UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FORTY-SIXTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 12, 2003

1)  What rational number is equidistant from and  – ?

Solution 1

2)  Express    as a rational number in lowest terms.

Solution 2

3)  Define the binary operation  □  by  a □ b =   + ab – .   Find all real numbers x such that  7 □ x = 59 .

Solution 3

4)  Find the positive real number x which satisfies the equation    = – .

Solution 4

5)  2000 different episodes of a talk show originally aired on television network A.  Cable channel B then rebroadcast 200 episodes of the show.  Cable channel C later rebroadcast 460 episodes of the show, including all but 20 of the episodes previously shown on cable channel B.  What percentage of the show's episodes have been rebroadcast by at least one of the cable channels ?

Solution 5

6)  On a balance scale, 3 green balls balance 6 blue balls, 2 yellow balls balance 5 blue balls and 6 blue balls balance 4 white balls.  How many blue balls are needed to balance a set of 4 green balls, 2 yellow balls and 2 white balls ?

Solution 6

7)  The semicircle has radius .  Chord has length and is parallel to .  If O is the center of the semicircle and is perpendicular to , find the length of .

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Solution 7

  8) There are two candy jars, each of which contains the same number of pieces of candy.  Bert takes 25 pieces of candy from the first jar and gives the rest to Karla, while Karla takes 17 pieces of candy from the second jar and gives the rest to Bert.  When they are done, Karla discovers that Bert has more pieces of candy than she does.  How many more pieces of candy does Bert have ?

Solution 8

  9)  Last week, Earl took three quizzes in his algebra class.  He earned 13 points out of 20 on the first quiz and 41 points out of 50 on the second quiz.  There were 30 points available on the third quiz.  When each quiz score was converted to a percent, the average percent score for the three quizzes was 79%.  How many points did Earl earn on the third quiz ?

Solution 9

10)  Find the value of   .

Solution 10

11) A square in the plane has a pair of opposite  vertices at the points (2 , 4) and (– 2 , 2).  If the points (a , b) and (c , d) are the other two vertices, determine the value of  a + b + c + d.

Solution 11

12)  If you first multiply a number x by 4 and then subtract 12, you get twice as much as you get when you first subtract 12 from x  and then multiply by 4.  What is x ?

Solution 12

13)

Solution 13

14)  Consider the arrangement AAABBBCC.  In how many ways can the eight letters be rearranged so that each position in the rearrangement is occupied by a type of letter which is different from the type of letter which occupied that position in the given arrangement?

Solution 14

15)  Statistics for road use in a certain county show that in the past year, there were 32 accidents per 100,000 miles driven on rural roads and 18 accidents per 100,000 miles driven on city roads.  Combined statistics for both rural and city roads show that there were 24 accidents per 100,000 miles driven.  Let x be the total number of accidents on rural roads and y be the total number of accidents on city roads.  Determine the value of .  Express your answer in the form where p and q are positive integers having no common divisor other than 1.  

Solution 15

16)  How many pairs of positive integers ( a , b ) with a + b ≤ 1000  satisfy   = 121 ?  

Solution 16

17)  Find the coordinates of the point on the circle   + = 10  that is closest to the line  y = 3x + 20 .

Solution 17

18)  If 4  balls are randomly drawn without replacement from a box containing 5 red balls, 4 blue balls, 4 green balls and 2 yellow balls,  what is the probability that for each color, the selection contains no more than 2 balls of that color ?

Solution 18

19)  Define    = 2 ,   = 8  and   =   for n ≥ 2.  Find .

Solution 19

20)  In the isosceles trapezoid ABCD, and are parallel.  If AD = 20, BC = 10 and AC = 25,    find CD.

Solution 20

21)  Seven congruent rectangles form a larger rectangle ABCD.  If the area of rectangle ABCD is 756 square units, what is the perimeter of ABCD ?

Solution 21

22)  Find the degree measure of the acute angle β which satisfies the equation   cos(81°) + cos(39°) = cos(β) .

Solution 22

23)  Suppose that and are concentric circles, with being the larger circle.  The length of a chord of which is tangent to is 28 cm.  What is the area of the  region which lies between and ?

Solution 23

24)  Suppose that S is a set of 5 distinct positive integers such that when any 4 of the integers are added together, the possible sums are
       169, 153, 182, 193 and 127.  What is the largest integer in S ?

