ⇒ a – 12 = 5a –
– 4a – 12 = 0 ⇒ (a – 6)(a + 2) = 0
⇒
a = 6 or a = – 2 and a positive ⇒ a = 6
Question 16
Solution 17
·
=
·
⇒
·
=
·
⇒
=
·
·
⇒
=
Question 17
Solution 18
2x +
= 40 ⇒
= 40 ⇒ x = 15.
Area =
=
= 225
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Question 18
Solution 19
+
+ 2004 =
=
=
= 2(20040 = 4008
Question 19
Solution 20
Since sin(x) =
=
s=
s=
s=1+1+1+1+1=5
Question 20
Solution 21
Area of an equilateral triangle of side 2 is
Area of the hexagon is
=
=
Area of sectors of 6 outer circles 6 ·
Area of inside circle = π.
Then shaded area =
– π – 2π =
=
.
.
·
= 2π
– 3π .
Question 21
Solution 22
tan(θ) = 3,
θ acute
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cos(2θ) = 2
=2·
–1=
–1=–
OR
cos(2θ) =
=
–
=
⇒
=–4
–
=–
= –
Question 22
Solution 23
=
(y) = –
–
=
(x) +
=1+
–
– 2(– 4) +
+
=1–
+8+
=
=
Question 23
Solution 24
–
x·
(1 +
x = 0 or
= 16x –
) – 16x (1 +
=
⇒ x·
– 16
)=0 ⇒ x·
– 16x +
=0
=0 ⇒
⇒ x = 0 or x =
Question 24
Solution 25
Given 3x +
=4 ⇒
= 64 ⇒
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+3·
·
+
+
·
+
+
= 64 –
+
= 64
= 64
(4) = 64 – 18 = 46
Question 25
Solution 26
f (x + y) = 4 f (x) f (y) for all real numbers x and y. f (3) = 32
f(3) = f(2 + 1) = 4 · f(2) · f(1)
f(2) = f(1 + 1) = 4 · f(1) · f(1)
32 = 4 · 4 · f(1) · f(1) · f(1)
= 2 ⇒ f(1) =
Question 26
Solution 27
=
=[
=
= 11 [ 5 +
– 11 ] [ 11 –
–
]
– 121 +
+7+5–
+ 7 ] – 4 – 121
= 11(24) – 125 = 264 – 125 = 139
OR
12 +
– 11 = 1 +
=
= 11 – (12 –
)=
–1
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=
= 4 · 35 – 1 = 139
Question 27
Solution 28
x y = 8 and
y
+ x + y = 144.
x(xy + 1) + y(xy + 1) = 144.
(xy + 1)(x + y) = 144 ⇒ x + y =
=
= 256
+ 2xy +
= 256
+
⇒ x+y=
= 16
= 256 – 2xy = 256 – 2(8) = 240
Question 28
Solution 29
There are 2 materials, 3 styles, 4 patterns and 5 colors. There are
= 6 ways that a sweater could differ from the wool, sleeveless,
triangle red sweater in exactly two ways. For each of the 6 pairs of "differences" we count the number of such sweaters and add for
the total.
material style pattern color # different
1
1
2
2
3
3
1
2
4
4
6
4
4
8
12
3
2
3
35
Question 29
Solution 30
–
=
⇒ (x – y)(x + y) =
(x – y)(x + y) = 1 ·
Since x and y are positive integers, the only possibilities are:
or
(x – y)(x + y) =
·
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Adding
2x = 1332
2x = 132
x = 666 and y = 665
x = 66 and y = 55
So the solution with minimal x is: (x , y) = (66 , 55)
Question 30
Solution 31
f (x) =
. Find the largest value of n such that f (n) = f (2n).
f (n) = f (2n) ⇒
=
n (9 +
) = 2n (9 +
)
n (9 +
– 18 –
n
= 0 ⇒ n = 0 or n =
)=0
Largest solution n =
Question 31
Solution 32
It takes 9 digits to number the pages 1 – 9, it takes 2(90) = 180 digits to number pages 10 – 99 leaving 2004 – 9 – 180 = 1815 digits
to number pages
100 – ??? Thus can number
= 605 three digit pages for a total of 605 + 99 = 704 pages.
