DISINI test_11_sol
UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-FOURTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 9, 2011
1) Evaluate 1 +
11
10
12
11
13
12
1
10
14
13
1+
=
14
10
1
11
=
1+
1
13
. Express your answer as a rational number in lowest terms.
2011! + 2012!
2011! + 2010! . Express your answer as a rational number in lowest terms.
2011
2011! 1 + 2012
2010! 2011+1
=
1+
7
5
2) Simplify the expression
2011! + 2012!
2011! + 2010!
2011
1
12
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
xy = 19
(1)
(2)
(1) ï y = x – 4
x=
4≤
16 + 76
2
=
(2) ï x(x – 4) = 19 ï x2 – 4x – 19 = 0
4≤
4 23
2
y=x–4 ï y=–2±
x+y= 2±
a+b
2
23
23
23 + (– 2 ±
4) Find the value of a + b
=2±
2
23 ) = 2
23
if ab = – 14 and a2 + b2 = 30 .
= a2 + 2 ab + b2 = 30 + 2(–14) = 2
5 Consider the circle with diameter AD. If — ABC = 130 °, —CDA = 50 °
and —BCA = 20 °, find —BAD. Express your answer in degrees.
C
20° 90°
B
130°
30°
A
50°
D
—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?
1
3x + 5 x = 176
16
x
5
= 176
x = 176 ·
5
16
x = 55
8) The two roots of the quadratic equation x2
85 x + c = 0 are prime numbers. What is the value of c?
Let a and b be the roots of the quadratic ï x2
85 x + 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, y and z, define F x, y, z = x y + y z + z x. For which real numbers a is F 2, a, a
F 2, a, a
1 = F 5, a, a + 1 ?
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
B
3
3
= 5 G and B + G = T
2
B
3
= 5G ï B =
3
2
B
3
+ 5 G = xT ï
6
G
5
3
=x·
19
G
10
3
2
·
3
G
5
3
G
5
ï x=
=
9
G
10
3
+ 5 G = x(B + G) = x
6
5
·
10
19
=
12
19
9
10
G+G
11) Find the integer value of the expression log 7
1
3
log8 25 = log23 25 =
1
8
log7
log2 52 =
ä log 8 25 + log 2 5 ä log 5 49 .
log2 5 ï log 8 25 + log 2 5 =
5
3
log2 5
= – log7 8 = – log7 23 = – 3 log7 2
log5 49 = log5 72 = 2 log5 7 =
log7
2
3
1
8
1
8
2
log7 5
· ( log 8 25 + log 2 5) · log5 49 = – 3log7 2 ·
5
3
2
log7 5
log2 5 ·
= – 10
log7 2
log7 5
· log2 5 = – 10
1
log2 5
log2 5
= – 10
12) If sin(x) + cos(x) = 2 , find the value of sin3 x + cos3 x . Express your answer as a rational number in lowest terms.
3
(sin(x) + cos x
1 3
2
=
1
8
= sin3 x + cos3 x + 3 sin x cos x [sin(x) + cos(x)]
2
(sin(x) + cos x
1 2
2
1
8
= sin3 x + 3 sin2 x cos x + 3 sin x cos2 x + cos3 x
= sin2 x + 2 sin x cos x + cos2 x
= sin2 x + cos2 x + 2 sin x cos x ï
= sin3 x + cos3 x + 3 ÿ –
3
8
1
2
1
4
= 1 + 2sin(x)cos(x) ï sin(x)cos(x) = –
ï sin3 x + cos3 x =
1
8
+
9
16
=
11
16
13 Find the area of the region bounded by the lines x + 2 y = 2, – 4 x + y = 1,
x + 2 y = 11 and x – y = 2.
–4x+y=1
x+2y=11
1,5
1
4
2
4
5,3
0,1
3
1
2,0
2
3
x+2y=2
x–y=2
3
8
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.