Solution 24

25)  Suppose  i = .  Define a sequence of complex numbers by   = 0 and   = + i  for n ≥ 1.  Find .

Solution 25

26)  If   = 81,  find  .

Solution 26

27)  Find the sum of the real solutions to the equation    + = 6 .

Solution 27

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Solution 28

29)  Recall that for a positive integer n,  n! = n · (n – 1) · (n – 2) · · · 3 · 2 · 1.   Find the number of zeros at the end of 2003! .

Solution 29

30)  On a test, the average score of those who passed was 75, the average score of those who failed was 35 and the average score of the entire class was 60.  What fraction of the class passed ?  Express your answer as a rational number in lowest terms.

Solution 30

31)  Find the sum of all of the real zeros of the function  f (x) = 2 cos(2x) + 1  in the interval  0 ≤ x ≤ 100 π .

Solution 31

32)  In Δ ABC, line segments and are parallel to and the three regions CED, DEGF and FGBA have equal areas.  Find the value of .

Solution 32

33)  The sum of all of the zeros of a cubic polynomial p (x) = + a + b x + c  with c ≠ 0   is equal to twice the product of all of its zeros.  The sum of the squares of all of the zeros of p(x) is equal to three times the product of all of its zeros.  If  p (1) = 1,  find c.

Solution 33

34)  Find the minimum value of    if   ≤ x ≤   and   ≤ y ≤ .

Solution 34

35)  A circle is inscribed in Δ ABC where AB = 4,       BC = 5 and AC = 3.  Let t be the length of the line segment joining the points where the circle is tangent to sides and .  Find .

Solution 35

36)  Suppose Δ ABC is inscribed in a circle whose radius is 10 cm.  If the perimeter of the triangle is 31 cm, determine the value of  sin(A) + sin(B) + sin(C) .

Solution 36

37)  In Δ ABC, the respective coordinates of A and B are (0 , 0) and (15 , 20) .  It is known that C has integer coordinates.  
       What is the minimum positive area of Δ ABC ?

Solution 37

38)  An obtuse isosceles triangle ABC with equal angles at  B and C is inscribed in a circle.  Tangents are drawn through B and C which intersect at a point D.   Suppose that  3 ∠ ABC = ∠ BDC.   If x is the radian measure of  ∠ BAC, determine the value of x.

Solution 38

39)  If a and b are decimal digits (not necessarily distinct),  let n(a,b) = aba + bab.  For example,  n(5,8) = 585 + 858 = 1443.
       Find  ∑ n(a,b), where the sum is computed over all ordered pairs of decimal digits a and b .

Solution 39

40)  If x is a real number, let  [ x ] be the largest integer less than or equal to x.  Define f (x) = [ x ] x .  What is the largest possible two digit integer value of f (x)?

Solution 40

41)  A circle is inscribed in an equilateral triangle with side length 1 unit.  Construct a circle that is tangent to and two sides of the triangle.  Then construct a circle that is tangent to and two sides of the triangle.  Continue constructing such circles indefinitely.  Find the sum of the areas of this infinite sequence of circles.

Solution 41

Solution 1

· = =

Question 1

Solution 2

= = = · = =

Question 2

Solution 3

a □ b =   + ab – .   7 □ x = 59 .

+ 7x – = 59

– 7x + 10 = 0

(x – 2)(x – 5) = 0

x = 2 , 5

Question 3

Solution 4

= – 2x (x + 2)

2x = 2(x + 2) – x(x + 2)

2x = 2x + 4 – – 2x

+ 2x – 4 = 0

x = = =

x = – 1 ±

x = – 1 +        (positive solution)

Question 4

Solution 5

Number rebroadcast by C = 460,  number rebroadcast by B and not by C = 20.

Total number rebroadcast = 460 + 20 = 480.

% = · 100 = 24%

Question 5

Solution 6

3G = 6B  ⇒ G = 2B

2Y = 5B  ⇒ Y = B

4W = 6B  ⇒  W= B

4G + 2Y + 2W = 4(2B) + + = 8B + 5B + 3B = 16B

Question 6

Solution 7

= – =   ⇒  x = r = =

Question 7

Solution 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.

