UNIVERSITY OF VERMONT DEPARTMENT OF MATHEMATICS AND STATISTICS

UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-FOURTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 9, 2011
1) Evaluate I1 +
1
1
1
1
M I1 + 11 M I1 + 12 M I1 + 13 M.
10
I 10 M I 11 M I 12 M I 13 M =
11
12
13
14
14
10
=
+ 2012!
I 2011!
M
2011! + 2010!
2011
7
5
+ 2012!
I 2011!
M
2011! + 2010!
2) Simplify the expression
2011
2011! H1 + 2012L
2010! H2011+1L
=
Express your answer as a rational number in lowest terms.
2011
. Express your answer as a rational number in lowest terms.
2011 H2013L
=
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 H23L
2
y=x–4 ï y=–2±
x+y= 2±
23 + (– 2 ±
=2±
23
23
23 ) = 2
23
4) Find the value of Ha + bL2 if ab = – 14 and a2 + b2 = 30 .
Ha + bL2 = a2 + 2 ab + b2 = 30 + 2(–14) = 2
5L 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 FHx, y, zL = x y + y z + z x. For which real numbers a is FH2, a, a - 1L = FH5, a, a + 1L?
FH2, a, a - 1L = FH5, a, a + 1L 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
9
+ 5 G = x(B + G) = xI 10 G + GM
6
5
·
10
19
=
12
19
11) Find the integer value of the expression log 7 I 8 M ä Ilog 8 25 + log 2 5M ä log 5 49 .
1
1
3
log8 25 = log23 25 =
log2 52 =
2
3
log2 5 ï log 8 25 + log 2 5 =
5
3
log2 5
log7 I 8 M = – log7 8 = – log7 23 = – 3 log7 2
1
log5 49 = log5 72 = 2 log5 7 =
2
log7 5
log7 I 8 M · ( log 8 25 + log 2 5) · log5 49 = – 3log7 2 ·
1
5
3
2
log7 5
log2 5 ·
= – 10
log7 2
log7 5
· log2 5 = – 10
log2 5
log2 5
= – 10
12) If sin(x) + cos(x) = 2 , find the value of sin3 HxL + cos3 HxL. Express your answer as a rational number in lowest terms.
1
(sin(x) + cosHxLL3 = sin3 HxL + 3 sin2 HxL cosHxL + 3 sinHxL cos2 HxL + cos3 HxL
I2M =
1 3
1
8
= sin3 HxL + cos3 HxL + 3 sinHxL cosHxL[sin(x) + cos(x)]
(sin(x) + cosHxLL2 = sin2 HxL + 2 sinHxL cosHxL + cos2 HxL
I 2 M = sin2 HxL + cos2 HxL + 2 sinHxL cosHxL ï
1 2
1
8
1
4
= 1 + 2sin(x)cos(x) ï sin(x)cos(x) = –
= sin3 HxL + cos3 HxL + 3 ÿ I–I 8 MMA 2 E ï sin3 HxL + cos3 HxL =
3
1
1
8
+
9
16
=
11
16
13L 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
H1,5L
1
4
2
H5,3L
4
H0,1L
3
1
2
H2,0L
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
14L Find the area of the circle that contains the point Q H9, 8L and that is tangent
to the line x – 2 y = 2 at the point P H6, 2L.
QH9,8L
CHa,bL
PH6,2L
b–2
a–6
= – 2 ï b – 2 = – 2a + 12 ï 2a +b = 14
(1)
(a – 9L2 + (b – 8L2 = (a – 6L2 + (b – 2L2
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 =
7
2
radius2 = I 2 – 9M + (7 – 8L2 = I–
7
area =
(2)
2
11 2
M
2
+ H–1L2 =
121
4
+1=
125
4
125
p
4
15) If log x Iy2 M = 3, determine the value of log y Ix2 M. Express your answer as a rational number in lowest terms.
log x Iy2 M = 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 = H–12L2 + 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 = K
20
O=
2
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 K
7
7
6
O + K O · K O = 15 + 49 = 64
2
1
1
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 - 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 Ia, a2 M, Ib, b2 M and Ic, c2 M.
The sum of the slopes is
b2 – a2
b–a
+
c2 – b2
a2 – c2
+ a–c
c–b
= b + a + c + a + a + c = 2(a + b + c)
x3 - 60 x2 + 153 x + 1026 = Hx – aL Hx – bL Hx – cL = x3 – Ha + b + cL 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 86, 3, 10, 9, x< be equal to the median?
1)
83, 6, 9, 10, x< ï 3 + 6 + 9 + 10 + x = 28 + x ï 9 =
2)
83, 6, 9, x, 10<
ï 9=
28 + x
5
ï 45 = 28 + x ï x = 17
3)
83, 6, x, 9, 10<
ï x=
28 + x
5
ï 5x = 28 + x ï 4x = 28 ï x = 7
4)
83, x, 6, 9, 10<
ï 6=
28 + x
5
ï 30 = 28 + x ï x = 2
5)
8x, 3, 6, 9, 10<
ï 6=
28 + x
5
ï 30 = 28 + x ï x = 2 ï x = 7
28 + x
5
ï 45 = 28 + x ï x = 17
â
â
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 sinHqL.
