2015 OHMIO Individual Competition
1. A sequence has 1 as the first term. Each term thereafter is the sum of all the values of
the previous terms. If the 2015th value in the sequence can be written as 2 n , find the
value of n.
2. As t takes on all real values, the set of points (x, y) is defined parametrically by
x = t 2 - 2 , y = t 3 - 9t + 5 . Find the ordered pair where the curve crosses itself.
cos 2x
sin 2x
3. For some angle x, suppose
.
= 2 . Find the exact value of
cos x
sin x
4. In the picture at right, two semicircles are drawn within a quarter circle.
What percentage of the area of region B is the area of region A?
x+ y
.
2
Suppose positive integers a, b, c are all less than or equal to 2017. What is the maximum
possible value of aV(bVc)- (aVb)Vc ?
5. For real numbers x and y, define the operation V as follows: xV y =
6. Find the area of the region above the graph of y = x -1 + x - 3 and below the graph of
y = 8.
7. Three vertices of a cube are connected to form a triangle. The quotient of the largest
and smallest possible perimeters of the triangle can be written in the form a 2 - b for
positive integers a, b. Find a + b.
8. The region enclosed by a trapezoid with bases of 20 units and 15 units and height 6
units is revolved around the longer base to create a three-dimensional solid. In terms of
p , what is the volume of the solid?
9. An unfair coin is flipped four times. The probability of obtaining 2 heads and 2 tails in
any order is equal to the probability of obtaining 3 heads and 1 tail in any order. If the
probability of either event is between 0 and 1, what is the probability of obtaining a head
in just one flip? Write as a simplified fraction.
10. Find the sum of the roots to the equation log x log x - log x 3 + 2 = 0 .
11. Three sides of a quadrilateral measure 5, 18, and 7 units. How many integer values
could be the length of the fourth side?
12. If x 3 + 5x 2 + 3x -1= A(x -1)3 + B(x -1)2 + C(x -1)+ D is an identity for constants
A, B, C, D, determine the value of B + D.
13. A pump can fill a pool at a rate Rt   8t 2 t 3  9 in cubic feet per minute. If the
pump takes 3 minutes to fill an empty pool with water, what is the volume of the pool in
cubic feet?
14. Let z be a complex number and z be its conjugate. When graphed, the equation
z × z +(-3+ 4i)z +(-3- 4i)z = 0 represents a circle centered at (a, b) with diameter c. Find
the value of a + b + c.
15. Find the coefficient of x 5 when (x 2 - x + 2)(x - 2)7 is expanded.
16. Brian throws a fair 6-sided die. If it comes up greater than 3, he wins. If not, he
throws it again and wins if it comes up greater than 4. If not, he throws it again and wins
only if it comes up greater than 5. Compute the probability Brian will win, written as a
simplified fraction.
17. When graphed in a polar coordinate system, how many petals are there in the graph of
r = cos ( 8q ) sin ( 8q ) ?
18. Katy Perry counts in a strange way: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, etc., where she starts
back over from 1 after reaching a new integer. How many total integers will she count
(including repeats) when she first counts 100?
19. Find the least solution to the equation (x - 3)3 + (x + 4)3 = (2x +1)3 .
20. The numbers 0, 1, 10, 11, 100, 101, 110, 111 represents the binary representation of
the integers from 0 to 7, requiring twelve 1’s. How many total 1’s are needed when the
integers 0 to 1023, inclusive, are written in binary?
TB1. Let f (x) = e xe
....
xexe
and g(x) = 2x + 2x + 2x +... . Find the value of f '(0)× g'(3) .
é 2 0 ù
é 3 -1 ù
TB2. Consider the following matrices: A = ê
ú, B = ê
ú . Find the
ë 1 5 û
ë 4 1 û
-1
ææ
-1 öT ö
T
æ
ö
T -1
÷
determinant of the following matrix: ççç ( AB)
÷ ÷÷ ÷ .