1
1
1
1
1
1
Area = 5(5) – 2 (1)(2) – 2 (3)(3) – 2 (4)(2) – 2 (1)(4) = 25 – 2 (2 + 9 + 8 + 4) = 25 – 2 (23) =
50 – 23
2
=
27
2
14 Find the area of the circle that contains the point Q 9, 8 and that is tangent
to the line x – 2 y = 2 at the point P 6, 2 .
Q 9,8
C a,b
P 6,2
b–2
a–6
= – 2 ï b – 2 = – 2a + 12 ï 2a +b = 14
(a – 9 2 + (b – 8 2 = (a – 6 2 + (b – 2
(1)
2
a2 – 18a + 81 + b2 – 16b + 64 = a2 – 12a + 36 + b2 – 4b + 4 ï – 6a – 12b = – 105 ï 2a + 4b = 35
2a +b = 14
2a + 4b = 35
subtract
3b = 21 fl b = 7 ï a =
radius2 =
area =
7
2
(2)
2
7
2
– 9 + (7 – 8 2 = –
11 2
2
+ –1 2 =
121
4
+1=
125
4
125
4
15) If log x y2 = 3, determine the value of log y x2 . Express your answer as a rational number in lowest terms.
log x y2 = 3 ï 2 logx y = 3 ï logx y =
3
2
ï log y x =
2
3
ï log y x2 =
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
z=a+b ï z + |z|=a+ bi+
a+
a2 + 81 = 3 ï
= 3 + 9 Â. Determine the value of z 2 .
a2 + b2 = 3 + 9 i ï b = 9 and a +
a2 + 81 = 3 – a
a2 + 81 = 9 – 6a + a2 ï 6a = – 72 ï a = – 12
a2 + 81 = 3
z
2
= a2 + b2 = –12 2 + 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.
20
2
The total number of pairs chosen from 20 =
=
20!
2! ÿ 18!
=
20 ÿ 19
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
The probability that the sum is divisible by 3 =
= 15 + 49 = 64
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
60 x2 + 153 x + 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
x3
b2 – a2
b–a
+
c2 – b2
a2 – c2
+ a–c
c–b
= b + a + c + a + a + c = 2(a + b + c)
60 x2 + 153 x + 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?
28 + x
5
1)
3, 6, 9, 10, x ï 3 + 6 + 9 + 10 + x = 28 + x ï 9 =
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 BC and CD respectively such that BM = ND = 1
and let q = — MAN. Find sin q .
N
D
C
1
2
2
3
N
D
C
1
2
2
2 2
3
10
M
10
1
θ
A
B
3
By the law of cosines
2
2
2
=
10
8 = 10 + 10 – 20 cos(q) ï cos(q) =
12
20
2
=
+
3
5
10
2
–2·
ï sin( ) =
10 ·
10 · cos(q)
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 + x 2 = (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
7
13
39
105
237
474
1
6
6
26
66
132
237
1
5
20
40
66
105
1
4
10
20
20
26
39
1
3
6
10
6
13
1
2
3
4
5
6
7
1
1
1
1
1
1
1
OR
6+6
6
4+2
4
–
2+4
2
–
2+4
2
23) If x and z are real number such that 2
4+2
4
x –3
= 474
z = – 9 and
x +
z = 23 , find x + z.
Adding the given equations:
y = –55 ï
–5
y = 11 ï
x = 12 ï x + y = 144 + 121 = 265
1
24) If sin(a) = 4 , find sin(3a). Express your answer as a rational number in lowest terms.
1
α
15
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 =
3–1
3+1
1
a2 =
2
1
2
=
–1
+1
– –1
3
1
– +1
–2 – 1
–2 + 1
–
1
16
=
1
4
44
ÿ 16 =
11
16
1
2
–1
=
2
3
=–
1
3
2
–4
=
2
a4 =
15
16
x– y
1
a3 =
3
= x + y . Define the sequence an by a1 = f (3,1) and an+1 = f an , 1 for n ¥ 1. Find a2011 .