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 I2
2 M = I 10 M + I 10 M – 2 ·
2
2
8 = 10 + 10 – 20 cos(q) ï cos(q) =
12
20
=
2
3
5
ï sin(q) =
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 + xL2 = (5 + x)(20 + x) ï 121 + 22x + x2 = 100 + 25x + x2 ï 21 = 3x ï x = 7
22L 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
K
6+6
4+2
2+4
2+4
4+2
O–K
OK
O–K
OK
O = 474
4
4
6
2
2
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:
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) + Icos2 HaL –sin2 IaMM sinHaL
sin(3a) = sinHaLA3 cos2 HaL – sin2 HaLE =
3–1
3+1
1
a2 =
2
1
2
=
–1
+1
– –1
3
1
– +1
–2 – 1
–2 + 1
–
1
M
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 =
I3 µ
= x + y . Define the sequence 8an < by a1 = f (3,1) and an+1 = f Han , 1L 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+
x =
x2 −4 = −3
x2 = 1
x = 1
7
7
27L Triangle ABC is a 3 - 4 - 5 right triangle with AB = 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,
⁄ xk .
¶
find
D3
x1
k= 1
x3
A
C
α
D1
α
D3
x1
α
x2
x3
A
x2
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 = I 5 M I
12
M
5
sin(a) =
4
5
sin(a) =
x4
x3
ï x4 = I 5 M I
12
M
5
In general xn = I 5 M
I
12
M
5
4 2
4 n – 1 12
I5M
4 3
D2
B
⁄ xk = 125 +
¶
k= 1
I5M I
4 1 12
M
5
⁄ xk = 125 (1 +
+ I5M I
4 2
12
M
5
+ I5M I
4 3 12
M
5
+· · ·
¶
I5M + I5M + I5M + · · · ) =
4 1
k= 1
4 2
4 3
12
1
5 1– 4
= 12
5
28) Let f (x) = 3 x2 – x. Find all values of x such that f (f(x)) = x .
f (f (x)) = x ï f I3 x2 –xM = x ï 3 I3 x2 –xM – I3 x2 –xM = x
2
ï
3 I9 x4 –6 x3 + x2 M – 3 x2 + x = x ï 9 x4 – 6 x3 = 0 ï x3 H9 x – 6L = 0 ï x = 0 ,
29L 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 + I5
3 M –2· 2· 5
2
BC2 = 4 + 75 – 20
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
D
= sin(30°) =
= cos(30°) =
BC2 = K
5
3
2
2
O +I
1
2
ï CD =
3
2
11 2
M
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 K
4
O 3! = 6 · 6 = 36
2
3 tied: pick 3 from 4 arrange in 2! ways K
4
O 2! = 4 · 2 = 8
3
4 tied: either all 4 or two and two
all four 1 way, two and two pick two from 4 to be first 1 + K
4
O=1+6
2
Total = 24 + 36 + 8 + 7 = 75
32L 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 = 2 p H2L2
1
A + B = 16
2A + B = 2p
A = 2p – 4
4 = 8p – 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 I 20 M = 8L +
1
t=
1
4 L+4S
=
1
4J
1
10
N+4J
1
20
N
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
f
d
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 = K
11
O=
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
K
4
O=4
3
$2 + $2 + $1
K
4
4
O · K O = 6(4) = 24
2
1
$2 + $2 + $.5
K
4
3
O · K O = 6(3) = 18
2
1
$2 + $1 + $1
K
4
4
O · K O = 4(6) = 24
1
2
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 Ix2 + x + 1 +
x2 Ix2 +
x2
x2 JIx +
1 2
M
x
1
1
x
1
+
x2
+ 2 +x +
+ Ix +
–1 ≤
1+4
x+
1
x
=
x+
1
x
>0 ï x+
2
1
M
x
– 1M = 0
– 1N = 0
–1 ≤ 5
2
=
1
x
1
x
M=0
=
–1 +
2
5
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 I1 – r3 M
Substituting r = 1 – a ï a3 = 3A1 – H1 – aL3 E
a3 = 3A1– 91 – 3 a + 3 a2 – a3 =E = 9a – 9 a2 + 3 a3
2 a3 – 9 a2 + 9a = 0
aI2 a2 – 9 a + 9M = 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
38L Points A = H0, 0L, B = H18, 24L and C = H11, 0L 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 Hx , yL.
y
BH18,24L
P
AH0,0L
25
BH18,24L
20
15
25
30
10
"h3
P
"h2
5
"h1
CH11,0L
AH0,0L
5
Area of DABC =
10
1
2
15
H11L H24L = 132 ï Area DABP = DAPC = DPBC =
132
3
= 44
Let P = ( xp , h1 )
1
2
H11L Hh1 L = 44 ï h1 = 8
1
2
H30L Hh3 L = 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
4
xp – 8
=
J N + 12
4 2
3
44
15
ï
3
xp – 8
5
3
=
44
15
x
CH11,0L
4
3
xp =
5
3
44
ÿ 15 + 8 =
(xp , h1 ) = I
29
,
3
116
9
ï xp =
3
4
ÿ
116
9
=
29
3
8M
39L 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
3
B
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
= I 2 , 2M
3
1
1
1
= 2 r(3) + 2 r(4) + 2 r(5) ï 12 = 12r ï r = 1
Thus E = (3 – 1 , 1) = (2 , 1).
DE2 = I2– 2 M + H1 – 2L2 =
3 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 H11 –1L
O=
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 K
Sum = (10) A20 + 21 + 21 + ÿ ÿ ÿ 210 E
Sum = H10L A211 – 1E
11 H10L
2
= 11(10).
Sum = H10L I211 –1M = (10)(2048 – 1) = 20470
ì
ì
41L As shown in the sketch, circular arcs A C and B C have respective centers at
C
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
From triangle AFG Hr + xL2 = r2 + 122
Substituting for x
Hr + 24 – 2 rL2 = r2 + 122 ï
H24 – rL2 = r2 + 122
242 – 48r + r2 = r2 + 122 ï 576 – 48r = 144 ï 48r = 432 ï r =
432
48
=9
B