Download 2013 - OCTM Tournament

2013 OHMIO Individual Competition: Solutions
1. Suppose that the numbers 1, 2, 4, 8, 16, 32, 64, 128, and 256 are placed into the nine
squares in such a way that the product of the numbers appearing in any row, column or
diagonal is the same. What is the value of this common product?
Answer: 4096
8
i
8
Solution: The product of all entries is
 2i  2 i0  2
9(8  0)
2
 2 36 . This is the product
i0
of all three rows, so any row would have a product of 2 36  3  212  4096
1
2. The region bounded by y  x  3 , 0  x  5 , y  0 , is rotated about the x-axis. What is the exact
volume of the solid, in terms of  ?
Answer: 35  or 11 2 
3
3
Solution 1: The region formed consists of two similar triangles. Using the disc
3
5
method, the required volume is:   (3  x)2 dx    (x  3)2 dx =
0


  9x3x
2
1 
 x3
3 
x3
1 3

x 3x 2 9x 
3

3
x5
 
x0
=  9  0    (
35
35
 9) 

3
3
x3
1

3
Solution 2: The volume of a cone of height 1 and radius 1 is  1  . The first revolution will be a
3
3

3
cone 3 times the size with volume 3   9 . Similarly (no pun intended), the second revolution will
3
8
35
8
3 
be a cone 2 times the size with volume 2   
, yielding a total volume of 9 

3
3
3
3
3. When Alice entered the Forest of Forgetfulness, she forgot the day of the week. She met the Lion and
Unicorn resting under a tree. The Lion lies on Mondays, Tuesdays, and Wednesdays and tells the truth on
the other days of the week. The Unicorn, on the other hand, lies on Thursdays, Fridays, and Saturdays,
but tells the truth on the other days of the week. They made the following statements:
Lion: “Yesterday was one of my lying days.”
Unicorn: “Yesterday was one of my lying days.”
From these two statements, Alice was able to deduce the day of the week. What day was it?
Answer: Thursday
Solution: If the Lion was telling the truth, the day of the week must be Thursday and if he is lying, then
the day of the week must be Monday. So the day of the week is either Monday or Thursday. If the
Unicorn is telling the truth, the day of the week must be Sunday. If he is lying, then the day of the week
must be Thursday. The intersection of the possible days is Thursday .
4. If sin x  cos x 
Answer:
1
, find the value of sin 3 x  cos3 x as a simplified fraction.
2
11
16
1
 (sin x  cos x)2  sin 2 x  cos2 x  2sin x cos x
4
1
3
 1  2sin x cos x  sin x cos x  
4
8
3
By factoring a sum of cubes, sin x  cos 3 x  (sin x  cos x)(sin 2 x  sin x cos x  cos2 x)
Solution: Square both sides:
3  11
 1 
Thus, sin 3 x  cos 3 x     1   
 2 
8  16
5. The longest diagonal of a cube is increasing at a rate of 5 feet per second. The simplified rate at which
the volume of the cube increases when the side length of the cube is 2 feet can be written as p q cubic
feet per second, where p and q are integers and q is a prime. Find the value of p + q.
Answer: 23
Solution: Letting s be the side length and D the diagonal, we have D  3s . Taking the derivative with
dD
ds
ds
5
 3 