25) Let f (x,y)
a1 =
1
4
3
2
=–2
3
=
–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
26) Find the sum of all of the positive real solutions of
x2 – 4 – 4 = 1 .
x2 −4 −4
= 1
x2 −4 −4 = 1
x2 −4 = 5
x2 −4 = 5
x2 = 9
x = 3
x2 −4 −4 = −1
x2 −4 = 3
x2 −4 = 3
x2 = 7
x2 −4 = −5
x2 = −1
Sum = 1 + 3 +
7 =4+
27 Triangle ABC is a 3
x =
x2 −4 = −3
x2 = 1
x = 1
7
7
5 right triangle with AB = 4.
4
C
Construct the perpendicular AD1 and let AD1 = x1 . Construct the
perpendicular D1 D2 and let D1 D2 = x2 . Construct the perpendicular
D1
D2 D3 and let D2 D3 = x3 . If this process is continued forever,
¶
find
D3
xk .
x1
k= 1
A
C
α
D1
α
D3
x1
α
x2
x3
A
x4 x
5
α
α
D2
B
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
sin(a) =
4
5
sin(a) =
x3
x2
ï x3 =
4 2
5
12
5
sin(a) =
4
5
sin(a) =
x4
x3
ï x4 =
4 3
5
12
5
In general xn =
4 n – 1 12
5
5
12
5
x2
D2
x3
B
¶
xk =
12
5
xk =
12
(1
5
k= 1
4 1 12
5
5
+
4 2
5
+
12
5
+
4 3 12
5
5
+· · ·
4 3
5
+· · · )=
12
1
5 1– 4
¶
k= 1
4 1
5
+
4 2
5
+
+
= 12
5
28) Let f (x) = 3 x2 – x. Find all values of x such that f (f(x)) = x .
2
f (f (x)) = x ï f 3 x2 –x = x ï 3 3 x2 –x – 3 x2 –x = x
ï
3 9 x4 –6 x3 + x2 – 3 x2 + x = x ï 9 x4 – 6 x3 = 0 ï x3 9 x – 6 = 0 ï x = 0 ,
29 In triangle ABC, —CAB = 30 °, AC = 2 and AB = 5
2
3
B
3
Find BC.
5 3
A
30°
2
C
By the law of cosines:
BC2 = 22 + 5
2
3
BC2 = 4 + 75 – 20
–2· 2· 5
3
2
3 ·
3 cos(30°)
= 79 – 30 = 49 ï BC = 7
OR
C
5 3
5
3
2
30°
A
x =
2
11
2
B
CD
5
3
2+x
5
3
BC2 =
D
= sin(30°) =
= cos(30°) =
5
3
2
2
+
1
2
ï CD =
3
2
11 2
2
5
3
2
ï 2+x=
=
75
4
+
121
4
=
15
2
ï x=
11
2
196
4
= 49 ï BC = 7
30) How many liters of a brine solution with a concentration of 30% salt should be added to 300 liters of brine with a concentration
of 23% salt so that the resulting solution has a concentration of 26% salt?
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
1
2A + B = 2 p 2
2
A + B = 16
2A + B = 2p
A = 2p – 4
4 = 8 – 16
33) Two large pumps and one small pump can fill a swimming pool in 4 hours. One large pump and three small pumps can
fill the same swimming pool in 4 hours. How many minutes will it take four large pumps and four small pumps, working
together, to fill the swimming pool? (Assume that all large pumps pump at the rate R and all small pumps pump at the rate r.)
Let L = rate of a large pump and S = rate for a small pump.
1
2L+S
1
L+3S
= 4 ï 8L + 4S = 1
(1)
= 4 ï 4L + 12S = 1
(2)
(1) – 2(2) ï 1 = 20S ï S =
1 = 8L + 4
t=
1
4 L+4S
1
20
= 8L +
1
=
4
1
10
+4
1
20
1
5
1
20
ï L=
=
1
2
5
+
1
=
5
1
3
1
10
=
5
3
hours = 100 minutes
5
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
a
b
c
e
d
f
g
e
England
From the given information:
a + b + c + d + e + f + g + e = 40
a + b + e + d =17
b + c + e + f = 28
d + e + f + g = 10
b + e = 12
g=3
a=4
(1)
(2)
(3)
(4)
(5)
(6)
(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
35) A bag contains 11 candy bars: three cost 50 cents each, four cost 1 dollar each and four cost 2 dollars each.
