UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS FIFTY-FOURTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 9, 2011
1) Evaluate 1+ 1 10 1+
1 11 1+
1 12 1+
1
13 . Express your answer as a rational number in lowest terms.
11
10 12 11
13 12
14 13 =
14 10 =
7 5
2) Simplify the expression
2011! +2012!
2011! +2010!
2011 . Express your answer as a rational number in lowest terms.
2011! +2012!
2011! +2010!
2011 =
2011!1+2012 2010!2011+1
2011 =
2011 2013 2012
2011 = 2013 2012
3) The difference of two positive numbers is 4 and the product of the two numbers is 19. Find the sum of the two numbers.
x – y = 4 (1) xy = 19 (2)
(1) ï y = x – 4 (2) ï x(x – 4) = 19 ï x2 – 4x – 19 = 0
x = 4≤ 16+76
2 = 4≤ 4 23
2 = 2 ± 23
y = x – 4 ï y = – 2 ± 23
x + y = 2 ± 23 + (– 2 ± 23 ) = 2 23
4) Find the value of a + b2 if ab =– 14 and a2 +b2 = 30 .
a + b2 = a2 +2ab +b2 = 30 + 2(–14) = 2
5 Consider the circle with diameterAD. If
—
ABC =130 °,—
CDA = 50 °and
—
BCA =20 °, find—
BAD. Express your answer in degrees.A
B
C
D
30° 130°
20° 90°
—
BAC = 180° – 130° – 20° = 30°
—
CAD = 180° – 90° – 50° = 40°
—
BAD = 30° + 40° = 70°6) If the length of a rectangle is increased by 40% and the width is decreased by 15%, what is the percentage change in the area of the rectangle?
(1.4L)(.85W) = 1.19LW
Change = 19%
7) One day last month, Ray's Reasonably Reliable Repair Service offered the following Saturday special: "Buy 3 shock absorbers at the regular price and receive an 80% discount on the fourth."
Jerome bought 4 shock absorbers on that day and paid a total of $176. What was the regular price of one shock absorber?
3x + 1
5x = 176
16
5x = 176
x = 176 · 5
16
x = 55
8) The two roots of the quadratic equation x2 85x+c=0 are prime numbers. What is the value of c?
Let a and b be the roots of the quadratic ï x2 85x+c = (x – a)(x – b) = x2 – (a + b)x + ab
Thus a + b = 85 and ab = c
a + b odd ï one is even and the other is odd . Thus one of a and b must be 2 ï the other is 83.
Hence ab = 2(83) = 166
9) For real numbers x, yand z, define F x, y,z =x y + y z+z x. For which real numbers a is F 2,a,a 1 =F 5,a,a+1 ?
F 2,a,a 1 =F 5,a,a+1 ï 2a + a(a – 1) + 2 (a – 1) = 5a + a(a + 1) + 5 (a + 1)
2a + a2 – a + 2a – 2 = 5a + a2 + a + 5a + 5 ï 3a – 2 = 11a + 5 ï 8a = – 7 ï a = – 7 8
10) Michelle has a collection of marbles, all of which are either blue or green. She is creating pairs of 1 blue marble and 1 green marble.
After a while, she notices that 2
3 of all the blue marbles are paired with 3
5 of all the green marbles. What fraction of Michelle’s
marble collection has been paired up? Express your answer as a rational number in lowest terms.
Let B = number of blue marbles, G = number of green marbles, T = total number of marbles and x = fraction of total paired.
2
3B = 3
5G and B + G = T
2
3B = 3
5G ï B = 3 2 ·
3 5G =
9 10G
2
3B + 3
5G = xT ï 3 5G +
3
5G = x(B + G) = x 9
10G+ G
6
5G = x · 19
10G ï x = 6 5 ·
10 19 =
11) Find the integer value of the expression log7 1
8 ä log825 + log25 ä log549 .
log825 = log2325 =
1 3log25
2 = 2
3 log25 ï log825 + log25 = 5 3 log25
log7 1
8 = – log78 = – log72
3 = – 3 log 72
log549 = log572 = 2 log 57 =
2 log75
log7 1
8 · ( log825 + log25) · log549 = – 3log72 · 5
3 log25 · 2
log75 = – 10 log72
log75 · log25 = – 10 log25 log25 = – 10
12) If sin(x) + cos(x) = 1
2, find the value of sin
3 x + cos3 x. Express your answer as a rational number in lowest terms.