respect to time, t:
.
dt
dt
dt
3
dV
5
60
dV
ds
 3(2)2

 20 3 ft 3 .
Since V  s 3 , we have
 3s 2 , thus,
dt
dt
dt
3
3
Thus, p + q = 20 + 3 = 23 .
6. On Pluto, the inhabitants use the same mathematical operators that we do (+, -, etc.), and they also use
an operator @ that we do not know. Scientists have determined that the following statements are true for
all real numbers x and y:
x@0  x
x@y  y@x
(x  1)@y  (x@y)  y  1
Find the value of 12@5.
Answer: 77
Solution: 12@5  5@12  4@12 13  3@12  26  2@12  39  1@12  52  0@12  65 
12@0  65  12  65  77
7. Point W is on the edge of a right cylinder. Point H is 72° away from point W
along the arc of the edge. Point T is on the other edge directly below point H.
Similarly, point A is 72° away from point T along that same edge. For a cylinder of
diameter and height equal to 5 , determine the exact area of quadrilateral WHAT.
Answer: 
Solution: When the lateral face is “unwrapped,” we get a rectangle of height
5
and width  5 . Quadrilateral WHAT is a parallelogram whose area is
72 1
 the area of the lateral
360 5
 5 5
  .
5
8. For what integral value of base B is 61B  51B  3731B ?
face, thus, the area is
Answer: 8
Solution: 6B  15B  1  3B 3  7B 2  3B  1 , thus 3B3  23B2  8B  0 .
1
B(3B2  23B  8)  B(3B  1)(B  8)  0  B  0 ,  , 8 .
3
9. All of the positive integers are written in a triangle pattern, beginning with the four lines and continuing
in the same way:
1
2 3 4
5 6 7 8 9
10 11 12 13 14 15 16
…
In this figure, which number appears directly below 2013?
Answer: 2103
Solution: Observe the following: The nth row has 2n – 1 numbers, that each number of the nth row is equal
to 2n – 2 plus the number directly above it, and that the last number in the nth row is n2. Noting 442 =
1936 and 452 = 2025, we can deduce 2013 is in the 45th row. The number below it equals
2013  2  46  2  2103
10. In the figure, CAPT is a square with diagonal AT = 2 2 and LAI is
equilateral. Points L and I lie on the perimeter of the square. The exact area of
LAI can be written as a 3  b where a and b are integers. Find a + b.
Answer: -4
Solution: The side length of the square is 2. Let LT = x. Then
2x  2 2  2  x  since the triangle is equilateral. Thus,
2
4  48
 2  2 3 , so LI  2x  2 2  2 6 . We
2
2
3
3
2 2  2 6 
32  16 3  8 3  12  a  b  4
then see area =
4
4
x 2  4x  8  0  x 




11. The zeroes of f (x)  x 3  6x 2  kx  64 can be arranged to form a geometric sequence of real
numbers. Find the value of k.
Answer: 24
Solution: Let a, a  r,a  r 2 denote the three roots. Then we can expand the factors to get:
(x  a)(x  ar)(x  ar 2 )  x 3  (a  ar  ar 2 )x 2  (a 2 r  a 2 r 2  a 2 r 3 )x  a 3r 3 .
We now have a system: a 1 r  r 2  6 , a2 r 1  r  r 2  k , and (ar)3  64 . From the last equation,
ar  4 . Using the first and second equations, we see that ar  a(1  r  r 2 )  k  4  6  k  k  24
12. For any real x, the function f (x) satisfies f (1  x)  (1  x) f (x)  5 . Find the value of f (5) .
20
Answer: 
21
 f (5)  5 f (4)  5
Solution: Substituting x = -4 and x = 5 respectively, 
.
 f (4)  4 f (5)  5
Multiplying the second equation by -5 and summing results in f (5)  
20
.
21
13. When three fair six-sided die are rolled, the probability that the sum of the three rolls will be at most 6
p
can be expressed as
where p and q are relatively prime integers. Find the value of p + q.
q
Answer: 59
Solution: There is 1 way to roll a 3 (1-1-1), 3 ways to roll a 4 (1-1-2, 1-2-1, or 2-1-1), 6 ways to roll a 5
(1-2-2, 2-1-2, 2-2-1, 3-1-1, 1-3-1, 1-1-3), and 10 ways to roll a 6
(1-1-4, 1-4-1, 4-1-1, 1-2-3, 1-3-2, 2-3-1, 2-1-3, 3-1-2, 3-2-1, or 2-2-2) . Thus, the desired probability is
1  3  6  10 20
5
. Hence, p + q = 59 .