Three candy bars are randomly chosen from the bag, without replacement. What is the probability that the total
cost of the three candy bars is 4 dollars or more? Express your answer as a rational number in lowest terms.
Number of ways to pick 3 bars out of 11 =
11
3
=
11!
3! ÿ 8!
=
11 ÿ 10 ÿ 9 ÿ 8!
3 ÿ 2 ÿ 8!
=
11 ÿ 10 ÿ 9
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
1
> 0, determine the value of x + .
x
x
x4 + x3 + x2 + x + 1 = 0
x2 x2 + x + 1 +
x2 x2 +
x2 x +
1
x2
1
x
1
+
+ 2 +x +
1 2
x
+ x +
–1 ≤
1+4
x+
1
x
=
x+
1
x
>0 ï x+
2
=
1
x
=0
x2
1
x
1
x
–1 =0
–1 =0
–1 ≤ 5
2
=
–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
a3 + a3 r3 + a3 r6 + a3 r9 + · · · =
a3
1 – r3
(1)
=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 – 3 a + 3 a2 – a3 = 9a – 9 a2 + 3 a3
2 a3 – 9 a2 + 9a = 0
a 2 a2 – 9 a + 9 = 0 ï a(2a – 3)(a – 3) = 0 ï a = 0,
3
2
or 3
×
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
y
Points A = 0, 0 , B = 18, 24 and C = 11, 0 are the vertices of
triangle ABC. Point P is chosen in the interior of this triangle so that the
area of triangles ABP, APC and PBC are all equal. Find the coordinates of P.
Express your answer as an ordered pair x , y .
38
B 18,24
P
A 0,0
25
B 18,24
20
15
25
30
10
"h3
P
"h2
5
"h1
C 11,0
A 0,0
5
Area of DABC =
10
1
2
15
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
4
Using the distance from a point to a line formula
3
x–y=0
xp – 8
4 2
3
+ 12
4
=
44
15
ï
3
xp – 8
5
3
=
44
15
x
C 11,0
4
3
xp =
5
3
44
ÿ 15 + 8 =
(xp , h1 ) =
29
,
3
116
9
ï xp =
3
4
ÿ
116
9
=
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.
C
A
B
C
D
4
5
E
r
A
B
3
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 =
The area of the triangle can be computed as
C–A
2
1
(3)(4)
2
3
,
2
=
1
2
1
1
= 2 r(3) + 2 r(4) + 2 r(5) ï 12 = 12r ï r = 1
Thus E = (3 – 1 , 1) = (2 , 1).
3 2
DE2 = 2– 2
+ 1–2
2
=
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.
11
11 11 –1
=
ways to place the two ones. The total number of ones is then 2·
2
2
The ones are equally distributed in the n possible places, so each place will have 10 ones.
There are
Sum = (10) 20 + 21 + 21 + ÿ ÿ ÿ 210
Sum = 10
211 – 1
11 10
2
= 11(10).
Sum = 10 211 –1 = (10)(2048 – 1) = 20470
C
41 As shown in the sketch, circular arcs A C and B C have respective centers at
B and A. Suppose that S is a circle that is tangent to each of these arcs and also
to the line segment joining A and B. Find the radius of S if AB = 24.
A
D
C
E
r
G
r
r
x
A
12
B
F
AE = AB = radius of circular arc BC = 24 ï 2r + x = 24 ï x = 24 – 2r
2
= r2 + 122
From triangle AFG
r+x
Substituting for x
r + 24 – 2 r 2 = r2 + 122 ï
24 – r 2 = r2 + 122
242 – 48r + r2 = r2 + 122 ï 576 – 48r = 144 ï 48r = 432 ï r =
432
48
=9
B
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-FOURTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 9, 2011
1) Evaluate 1 +
11
10
12
11
13
12
1
10
14
13
1+
=
14
10
1
11
=
1+
1
13
. Express your answer as a rational number in lowest terms.