ççè
ø ø
èè
ø
((
)
)
2015 OHMIO Individual Competition Answers
1. 2013
11. 23
2. (7, 5)
12. 16
3. 1- 3 or - 3 +1
13. 336
4. 100%
14. 9
5. 504
15. 1008
6. 30
13
16.
18
7. 6
17. 32
8. 600 p
18. 5050
9.
3
5
10. 110
19. -4
20. 5120
2
TB1.
5
1
TB2.
70
2015 OHMIO Individual Competition Solutions
1. The first few terms are: 1, 1, 2, 4, 8, 16, 32. For m >1, it appears the mth term is 2 m-2 .
Thus, the 2015th term is 2 2013 so n = 2013.
2. There should be two values t1 ¹ t 2 where t12 - 2 = t2 2 - 2 and t13 - 9t1 + 5 = t2 3 - 9t2 + 5 .
The first equation implies t1 = ±t2 so t1 = -t 2 . Substituting into the second equation,
t13 - 9t1 + 5 = -t13 + 9t1 + 5 . This simplifies to t13 = 9t1 for which t1 = 0, 3,-3. Since we
require two different values of t, we substitute either 3 or -3 to find the ordered pair to be
(7, 5).
2 ± 12 1± 3
cos 2x 2 cos2 x -1
=
3.
.
=
= 2 ® 2 cos2 x - 2 cos x -1 = 0 . Thus, cos x =
4
2
cos x
cos x
1- 3
sin 2x 2sin x cos x
Since cos x £ 1, cos x =
and
=
= 2 cos x = 1- 3 .
2
sin x
sin x
4. Without loss of generality, assume the quarter circle has radius 2. Its area is
1
1
p
p (2)2 = p . The area of either semicircle is p (1)2 = . The area of region A is
4
2
2
p 1
p
p p
2( - ×1×1) = -1. Either region adjacent to region A has area - ( -1) = 1 . The
4 2
2
2 2
æp
ö p
area of region B is p - ç -1+ 2 ×1÷ = -1 . Since A and B have the same area, the area
è2
ø 2
of region A is 100% of the area of region B.
b+c a+b
a+
+ c 2a + b + c - (a + b + 2c) a - c
2 - 2
5. aV(bVc) - (aVb)Vc =
. The greatest
=
=
2
2
4
4
2017 -1
possible value is
= 504 .
4
ì 4 - 2x
x £1
ï
6. y = x -1 + x - 3 = í
2
1 < x < 3 . Graphing these lines coupled with y = 8
ï 2x - 4
x³3
î
yields a trapezoidal region with height 6 and base lengths of 8 and 2. The area is
1
(8 + 2)(6) = 30 .
2
7. Without loss of generality, let the edge of the cube be 1 unit. The smallest possible
perimeter is 1+1+ 2 = 2 + 2 when the triangle has two sides as edges of the cube. The
largest possible perimeter is 2 + 2 + 2 = 3 2 when the triangle has all three sides as
diagonals of the cube. The ratio is
3 2
3
=
= 3 2 - 3. So a + b = 6.
2+ 2
2 +1
8. Without loss of generality, consider a right trapezoid as shown below. When revolved
about the larger base, two disjoint solids will be formed: a cylinder of radius 6 and height
1
15 and a cone of radius 6 and height 5. The total volume is p 6 2 ×15 + p 6 2 × 5 = 600 p .
3
15
6
5
15
9. Let 0 < p < 1 be the probability of a head in one flip. Then
æ 4 ö 2
æ 4 ö 3
3
ç
÷ p (1- p)2 = ç
÷ p (1- p) . Simplifying, 6(1- p) = 4 p and thus p = .
5
è 2 ø
è 1 ø
10. Use the power property of logs to write ( log x ) - 3log x + 2 = 0 . Factoring,
(log x - 2)(log x -1) = 0 so x =100,10 and the sum is 110 .
11. Let x be the length of the fourth side. If x is the largest side, then 5 + 7 +18 > x so
x < 30 . If x is the smallest side, then x + 5 + 7 >18 so x > 6 . Thus x can be any integer
from 7 to 29, inclusive, yielding 29 – 7 + 1 = 23 possible values.