(sin(x) + cos x 3 = sin3 x + 3 sin2 x cos x + 3 sinx cos2 x + cos3 x
1
2 3
= 1
8 = sin
3 x + cos3 x + 3 sinx cos x[sin(x) + cos(x)]
(sin(x) + cos x 2 = sin2 x + 2 sinx cosx + cos2 x
1
2 2
= sin2 x + cos2 x + 2 sinx cosx ï 1
4 = 1 + 2sin(x)cos(x) ï sin(x)cos(x) = – 3 8
1
8 = sin
3 x + cos3 x + 3ÿ – 3 8
1
2 ï sin
3 x + cos3 x = 1 8 +
9 16 =
11 16
13 Find the area of the region bounded by the lines x+2y=2, – 4x+ y=1, x +2y= 11 andx–y=2.
x+2y=11
–4x+y=1
x+2y=2
x–y=2 0,1
2,0
5,3 1,5
4
1 4
2
3
3 2
Solving the equations for the straight lines in pairs gives the intersection points (0,1), (2,0), (5,3) and (1,5).
Construct the rectangle with vertices (0,0), (5,0), (5,5) and (0,5).
The desired are is the area of the rectangle minus the area of the four right triangles as indicated.
Area = 5(5) – 1
2(1)(2) – 1
2(3)(3) – 1
2(4)(2) – 1
2(1)(4) = 25 – 1
2(2 + 9 + 8 + 4) = 25 – 1 2(23) =
50 – 23 2 =
27 2
14 Find the area of the circle that contains the pointQ 9, 8 and that is tangent to the line x– 2y =2 at the pointP 6, 2 .
Q 9,8
P 6,2 C a,b
b– 2
a– 6 = – 2 ï b – 2 = – 2a + 12 ï 2a +b = 14 (1)
(a – 92 + (b – 82 = (a – 6 2 + (b – 22
a2 – 18a + 81 + b2 – 16b + 64 = a2 – 12a + 36 + b2 – 4b + 4 ï – 6a – 12b = – 105 ï 2a + 4b = 35 (2)
2a +b = 14
2a + 4b = 35 subtract
3b = 21 flb = 7 ï a = 7
2
radius2 = 7 2 – 9
2
+ (7 – 82 = –11 2
2
+ –12 = 121 4 + 1 =
125 4
area = 125
4
15) If logx y2 = 3, determine the value of log y x
2 . Express your answer as a rational number in lowest terms.
logx y2 = 3 ï 2 log
xy = 3 ï logxy = 3
2 ï logyx = 2
3 ï logyx 2 = 4
3
16) When a complex number z is expressed the form z=a+b where a and b are real numbers, the modulus (or absolute value)
of z is defined by z = a2 +b2 . Suppose that z + z = 3 +9Â. Determine the value of z 2.
z = a + b ï z + | z | = a + b i + a2+b2 = 3 + 9 i ï b = 9 and a + a2 +81 = 3
z 2 = a2 + b2 = –122 + 92 = 144 + 81 = 225
17) Twenty balls numbered 1 to 20 are placed in a jar. Larry reaches into the jar and randomly removes two of the balls. What is the probability that the sum of the numbers on the two removed balls is a multiple of 3? Express your answer as a rational number in lowest terms.
The total number of pairs chosen from 20 = 20
2 = 20! 2! ÿ18! =
20ÿ19 2 = 190
i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
i mod 3 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2
The sum of the two number chosen is divisible by three if the sum = 0 mod 3. This can happen if both numbers are 0 mod 3 or one is 1 mod 3 and the other is 2 mod 3.
Thus the number of pairs is 6
2 + 7
1 · 7
1 = 15 + 49 = 64
The probability that the sum is divisible by 3 = 64
190 = 32 95
18) The three vertices of a triangle are points on the graph of the parabola y=x2. If the x-coordinates of the vertices are the roots
of the cubic equation x3 60x2+153x+1026=0, find the sum of the slopes of the three sides of the triangle.
Let a, b and c be the roots. The corresponding point on the parabola are a,a2 , b,b2 and c,c2 .
The sum of the slopes is b2–a2
b–a + c2–b2
c–b + a2–c2
a–c = b + a + c + a + a + c = 2(a + b + c)
x3 60x2+153x+1026= x–a x–b x–c = x3 – a+b+c x2 + (ab + ac + bc)x – abc ï a + b + c = 60 Sum of the slopes = 2(a + b + c) = 2(60) = 120
19) For how many real numbers x will the mean of the set 6, 3, 10, 9,x be equal to the median?