2011! + 2012!
2011! + 2010! . Express your answer as a rational number in lowest terms.
2011
2011! 1 + 2012
2010! 2011+1
=
1+
7
5
2) Simplify the expression
2011! + 2012!
2011! + 2010!
2011
1
12
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
xy = 19
(1)
(2)
(1) ï y = x – 4
x=
4≤
16 + 76
2
=
(2) ï x(x – 4) = 19 ï x2 – 4x – 19 = 0
4≤
4 23
2
y=x–4 ï y=–2±
x+y= 2±
a+b
2
23
23
23 + (– 2 ±
4) Find the value of a + b
=2±
2
23 ) = 2
23
if ab = – 14 and a2 + b2 = 30 .
= a2 + 2 ab + b2 = 30 + 2(–14) = 2
5 Consider the circle with diameter AD. If — ABC = 130 °, —CDA = 50 °
and —BCA = 20 °, find —BAD. Express your answer in degrees.
C
20° 90°
B
130°
30°
A
50°
D
—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?
1
3x + 5 x = 176
16
x
5
= 176
x = 176 ·
5
16
x = 55
8) The two roots of the quadratic equation x2
85 x + c = 0 are prime numbers. What is the value of c?
Let a and b be the roots of the quadratic ï x2
85 x + 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, y and z, define F x, y, z = x y + y z + z x. For which real numbers a is F 2, a, a
F 2, a, a
1 = F 5, a, a + 1 ?
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
B
3
3
= 5 G and B + G = T
2
B
3
= 5G ï B =
3
2
B
3
+ 5 G = xT ï
6
G
5
3
=x·
19
G
10
3
2
·
3
G
5
3
G
5
ï x=
=
9
G
10
3
+ 5 G = x(B + G) = x
6
5
·
10
19
=
12
19
9
10
G+G
11) Find the integer value of the expression log 7
1
3
log8 25 = log23 25 =
1
8
log7
log2 52 =
ä log 8 25 + log 2 5 ä log 5 49 .
log2 5 ï log 8 25 + log 2 5 =
5
3
log2 5
= – log7 8 = – log7 23 = – 3 log7 2
log5 49 = log5 72 = 2 log5 7 =
log7
2
3
1
8
1
8
2
log7 5
· ( log 8 25 + log 2 5) · log5 49 = – 3log7 2 ·
5
3
2
log7 5
log2 5 ·
= – 10
log7 2
log7 5
· log2 5 = – 10
1
log2 5
log2 5
= – 10
12) If sin(x) + cos(x) = 2 , find the value of sin3 x + cos3 x . Express your answer as a rational number in lowest terms.
3
(sin(x) + cos x
1 3
2
=
1
8
= sin3 x + cos3 x + 3 sin x cos x [sin(x) + cos(x)]
2
(sin(x) + cos x
1 2
2
1
8
= sin3 x + 3 sin2 x cos x + 3 sin x cos2 x + cos3 x
= sin2 x + 2 sin x cos x + cos2 x
= sin2 x + cos2 x + 2 sin x cos x ï
= sin3 x + cos3 x + 3 ÿ –
3
8
1
2
1
4
= 1 + 2sin(x)cos(x) ï sin(x)cos(x) = –
ï sin3 x + cos3 x =
1
8
+
9
16
=
11
16
13 Find the area of the region bounded by the lines x + 2 y = 2, – 4 x + y = 1,
x + 2 y = 11 and x – y = 2.
–4x+y=1
x+2y=11
1,5
1
4
2
4
5,3
0,1
3
1
2,0
2
3
x+2y=2
x–y=2
3
8
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.
