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

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

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

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  y + x + y = 144.  
       Determine the value of   + .

Solution 28

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   .

Solution 36

37)  Let ABC be a right triangle with        hypotenuse 4.  On each side of Δ 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

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

Question 6

Solution 7

Given  γ = α +    and  α – β = 40°

(A) – (B)      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     Complete the square

+ 2x + 1 + + 6y + 9 = 15 + 1 + 9   ⇒   + = 25  ⇒  center = ( – 1 , – 3)

Distance between  ( – 4 , 1 ) and   ( – 1 , – 3)    d = = = 5

Solution 10

+ – = 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 = · 5

Thus the least common multiple = 2 · · 5 · 7 = 630

Question 13

Solution 14

+ > +   ⇒   > + –   ⇒   >   ⇒  n >   ⇒  n > 6

Question 14

Solution 15

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

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 · · = 2π Area of inside circle = π.   Then shaded area = – π  – 2π  =   – 3π .

Question 21

Solution 22

tan(θ) = 3,    θ acute

cos(2θ) = 2 = 2 · – 1 = – 1 = –

OR

cos(2θ) = =   –   = – = – =  –

Question 22

Solution 23

=         (y) = –    ⇒   = – 4

–   = (x) + – +            

                                         = 1 + – 2(– 4) + = 1 – + 8 + = =

Question 23

Solution 24

– = 16x –   ⇒  x · – 16 – 16x + = 0

x · (1 + ) – 16x (1 + ) = 0  ⇒  x · = 0  ⇒

x = 0  or   =   ⇒  x = 0 or x =

Question 24

Solution 25

Given    3x + = 4  ⇒    = 64  ⇒

+ 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 ] [ 11 – ]

= – – 121 +

= 11 [ 5 + + 7 + 5 – + 7 ] – 4 – 121

= 11(24) – 125 = 264 – 125 = 139

OR

12 + – 11 = 1 +

=

                                                                               = 11 – (12 – ) = – 1

=

                                                                                                                                                          = 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 =   ⇒  x + y = = 16

= = 256

+ 2xy + = 256

+ = 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.

Question 29

Solution 30

– =   ⇒  (x – y)(x + y) =    Since x and y are positive integers, the only possibilities are:

                                      (x – y)(x + y) = 1 ·               or                 (x – y)(x + y) = ·   

                                                                                                                  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 – ) = 0

                             n = 0  ⇒  n = 0  or n =       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

=    =   =

                          =   =            Thus                

· · · · ·   = · · · · · · = 1003

Question 33

Solution 34

There are   = 5 · 5 · 7 · 8 · 9  ways total to arrange the 10 letters.

Now arrange  BBBCCD  in   = 2 · 5 · 6 ways.  For any such arrangement, e.g.   BCBCDB we can place the 4 A's in 4 of the 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 = = 171

The minimum score is 1 + 2 + 3 + · · · + 9 = = 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 and will increase the winning score
by one.  Suppose that no such i and j exist.  Then we must have
= + 1,   = + 1 = + 2 ,  · · ·   = + 8.  Now the only way that   + 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

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 .  Thus the of the areas of the three triangles is  A =

From the right triangle   + =   ⇒  A = (16 + 16) = 8

Question 37

Solution 38

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,    = + – 2(10)(20) cos(60°) = 100 + 400 – 200 = 300

Since = – ,  triangle PCB must be a right triangle with right angle at C.

From right triangle  CDB,   = 300 + 4  ⇒   = 304  ⇒   = = 76

Question 39

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     –