1) 3, 6, 9, 10,x ï 3 + 6 + 9 + 10 + x = 28 + x ï 9 = 28+x
5 ï 45 = 28 + x ï x = 17
2) 3, 6, 9,x, 10 ï 9 = 28+x
5 ï 45 = 28 + x ï x = 17
3) 3, 6,x, 9, 10 ï x = 28+x
5 ï 5x = 28 + x ï 4x = 28 ï x = 7
4) 3,x, 6, 9, 10 ï 6 = 28+x
5 ï 30 = 28 + x ï x = 2
5) x, 3, 6, 9, 10 ï 6 = 28+x
5 ï 30 = 28 + x ï x = 2 ï x = 7
Thus there are 3 possible values of x.
20) Each side of square ABCD has length 3. Let M and N be points on sides BCand CD respectively such that BM = ND =1
and let q =
—
MAN. Find sinq.C D
N
3
1 2
A B
C D
M N
3 3
1
1 2
2
θ
10 10
2 2
By the law of cosines 2 2 2 = 10 2 + 10 2 – 2 · 10 · 10 · cos(q)
8 = 10 + 10 – 20 cos(q) ï cos(q) = 12 20 =
3
5 ï sin( ) = 4 5
21) Find the value of the real number x such that 5 + x, 11 + x and 20 + x form a geometric progression in the given order.
11+x
5+x = 20+x 11+x
(11 + x2 = (5 + x)(20 + x) ï 121 + 22x + x2 = 100 + 25x + x2 ï 21 = 3x ï x = 7
22 Find the number of paths from the lower left corner to the upper right corner of the given grid, if the only allowable moves are along grid lines upward or to the right.
One such path is shown.
1
1
1
1
1
1
1
1
2
3
4
5
6
7
1
3
6
10
6
13
1
4
10
20
20
26
39
1
5
20
40
66
105
1
6
6
26
66
132
237
1
7
13
39
105
237
474
6+6
6 – 4+2
4
2+4 2 –
2+4 2
4+2 4 =474
23) If x and z are real number such that 2 x – 3 z = – 9 and x + z = 23 , find x + z .
Adding the given equations:
– 5 y = –55 ï y = 11 ï x = 12 ï x + y = 144 + 121 = 265
24) If sin(a) = 1
4, find sin(3a). Express your answer as a rational number in lowest terms.
α
15
1
sin(3a) = sin(2a)cos(a) + cos(2a)sin(a) = 2sin(a)cos(a)cos(a) + cos2 a –sin2 a sin a
sin(3a) = sin a 3 cos2 a – sin2 a = 1 4 3
15 16–
1 16 =
1 4ÿ
44 16 =
11 16
25) Let f(x,y)
=
x–yx+y. Define the sequence an by a1 = f(3,1) and an+1 = f an, 1 forn ¥ 1. Find a2011.
a1 = 3 – 1 3+1 =
1 2
a2 =
1 2– 1 1 2+1
=
–1 2 3 2
= – 1
3
a3 = –1
3– 1
–1
2+1
=
–4 3 2 3
= – 2
a4 = –2 – 1 –2+1 =
–3 –1 = 3
k = 0 mod 4 ï ak = 3
k = 1 mod 4 ï ak = 1 2
k = 2 mod 4 ï ak = –1 3
k = 3 mod 4 ï ak = –2
2011 mod 4 = 3 ï a2011 = – 2
x2−4 −4 = 1
x2−4 −4 = 1 x2−4 = 5
x2−4 −4 = −1 x2−4 = 3
x2−4 = 5 x2 = 9
x = 3
x2−4 = −5
x2 = −1
x2−4 = 3
x2 = 7
x = 7
x2−4 = −3 x2 = 1
x = 1
Sum = 1 + 3 + 7 =4 + 7
27 TriangleABCis a 3 4 5 right triangle withAB=4.
Construct the perpendicularAD1and letAD1 =x1. Construct the
perpendicular D1D2and letD1D2 =x2. Construct the perpendicular D2D3and letD2D3 =x3. If this process is continued forever,
find
k=1 ¶
xk.
x2 x1
x3
A B
C
D1
D2
D3
x2 x1
x3 x 4 x
5 α
α
α
α
α
A B
C
D1
D2
D3
D4
cos(a) = 3
5 cos(a) = x1
4 ï x1 = 12
5
sin(a) = 4
5 sin(a) = x2
x1
ï x2 = 4 5
12 5
sin(a) = 4
5 sin(a) = x3
x2
ï x3 = 4 5
2 12 5
sin(a) = 4
5 sin(a) = x4
x3
ï x4 = 4 5
3 12 5
In general xn = 4
5 n– 1 12
k=1 ¶
xk = 12
5 + 4 5
1 12 5 +
4 5
2 12 5 +
4 5
3 12 5 + · · ·
k=1 ¶
xk = 12
5(1 + 4 5
1
+ 4
5 2
+ 4
5 3
+ · · · ) = 12
5 1 1 –4
5
=12
28) Let f(x)=3x2 – x. Find all values of x such that f(f(x))=x .