1
1
1
1
1
1
Area = 5(5) – 2 (1)(2) – 2 (3)(3) – 2 (4)(2) – 2 (1)(4) = 25 – 2 (2 + 9 + 8 + 4) = 25 – 2 (23) =
50 – 23
2
=
27
2
14 Find the area of the circle that contains the point Q 9, 8 and that is tangent
to the line x – 2 y = 2 at the point P 6, 2 .
Q 9,8
C a,b
P 6,2
b–2
a–6
= – 2 ï b – 2 = – 2a + 12 ï 2a +b = 14
(a – 9 2 + (b – 8 2 = (a – 6 2 + (b – 2
(1)
2
a2 – 18a + 81 + b2 – 16b + 64 = a2 – 12a + 36 + b2 – 4b + 4 ï – 6a – 12b = – 105 ï 2a + 4b = 35
2a +b = 14
2a + 4b = 35
subtract
3b = 21 fl b = 7 ï a =
radius2 =
area =
7
2
(2)
2
7
2
– 9 + (7 – 8 2 = –
11 2
2
+ –1 2 =
121
4
+1=
125
4
125
4
15) If log x y2 = 3, determine the value of log y x2 . Express your answer as a rational number in lowest terms.
log x y2 = 3 ï 2 logx y = 3 ï logx y =
3
2
ï log y x =
2
3
ï log y x2 =
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
z=a+b ï z + |z|=a+ bi+
a+
a2 + 81 = 3 ï
= 3 + 9 Â. Determine the value of z 2 .
a2 + b2 = 3 + 9 i ï b = 9 and a +
a2 + 81 = 3 – a
a2 + 81 = 9 – 6a + a2 ï 6a = – 72 ï a = – 12
a2 + 81 = 3
z
2
= a2 + b2 = –12 2 + 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.
20
2
The total number of pairs chosen from 20 =
=
20!
2! ÿ 18!
=
20 ÿ 19
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
The probability that the sum is divisible by 3 =
= 15 + 49 = 64
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
60 x2 + 153 x + 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
x3
b2 – a2
b–a
+
c2 – b2
a2 – c2
+ a–c
c–b
= b + a + c + a + a + c = 2(a + b + c)
60 x2 + 153 x + 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?
28 + x
5
1)
3, 6, 9, 10, x ï 3 + 6 + 9 + 10 + x = 28 + x ï 9 =
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 BC and CD respectively such that BM = ND = 1
and let q = — MAN. Find sin q .
N
D
C
1
2
2
3
N
D
C
1
2
2
2 2
3
10
M
10
1
θ
A
B
3
By the law of cosines
2
2
2
=
10
8 = 10 + 10 – 20 cos(q) ï cos(q) =
12
20
2
=
+
3
5
10
2
–2·
ï sin( ) =
10 ·
10 · cos(q)
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 + x 2 = (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
7
13
39
105
237
474
1
6
6
26
66
132
237
1
5
20
40
66
105
1
4
10
20
20
26
39
1
3
6
10
6
13
1
2
3
4
5
6
7
1
1
1
1
1
1
1
OR
6+6
6
4+2
4
–
2+4
2
–
2+4
2
23) If x and z are real number such that 2
4+2
4
x –3
= 474
z = – 9 and
x +
z = 23 , find x + z.
Adding the given equations:
y = –55 ï
–5
y = 11 ï
x = 12 ï x + y = 144 + 121 = 265
1
24) If sin(a) = 4 , find sin(3a). Express your answer as a rational number in lowest terms.
1
α
15
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 =
3–1
3+1
1
a2 =
2
1
2
=
–1
+1
– –1
3
1
– +1
–2 – 1
–2 + 1
–
1
16
=
1
4
44
ÿ 16 =
11
16
1
2
–1
=
2
3
=–
1
3
2
–4
=
2
a4 =
15
16
x– y
1
a3 =
3
= x + y . Define the sequence an by a1 = f (3,1) and an+1 = f an , 1 for n ¥ 1. Find a2011 .