Question 32
Solution 33
·
·
···
Note that each factor in the product is of the form
=
=
for n = 1, 2, 3, · · · 1002
=
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=
·
=
Thus
·
···
=
·
·
·
···
= 1003
Question 33
Solution 34
There are
= 5 · 5 · 7 · 8 · 9 ways total to arrange the 10 letters.
= 2 · 5 · 6 ways. For any such arrangement, e.g. BCBCDB we can place the 4 A's in 4 of the
Now arrange BBBCCD in
7
indicated places
__B__C__B__C__D__B__. There are
=
= 5 · 7 choices. Thus the probability is
=
Question 34
Solution 35
The total number of points is 1 + 2 + 3 + 4 + · · · + 18 =
The minimum score is 1 + 2 + 3 + · · · + 9 =
= 171
= 45 ⇒ the maximum score is 171 – 45 = 126
Thus the number of possible scores for one team is 126 – 45 + 1 = 82. Of these scores
= 41 are winning scores.
The minimum winning score is 45 and the maximum willing score is 85. To see that every score between 45 and 85 is attainable, let
<
<
<
<
<
<
<
<
be any set of rankings which add to a winning score and let
<
<
<
<
<
<
<
<
the rankings which add to the corresponding losing score. If we can show that
for any
such arrangement, there must be some values of i and j such that
=
+ 1, then swapping
score
by one. Suppose that no such i and j exist. Then we must have
=
+ 1,
=
+1=
+2, · ··
=
+ 8. Now the only way that
and
will increase the winning
+ 8 could fail to be one of the
's is if
= 18. This
would result in
= 17,
= 16 , · · ·
= 10 which is not a winning score. Now since 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 is
attainable,
we can continue to increase this score by 1 as above, until we reach the maximum winning score of 85.
Question 35
Solution 36
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Considering one step in the process. If the larger square has side s and the smaller square has side x, then x =
s. See the figure.
Thus if the original square has side length 1,
A=
+
+
+
+· · · =1+
+
+
+· · ·=
=
Question 36
Solution 37
The area of an equilateral triangle of side s is
From the right triangle
+
=
⇒ A=
. Thus the of the areas of the three triangles is A =
(16 + 16) = 8
Question 37
Solution 38
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By division
=n–6+
Thus the original fraction will NOT be in lowest terms when n + 6 is an integer multiple
of the prime number 47.
For the values n = 1, 2, 3, · · · 2004,
n + 6 is a multiple of 47
= 42 times.
Question 38
Solution 39
It is a fact that secant lines to a circle from a common point satisfy a (a + b) = c (c + d). See the above figure.
Thus x (x + 2) = 6 (6 + 14) ⇒
+ 2x – 120 = 0 ⇒ (x – 10)(x + 12) = 0 ⇒ x = 10 or x = – 12. So PC = 10.
By the Law of Cosines in triangle PCB,
Since
=
–
=
+
– 2(10)(20) cos(60°) = 100 + 400 – 200 = 300
, triangle PCB must be a right triangle with right angle at C.
From right triangle CDB,
= 300 + 4 ⇒
= 304 ⇒
=
= 76
Question 39
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Solution 40
Solving the appropriate pairs of equations gives the points
,
,
,
.
The area bounded by the four straight lines can be computed by finding the area of the "dashed" rectangle and subtracting the areas of
the four right triangles.
–
A=
=
·
–
=
–
=
–
=
–
=
Question 40
Solution 41
Compute (1) + (2) and (1) – (2)
Compute x (A) and y (B)
Compute
⇒
+
and
–
adding and subtracting ⇒ 2x = 3, 2y = 1 ⇒ x =
,y=
(x , y) =
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Question 41
Created by Mathematica (April 10, 2004)
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