f(f(x)) = x ï f 3x2–x = x ï 3 3x2–x2– 3x2–x = x ï
3 9x4–6x3+x2 – 3x2 + x = x ï 9x4– 6x3 = 0 ï x3 9x– 6 = 0 ï x = 0 , 2 3
29 In triangleABC,
—
CAB=30 °,AC=2 and AB=5 3Find BC.
30° A B C 2 5 3
By the law of cosines:
BC2 = 22 + 5 3 2 – 2 · 2 · 5 3 cos(30°)
BC2 = 4 + 75 – 20 3 · 3
2 = 79 – 30 = 49 ï BC = 7
OR A B C D 30° 5 3 2 x =11 2 5 3 2 CD
5 3 = sin(30°) = 1
2 ï CD = 5 3
2
2+x
5 3 = cos(30°) = 3
2 ï 2 + x = 15
2 ï x = 11
2
BC2 = 5 3 2
2
+ 11
2 2
= 75
4 + 121
4 = 196
4 = 49 ï BC = 7
0.3 B + 300(0.23) = (B + 300)(0.26) ï 0.3 B + 69 = 0.26 B + 78 ï 0.04 B = 9 ï B = 9 0.04 =
900 4 = 225
31) Four horses compete in a race. In how many different orders can the horses cross the finish line, assuming that all four horses finish the race and that ties are possible?
0 tied: 4! = 24
2 tied: pick 2 from 4 arrange in 3! ways 4
2 3! = 6 · 6 = 36
3 tied: pick 3 from 4 arrange in 2! ways 4
3 2! = 4 · 2 = 8
4 tied: either all 4 or two and two all four 1 way, two and two pick two from 4 to be first 1 + 4
2 = 1 + 6
Total = 24 + 36 + 8 + 7 = 75
32 As shown in the sketch, on each side of a square with side length 4, an interior semicircle is drawn using that side as a diameter. Find the area of the shaded region.
4A + 4B = 42
2A + B = 1
2p 2 2
A + B = 16 2A + B = 2p
A = 2p – 4
4 = 8 – 16
Let L = rate of a large pump and S = rate for a small pump.
1
2L+S = 4 ï 8L + 4S = 1 (1)
1
L+3S = 4 ï 4L + 12S = 1 (2)
(1) – 2(2) ï 1 = 20S ï S = 1 20 1 = 8L + 4 1 20 = 8L + 1 5 ï L = 1 10 t = 1 4L+4S = 1 4 101 +4 201 = 21 5+ 1 5 = 13 5 = 5 3 hours = 100 minutes 34) There are 40 students in the Travel Club. They discovered that 17 members have visited Mexico, 28 have visited Canada, 10 have been to England, 12 have visited both Mexico and Canada, 3 have been only to England and 4 have been only to Mexico. Some club members have not been to any of the three foreign countries and an equal number have been to all three countries. How many students have been to all three countries? Mexico Canada England
a
b
c
d
e
e
f
g
From the given information: a + b + c + d + e + f + g + e = 40 (1)a + b + e + d =17 (2)
b + c + e + f = 28 (3)
d + e + f + g = 10 (4)
b + e = 12 (5)
g = 3 (6)
a = 4 (7)
(2), (5) and (7) ï 4 + 12 + d = 17 ï d = 1 (8)
(3) and (5) ï 12 + c + f = 28 ï c + f = 16 (9)
(1), (7), (5), (9), (8) and (6) ï 4 + 12 + 16 + 1 + 3 + e = 40 ï e = 4
Number of ways to pick 3 bars out of 11 = 11
3 = 11! 3! ÿ8! =
11ÿ10ÿ9ÿ8! 3ÿ2ÿ8! =
11ÿ10ÿ9
3ÿ2 = 11 · 5 · 3 = 165
Ways to get 3 bars costing $4 or more:
$2 + $2 + $2 4
3 = 4
$2 + $2 + $1 4
2 · 4
1 = 6(4) = 24
$2 + $2 + $.5 4
2 · 3
1 = 6(3) = 18
$2 + $1 + $1 4
1 · 4
2 = 4(6) = 24
Total number = 4 + 24 + 18 + 24 = 70
p = 70
165 = 14 33
36) If x4 + x3 + x2 + x + 1=0 and x + 1
x > 0, determine the value of x + 1 x.