25) Let f (x,y)
a1 =
1
4
3
2
=–2
3
=
–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
26) Find the sum of all of the positive real solutions of
x2 – 4 – 4 = 1 .
x2 −4 −4
= 1
x2 −4 −4 = 1
x2 −4 = 5
x2 −4 = 5
x2 = 9
x = 3
x2 −4 −4 = −1
x2 −4 = 3
x2 −4 = 3
x2 = 7
x2 −4 = −5
x2 = −1
Sum = 1 + 3 +
7 =4+
27 Triangle ABC is a 3
x =
x2 −4 = −3
x2 = 1
x = 1
7
7
5 right triangle with AB = 4.
4
C
Construct the perpendicular AD1 and let AD1 = x1 . Construct the
perpendicular D1 D2 and let D1 D2 = x2 . Construct the perpendicular
D1
D2 D3 and let D2 D3 = x3 . If this process is continued forever,
¶
find
D3
xk .
x1
k= 1
A
C
α
D1
α
D3
x1
α
x2
x3
A
x4 x
5
α
α
D2
B
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
sin(a) =
4
5
sin(a) =
x3
x2
ï x3 =
4 2
5
12
5
sin(a) =
4
5
sin(a) =
x4
x3
ï x4 =
4 3
5
12
5
In general xn =
4 n – 1 12
5
5
12
5
x2
D2
x3
B
¶
xk =
12
5
xk =
12
(1
5
k= 1
4 1 12
5
5
+
4 2
5
+
12
5
+
4 3 12
5
5
+· · ·
4 3
5
+· · · )=
12
1
5 1– 4
¶
k= 1
4 1
5
+
4 2
5
+
+
= 12
5
28) Let f (x) = 3 x2 – x. Find all values of x such that f (f(x)) = x .
2
f (f (x)) = x ï f 3 x2 –x = x ï 3 3 x2 –x – 3 x2 –x = x
ï
3 9 x4 –6 x3 + x2 – 3 x2 + x = x ï 9 x4 – 6 x3 = 0 ï x3 9 x – 6 = 0 ï x = 0 ,
29 In triangle ABC, —CAB = 30 °, AC = 2 and AB = 5
2
3
B
3
Find BC.
5 3
A
30°
2
C
By the law of cosines:
BC2 = 22 + 5
2
3
BC2 = 4 + 75 – 20
–2· 2· 5
3
2
3 ·
3 cos(30°)
= 79 – 30 = 49 ï BC = 7
OR
C
5 3
5
3
2
30°
A
x =
2
11
2
B
CD
5
3
2+x
5
3
BC2 =
D
= sin(30°) =
= cos(30°) =
5
3
2
2
+
1
2
ï CD =
3
2
11 2
2
5
3
2
ï 2+x=
=
75
4
+
121
4
=
15
2
ï x=
11
2
196
4
= 49 ï BC = 7
30) How many liters of a brine solution with a concentration of 30% salt should be added to 300 liters of brine with a concentration
of 23% salt so that the resulting solution has a concentration of 26% salt?
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
1
2A + B = 2 p 2
2
A + B = 16
2A + B = 2p
A = 2p – 4
4 = 8 – 16
33) Two large pumps and one small pump can fill a swimming pool in 4 hours. One large pump and three small pumps can
fill the same swimming pool in 4 hours. How many minutes will it take four large pumps and four small pumps, working
together, to fill the swimming pool? (Assume that all large pumps pump at the rate R and all small pumps pump at the rate r.)
Let L = rate of a large pump and S = rate for a small pump.
1
2L+S
1
L+3S
= 4 ï 8L + 4S = 1
(1)
= 4 ï 4L + 12S = 1
(2)
(1) – 2(2) ï 1 = 20S ï S =
1 = 8L + 4
t=
1
4 L+4S
1
20
= 8L +
1
=
4
1
10
+4
1
20
1
5
1
20
ï L=
=
1
2
5
+
1
=
5
1
3
1
10
=
5
3
hours = 100 minutes
5
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
a
b
c
e
d
f
g
e
England
From the given information:
a + b + c + d + e + f + g + e = 40
a + b + e + d =17
b + c + e + f = 28
d + e + f + g = 10
b + e = 12
g=3
a=4
(1)
(2)
(3)
(4)
(5)
(6)
(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
35) A bag contains 11 candy bars: three cost 50 cents each, four cost 1 dollar each and four cost 2 dollars each.