Question 8

Solution 9

Quiz 1   = 65%    – 14% ,  below the 79% average

Quiz 2   = 82%      + 3% ,  above the 79% average

Quiz 3    must be  + 11% ,  above the 79% average or · 100 = 90%

Thus  x = 27

Question 9

Solution 10

= = = 2 · 125 = 250

Question 10

Solution 11

Equating the changes in the x and y coordinates of opposite sides of the square;

Adding these four equations;

2a + 2b – 6 = – 2c – 2d + 6  ⇒  2a + 2b + 2c + 2d = 12

Thus  a + b + c + d = 6

Question 11

Solution 12

4x – 12 = 2(x – 12) · 4

4x – 12 = 8x – 96

4x = 84  ⇒  x = 21

Question 12

Solution 13

By the law of sines

=                                    (1)

= =                (2)

Solving  (1) and (2)  for  

= =

12x = 112 – 7x   ⇒  19x = 112  ⇒  x =

Question 13

Solution 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.

Question 14

Solution 15

Let x = number of accidents on rural roads, = number of 100,000 miles driven on rural roads, y = number of accidents on city roads and = number of 100,000 miles driven on city roads.  The the given information is:

= 3.2                             (1)

= 1.8                             (2)

= 2.4                      (3)

Solve (1) and (2) for and and substitute into (3)

= 2.4

x + y = x + y

x + y = x + y

x = y  ⇒   =

Question 15

Solution 16

= 121     Factor as

= = 121

Thus   = 11  ⇒  a = 11b

a + b = 11b + b = 12 b ≤ 1000

= 83 +   Thus there are 83 choices for the pair (a , b).

Question 16

Solution 17

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 = – (x + 1).

Thus, we solve simultaneously:

Substituting from (1) into (2)

+ = 10  ⇒   = 10  ⇒   = 9

x + 1 = ± 3  ⇒ 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)

Question 17

Solution 18

The total number of balls is 15.  The total number of ways to pick 4 of the 15 balls is = 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 ways,  blue in 1 way and green in 1 way for a total of 7 ways.

3 of the same color:  Red in = 100 ways,  blue in = 44 ways and green in   = 44 ways for a total of 188 ways.

Then the desired count is  1365 – 7 – 188 = 1170

Thus the required probability  is   = .

Question 18

Solution 19

   = 2 ,   = 8  and   =

= = = 4

= = =

= = =

= = =

= = = 2

= = = 8

Thus the sequence is periodic with period 6.   2003 ≡ 5 mod 6 .  Thus = =

Question 19

Solution 20

Draw perpendiculars from vertices B and C to side AD.  

From triangle ACF   = – = 400  ⇒ h = 20

From triangle CFD   = + = 425  ⇒ x = .  Thus  CD =

Question 20

Solution 21

Area of each small rectangle is xy.  The total area is 7xy = 756.  By construction,  3x = 4y  ⇒  y = x.

Thus  756 = 7x

           756 =

             =   = 144  ⇒  x = 12  ⇒ y = · 12 = 9

The perimeter of ABCD is  5x + 6y = 5(12) + 6(9) = 114

Question 21

Solution 22

cos(81°) + cos(39°) = cos(β)

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°

Question 22

Solution 23

If the radii of the circles and are R and r, the area between the circles if π ( – ).  From the figure   – = = 196.

Thus the required area is 196 π.

Question 23

Solution 24

Let  S = { , , , , }  with   < < < < .

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(   + + + + ) = 169 + 153 + 182 + 193 + 127 = 824  ⇒ + + + + = 206

By construction, the smallest sum is + + + = 127.  Thus = 206 – 127 = 79

Question 24

Solution 25

= 0 and   = + i

= + i = i

= + i = – 1 + i

= + i = 1 – 2i – 1 + i = – i

= + i = – 1 + i

= + i = 1 – 2i – 1 + i = – i

Thus for n > 2   if n ≡ 0 mod, = – i  and if n mod 2 = 1,   = – 1 + i.  Then   = – 1 + i .

Question 25

Solution 26

= 81  ⇒   =   and    =   ⇒   

= = = = 16

Question 26

Solution 27

+ = 6

Let  t = .  Then the equation is

t + = 6  ⇒   – 6t + 8 = 0

(t – 2)(t – 4) = 0  ⇒  t = 2  or  t = 4

= 2  or   = 4  ⇒  x =   or  x = 1  

Thus the sum of the real solutions is   + 1 = .