12. Let x = 0 to find -1= -A+ B -C + D . Let x = 2 to find 33 = A+ B+C + D . Add these
two equations to find 32 = 2(B+ D) so B+ D =16. [Note you could also use derivatives].
2
13. V =
36
3
ò 8t
0
2
8 36 12
16 23
16
t + 9dt = ò u du = u
= ( 216 - 27 ) = 336 cubic feet.
39
9
9
9
3
14. Let z = x + yi . Substituting and simplifying, we find x 2 + y2 - 6x + 8y = 0 or
(x - 3)2 + (y + 4)2 = 25 after completing the square. We find the center to be (3, -4) and
the diameter to be 10 so a + b + c = 3 + -4 + 10 = 9.
15. By the Binomial Theorem, we have
æ 7 ö 6
æ 7 ö 5
æ 7 ö 4
æ 7 ö 3
(x 2 - x + 2)(x 7 -2 × ç
÷x + 4 ×ç
÷x - 8 ×ç
÷ x +16 × ç
÷ x -+...-128).
è 1 ø
è 2 ø
è 3 ø
è 4 ø
The coefficient of x 5 is
æ 7 ö
æ 7 ö
æ 7 ö
2 × 4 ×ç
÷ + (-1)(-8)× ç
÷ +1×16 × ç
÷ = 8 × 21+ 8 × 35 +16 × 35 = 8(21+ 35 + 70) = 8(126) = 1008
è 2 ø
è 3 ø
è 4 ø
16. The desired probability is
3 3 2 3 4 1 156 13
+ × + × × =
= .
6 6 6 6 6 6 216 18
1
17. Using a double angle identity, r = sin (16q ) . This graph has 32 petals.
2
18. The first position when the new integer n is reached is 1+ 2 + 3+...+ n =
The first time 100 is counted requires
n(n +1)
.
2
100(101)
= 5050 total numbers.
2
19. Let a = x – 3, b = x + 4. The equation becomes a 3 + b 3 = (a + b)3 . Expanding the
right side, a 3 + b 3 = a 3 + 3ab(a + b)+ b 3 so a = 0,b = 0, or a = -b . We find x = 3, x = -4, or
1
x = - , the least of which is -4.
2
20. Noting 1024 = 210 we require arrays of 10 bits, each being a 0 or 1. This accounts for
10 × 210 such bits used in total, half of which are 1’s by symmetry. The total number of 1’s
required is then 10 × 29 = 5120 .
xf ( x)
so f (0) = e0× f (0) =1. Also g(x) = 2x + g(x) and g(3) = 6 + g(3) .
TB1. Note f (x) = e
This means g2 (3)- g(3)- 6 = 0 so g(3) = -2, 3, the former being extraneous. So g(3) = 3 .
Differentiating, f '(x) = ( f (x)+ xf '(x))exf (x) so f '(0) = ( f (0)+ 0 × f '(0))e0× f (0) = f (0) =1
1
1
1
1
2
2
g'(x)
=
(2x
+
g(x))
(2
+
g'(x))
g'(3)
=
(2
×
3+
g(3))
(2 + g'(3)) . This means
Also,
and
2
2
1 - 12
2
2 2
g'(3) = (9) (2 + g'(3)) and so g'(3) = . Hence, f '(0)× g'(3) = 1× = .
2
5
5 5
é 2 0 ù
é 3 -1 ù
é 6 -2 ù
ú, B = ê
ú First, AB = ê
ú and its determinant is 70.
TB2. A = ê
ë 1 5 û
ë 4 1 û
ë 23 4 û
T ö-1
ææ
T ö-1 ö
æ
-1
-1
1
T
çç
÷ ÷ = ( AB)T . Since det(U T ) =U and det(U -1 ) =
for some
AB
(
)
ç
÷
÷ ÷
ççè
det(U)
ø
ø ø
èè
1
matrix U, then this determinant is
70
((
)
)
(
)