x4 + x3 + x2 + x + 1=0
x2 x2 + x +1 +1 x +
1 x2 = 0
x2 x2 + 1
x2 +2 +x+
1
x– 1 = 0
x2 x+ 1 x 2
+ x + 1
x – 1 = 0
x + 1
x =
– 1≤ 1+4 2 =
– 1≤ 5 2
x + 1
x > 0 ï x + 1
x =
– 1 5 2
37) Suppose that a and r are real numbers such that the geometric series whose first term is a and whose ratio is r has a sum of 1 and the geometric series whose first term is a3 and whose ratio is r3 has a sum of 3. Find a.
a + a r + a r2 + a r3 + · · · = a
1 –r = 1 (1)
a3 + a3r3 + a3r6 + a3r9 + · · · = a3
1 –r3 = 3 (2)
(1) ï a = 1 – r ï r = 1 – a
(2) ï a3 = 3 1 –r3
Substituting r = 1 – a ï a3 = 3 1 – 1 –a 3
a3 = 3 1– 1 – 3a+3a2–a3 = 9a – 9a2 + 3a3 2a3 – 9a2 + 9a = 0
a = 0 ï series sums to 0
×
a = 3
2 ï r = – 1 2
a = 3 ï r = – 2 series is divergent
×
Thus a = 3
2
38 PointsA= 0, 0 ,B= 18, 24 andC= 11, 0 are the vertices of
triangleABC. PointPis chosen in the interior of this triangle so that the area of trianglesABP,APCandPBCare all equal. Find the coordinates ofP.
Express your answer as an ordered pair x, y.
A 0,0
B 18,24
C 11,0
P
x
y
A 0,0
B 18,24
C 11,0 P
"h1 "h2 "h3
30 25
5 10 15
5 10 15 20 25
Area of DABC = 1
2 11 24 = 132 ï Area DABP = DAPC = DPBC = 132
3 = 44
Let P = ( xp , h1)
1
2 11 h1 = 44 ï h1 = 8
1
2 30 h3 = 44 ï h3 = 44 15
The line through A and B has equation y = 24
18 x or 4
3 x – y = 0
Using the distance from a point to a line formula
4 3xp – 8
4 3 2
+12
= 44
15 ï
4 3xp – 8
5 3
= 44
4
3 xp = 5 3ÿ
44 15 + 8 =
116
9 ï xp = 3 4ÿ
116 9 =
29 3
(xp , h1) = 29
3, 8
39 Find the distance between the centers of the inscribed and circumscribed circles of a right triangle with sides of length 3, 4 and 5.
A B
C
A B
C
D
3
4
5
E
r
Let D be the center of the circumscribed circle, E the center of the inscribed circle and r the radius of the inscribed circle.
If A = (0,0), B = (3,0) and C = (3,4), then D = C – A
2 = 3 2, 2
The area of the triangle can be computed as 1
2(3)(4) = 1 2r(3) +
1 2r(4) +
1
2r(5) ï 12 = 12r ï r = 1
Thus E = (3 – 1 , 1) = (2 , 1).
DE2 = 2–3
2 2
+ 1 – 22 = 1 4 + 1 =
5
4 ï DE = 5 2
40) Let S be the set of all 11–digit binary sequences consisting of exactly two ones and nine zeros. For example, 00100000100 and 10000100000 are two of the elements of S. If each element of S is converted to a decimal integer and all of these decimal integers are summed, what is the value of the sum? Express your answer as an integer in base 10.
There are 11
2 = 11 11 –1
2 ways to place the two ones. The total number of ones is then 2· 11 10
2 = 11(10).
The ones are equally distributed in the n possible places, so each place will have 10 ones.
Sum = (10) 20 +21 +21 + ÿ ÿ ÿ210
Sum = 10 211–1 = (10)(2048 – 1) = 20470
41 As shown in the sketch, circular arcsA C and B Chave respective centers at BandA. Suppose thatSis a circle that is tangent to each of these arcs and also to the line segment joiningAandB. Find the radius ofSif AB=24.
A B
C
A B
C D
E
G
A
r
r
x
12
F r
AE = AB = radius of circular arc BC = 24 ï 2r + x = 24 ï x = 24 – 2r
From triangle AFG r+x2 = r2 + 122
Substituting for x r+24 – 2r2 = r2 + 122 ï 24 –r2 = r2 + 122