Question 27

Solution 28

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  ⇒  x + y = = 20

Then the perimeter of triangle ABC is  2x + 2y + 2r = 12 + 16 + 20 = 48  ⇒  x + y + r = 24

Now  x + y + r = 24  and  x + r = 12  ⇒  y = 12

Now  x + y + r = 24  and  y + r = 16  ⇒  x = 8

If the center of the circle is (h , k), then  (h , k) = (12 + 8 , 16 + 12) = (20 , 28)

Question 28

Solution 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:

+ + + = 400 + 80 + 16 + 3 = 499

Question 29

Solution 30

Let P = number who passed and F = number who failed.  Find  

From the given information  75P + 35F = 60(F + P)

75P + 35F = 60F + 60P  ⇒  15P = 25F  ⇒  F = P

= = =

Question 30

Solution 31

2 cos(2x) + 1 = 0  ⇒  cos(2x) = –

Thus  2x = + 2kπ   or   + 2kπ     for  k = 0, 1, 2, · · · 99

x = + kπ   or   + kπ       for  k = 0, 1, 2, · · · 99

For each k,   + kπ + + kπ = (2k + 1) π

Thus the sum of all the real zeros in [0 , 100π]  is:

π + 3π + 5π + · · · + 199π = π(1 + 3 + 5 + · · · + 199) = π · = 10000 π

Question 31

Solution 32

Area triangle CDE = CD cos(α) DE = β                        (1)

Area triangle CFG = CF cos(α) DE = 2β                       (2)

Area triangle CAB = CA cos(α) AB = 3β                      (3)

= =        By similar triangles   =   ⇒   =   ⇒   =   ⇒  CF = CD

= =        By similar triangles   =   ⇒   =   ⇒   =   ⇒  CA = CD

= = =      Rationalizing the denominator

= –

Question 32

Solution 33

Let ,   and   be the zeros of the   + a + b x + c.  Then

+ a + b x + c = (x – )(x – )(x – )

                                  = – + x +

Thus  a = –     b =     c = –

Given   = 2    ⇒  a = 2c

            + + = 3  ⇒   + + = – 3c

           p(1) = 1  ⇒ a + b + c = 0  ⇒  b = – a – c

=   + + + 2     Using the above:

= – 3c + 2(b) = – 3c + 2(– 2c – c)

= – 9c  ⇒  c(4c + 9) = 0  ⇒  c = 0  or  c = –     Since c cannot be zero (c = 0 would imply a = b = 0)   c = –

Question 33

Solution 34

Let  f =   · =             if   ≤ x ≤   and   ≤ y ≤

Then the minimum value of f is the maximum value of   = a +   where a = .

≤ x ≤   and   ≤ y ≤   ⇒   ≤ a ≤   ⇒   ≤ a ≤

Since a + is increasing on   ≤ a ≤ , the maximum occurs when a = .

Thus the maximum of a + = + = and the minimum value of is .

Question 34

Solution 35

The area of the triangle is · 4 · 3 = · 5 · r + · 3 · r + · 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 β is the angle at C,   =

Then  t = 4 · sin  ⇒   = 16 · = 16 · = 8(1 – )    since  cos(β) = from the given right triangle.

= 8 · =

Question 35

Solution 36

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) = .  Similarly,

sin(B) = and  sin(C) =

sin(A) + sin(B) + sin(C) = =

Question 36

Solution 37

The length of AB is = 25  and an equation of the line containing AB is
y – 20 = (x – 15)  or  4x – 3y = 0.

The height h of the triangle is distance from C to this line.   h = =

If m and n are integers and C is not on the line, ≥ 1  ⇒  h ≥

Thus the minimum occurs when h = and the minimum area is · 25 · =

Question 37

Solution 38

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,  3α + π + β = 2π.  Also by comparing arcs,  β = 2π – 2x.  These together give  2x – 3α = π.

From triangle BAC,  x + 2α = π.  Solve    ⇒  

⇒  7x = 5π  ⇒  x =

Question 38

Solution 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

Question 39

Solution 40

Let x = α + h  where α is an integer and 0 ≤ h < 1.

Then f(x) =  [ x ] x = α (α + h) = + α h ≥ .  Thus the largest possible value for α is 9.

Let + α h = β  ⇒  h = < 1

So with α = 9   < 1  ⇒&nb