3
6
216 54
14. A wooden block is in the shape of a right rectangular prism whose dimensions are n cm by n + 3 cm
by n + 9 cm for some integer n. The surface of the block is painted and the block is then cut into 1 cm
cubes by cuts parallel to the faces. If exactly half of these cubes have no paint on them, find the volume
of the original block.
Answer: 1120
Solution: There will be (n  2)(n  1)(n  7) blocks that will have no painted faces. Thus,
2(n  2)(n  1)(n  7)  n(n  3)(n  9) , hence 2n3  12n2  18n  28  n3  12n2  27n .
We can say n3  45n  28  0 . By the Rational Root Theorem, we wish to try n  1,2,4,7,14, 28
as possible roots. By trial and error, n = 7, so the volume of the original block is (7)(10)(16) = 1120 .
15. Let R  (x, y) : x  y  2and S  (x, y) : x  y  2. Determine the exact value of the area of the
region formed by R S .
Answer: 8
Solution: By the definition of absolute value, R consists of the points between
the lines y  x  2 and y  x  2 while S consists of the points between the lines
y  x  2 and y  x  2 . The intersection of these regions will form a square
centered at the origin with vertices at (0, 2) & (2,0) yielding side length 2 2 .
  8 .
Thus the desired area is 2 2
2
16. Itchy and Scratchy are playing a dice game with a fair 12-sided die. Itchy goes first and they alternate
rolls. The first to roll a prime number wins. Find the probability that Itchy wins as a simplified fraction.
Answer:
12
19
Solution: There are 5 possible prime faces. Let p 
5
be the probability of rolling a prime face on a roll
12
7
be the probability otherwise. Then Itchy can win with probability
12


 1  5  1  12
2
4
p  q p  q p  ...  p 


.
 1  q 2  12  1  49  19


144 
m
4
17. Find all positive integer pairs (m, n) such that
.

5n  69 n  12
Answer: (22, 6) and (21, 24)
and q 
Solution: mn  12m  20n  276  (m  20)(n  12)  36 . Since m and n are positive and exploring
m  20  1
 m  20  2
factors of 36, we have two cases: 
so (m, n) = 21, 24  or 
so (m, n) = 22,6  .
n  12  36
n  12  18
18. The measures of the interior angles in a convex hexagon form an arithmetic sequence. If the
difference between the second largest and second smallest angle is 36 degrees, what is the measure of the
smallest angle in degrees?
Answer: 90
Solution: The difference between the specified angles is 3d, where d is the common difference. Thus d =
720
 5
12. The angles averaged to
 120 so the smallest angle is 120    (12)  90 .
 2
6
19. Consider polynomials f, g, and h, where f is an even function, and g and h are odd functions. These
functions satisfy the following:
f (1)  0, f (4)  2, f (3)  6.
g(1)  1, g(2)  4, g(5)  3.
h(1)  2, h(3)  5, h(6)  3.
Find the value of g 1 (h( f (3)))  h( f (g(1)))  f (g(h(1))) .
Answer: 7
Solution: f (g(2))  g 1 (h(6))  h( f (1)) = f (4)  g 1 (3)  h(0) = 2  5  0  7 .
20. Right triangle ABC is inscribed in equilateral triangle PQR as shown. Given PC = 3, BP = CQ = 2,
find AQ.
Answer:
8
3
or 1 or 1.6
5
5
Solution: Let AQ = x, RA = 5 – x and note that BR = 3. Using the Law of Cosines:
1
BC 2  2 2  32  2(2)(3)  BC 2  7
2
1
AC 2  x 2  2 2  2(x)(2)  AC 2  x 2  2x  4
2
1
BA2  32  (5  x)2  2(3)(5  x)  BC 2  x 2  7x  19
2
By the Pythagorean Theorem, 7  x 2  2x  4  x 2  7x  19  x 
8
5
TB 1. Suppose that f (n)  log 2 3 log 3 4 ... logn1 n . Determine the value of
10
 f (2 ) .
k
k2
Answer: 54
Solution: By the Change of Base Formula, f (n) 
Thus, f (2 )  k and
k
10
k 
k 2
log 3 log 4
log n
log n

...

 log 2 n .
log 2 log 3 log(n  1) log 2
(2  10)
(9)  54 .
2
TB 2. Mitt is walking down a stairway with 10 steps. He can only take one or two steps at a time. For
example, for 3 steps, he can either take three single steps, one step then two steps, or two steps then one
step. In how many ways can Mitt walk down these 10 steps?
Answer: 89
Solution: This pattern follows recursively. Let fn denote the number of ways he can walk down n steps.
Note that for stairways of one or two steps, f1  1 (just one step) and f2  2 (either two single steps or one
two step). For fn with n  3 , he could either be on the (n – 1)st step or the (n – 2)nd step to get to the end
of the stairway. Thus, we have the recursive relation fn  fn1  fn 2 , which defines the Fibonacci
Numbers. Thus,
fn  1, 2, 3, 5, 8,13, 21, 34, 55, 89,... and there are
89 ways.