Three candy bars are randomly chosen from the bag, without replacement. What is the probability that the total
cost of the three candy bars is 4 dollars or more? Express your answer as a rational number in lowest terms.
Number of ways to pick 3 bars out of 11 =
11
3
=
11!
3! ÿ 8!
=
11 ÿ 10 ÿ 9 ÿ 8!
3 ÿ 2 ÿ 8!
=
11 ÿ 10 ÿ 9
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
1
> 0, determine the value of x + .
x
x
x4 + x3 + x2 + x + 1 = 0
x2 x2 + x + 1 +
x2 x2 +
x2 x +
1
x2
1
x
1
+
+ 2 +x +
1 2
x
+ x +
–1 ≤
1+4
x+
1
x
=
x+
1
x
>0 ï x+
2
=
1
x
=0
x2
1
x
1
x
–1 =0
–1 =0
–1 ≤ 5
2
=
–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
a3 + a3 r3 + a3 r6 + a3 r9 + · · · =
a3
1 – r3
(1)
=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 – 3 a + 3 a2 – a3 = 9a – 9 a2 + 3 a3
2 a3 – 9 a2 + 9a = 0
a 2 a2 – 9 a + 9 = 0 ï a(2a – 3)(a – 3) = 0 ï a = 0,
3
2
or 3
×
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
y
Points A = 0, 0 , B = 18, 24 and C = 11, 0 are the vertices of
triangle ABC. Point P is chosen in the interior of this triangle so that the
area of triangles ABP, APC and PBC are all equal. Find the coordinates of P.
Express your answer as an ordered pair x , y .
38
B 18,24
P
A 0,0
25
B 18,24
20
15
25
30
10
"h3
P
"h2
5
"h1
C 11,0
A 0,0
5
Area of DABC =
10
1
2
15
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
4
Using the distance from a point to a line formula
3
x–y=0
xp – 8
4 2
3
+ 12
4
=
44
15
ï
3
xp – 8
5
3
=
44
15
x
C 11,0
4
3
xp =
5
3
44
ÿ 15 + 8 =
(xp , h1 ) =
29
,
3
116
9
ï xp =
3
4
ÿ
116
9
=
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.
C
A
B
C
D
4
5
E
r
A
B
3
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 =
The area of the triangle can be computed as
C–A
2
1
(3)(4)
2
3
,
2
=
1
2
1
1
= 2 r(3) + 2 r(4) + 2 r(5) ï 12 = 12r ï r = 1
Thus E = (3 – 1 , 1) = (2 , 1).
3 2
DE2 = 2– 2
+ 1–2
2
=
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.
11
11 11 –1
=
ways to place the two ones. The total number of ones is then 2·
2
2
The ones are equally distributed in the n possible places, so each place will have 10 ones.
There are
Sum = (10) 20 + 21 + 21 + ÿ ÿ ÿ 210
Sum = 10
211 – 1
11 10
2
= 11(10).
Sum = 10 211 –1 = (10)(2048 – 1) = 20470
C
41 As shown in the sketch, circular arcs A C and B C have respective centers at
B and A. Suppose that S is a circle that is tangent to each of these arcs and also
to the line segment joining A and B. Find the radius of S if AB = 24.
A
D
C
E
r
G
r
r
x
A
12
B
F
AE = AB = radius of circular arc BC = 24 ï 2r + x = 24 ï x = 24 – 2r
2
= r2 + 122
From triangle AFG
r+x
Substituting for x
r + 24 – 2 r 2 = r2 + 122 ï
24 – r 2 = r2 + 122
242 – 48r + r2 = r2 + 122 ï 576 – 48r = 144 ï 48r = 432 ï r =
432
48
=9
B