Download 2010 - OCTM Tournament

OHMIO 2010 Ciphering and Solutions
Practice.
An employee at a car dealership is promised a car, a computer, and $18,000 in cash for her year's work at the
dealership, with one-twelfth of the total value of the car, computer, and cash earned each month that she works.
She is terminated after seven months and is given the car and $5,000 in cash. Had she worked another month,
she would have received $1,000 more in cash and the computer in addition to the car. To the nearest dollar,
what is the value of the car.
Solution:
Let a = value of the car and c = value of the computer
7(a  c  18000)
8(a  c  18000)
 a  5000
and
 a  c  6000
12
12
solving the 2 equations we get a  16, 000 or $16,000
!. The number 1234567891011...585960, which consists of the first 60 positive integers written in order to form a
single number with 111 digits is modified by removing 100 of its digits. What is the smallest number which can be
obtained in this way? (If the number begins with some 0 digits, you may write it either with or without the leading
0's). Credit will be given for both answers.
Solution:
We can only get five 0's at the beginning of the number and the last one must come at the end. Using the
smallest numbers first, the number is 00000123450.
2. In triangle ABC, angle A is 120 degrees, BC + AB = 21, and BC + AC = 20. What is the length of BC?
Solution:
Let b=AC. Then BC = 20  b and AB = b  1. By the Law of Cosines,
(20  b)2  b2   b  1  2b  b  1 cos 120  .and therefore,
2
400-40b+b2  2b2  2b  1  b  b  1 or 2b 2  43b  399  0.
Factoring we get  b-7  2b  57   0
and b  7 and therefore, BC = 13
3. The sums of 5 numbers a, b, c, d, and e taken in pairs are 183, 186, 187, 190, 191, 192 193, 194, 196, and
200. If a<b<c<d<e, determine the value of a.
Solution: The sum of all pairs uses each number 4 times giving
4a  4b  4c  4d  4e  1912
or a  b  c  d  e  478
If a and b are the smallest numbers and d and e are the largest, then
183 + c + 200 = 478 and c = 95, since a + c is the second largest sum
a = 186 - 95 = 91
4.
In triangle ABC, medians AE and BD intersect at F. If mBAC  mAFB  90 , and AB = 12,
find the length of BC.
Solution:
If we place the triangle on the coordianate axes so that A is at the origin, B is at (12,0) and C is at
(0,2b), then D is at (0,b) and E is a (6,b). Since AE and BD are perpendicular, their slopes are
 b  b 
negative recirocals. Therefore,     1
 6  12 
 BC 
2
b 2  72 In right triangle ABC,
 4b 2  144  4  72   144  432, BC= 432  12 3
5.
Find the coordinates of the point, other than the origin, on the curve y  3x 2  2 x3 , where
the tangent that passes through the origin intersects the curve.
Solution:
dy
 6 x  6 x 2 . The point we are seeking is  a,3a 2  2a3 
dx
and the tangent line has the equation
3a2  2a3  0    6a  6a 2   a  0  or
The slope of tangents to the curve is
4a3  3a 2  0
.
a 2  4a  3  0,
a  0, a 
3
 3 27 
, therefore the point is  , 
4
 4 32 
6.
If cats consider one rat worth three mice, a squirrel worth as much as one rat and
one mouse, and five chipmunks worth three rats, how many chipmunks would a cat pay
for ninety-seven mice and thirty-two squirrels?
Solution:
Let r, m, s, and c represent the values of one rat, one mouse, one squirrel, and one
chipmunk in that order. Then
r = 3m, s = r + m, and 5c = 3r Solving for each in terms of c, we get
5c
r
5c
20c
r=
, m= =
, and s = r + m =
97m + 32s =
3
3
9
9
 5c 
 20c 
97    32 
  125c
 9
 9 
The cat would pay 125 chipmunks for 97 mice and 32 squirrels
7.
The sum of n consecutive positive odd integers, starting with the least one, a, is 105.
Find the smallest possible value for a.
Solution:
The arithmetic progression a   a  2    a  4   ...   a  2  n  1   105  3  5  7
n
or the sum    2a  2n  2   n  a  n  1  3  5  7
2
n could have the values of 1, 3, 5, 7, 15, 21, 35, or 105. n values of 15, 21, 35, and 105 would
give a negative value to a. The n values of 1, 3, 5, and 7, give values to a of
105, 33, 17, and 9 respectively. The smallest is 9.
8..
Determine an expression for b in terms of a by eliminating x from the
following equations:
cos x  sin x  a
cos 2 x  b
Solution:
Squaring the first equation produces
sin 2 x  a 2  1
a 4  2a 2  1  1  b 2
cos 2 x  2sin x cos x  sin 2 x  a 2 or
 sin2x    cos 2 x 
2
or
2
  a 2  1  b2  1
b2 = 2a 2  a 4 or
2
b   a 2  a2
Tie Breaker 1
The equation x 3  ax2  bx  c  0 has the root
3
28  6  3 28  6.
If a, b, and c are integers, find the ordered triple  a, b, c  .
Solution:
Let x 
28  6  3 28  6. Then x3 =
3
and x3  12  3 3 8  x or


28  6 -

28  6 -3 3 28-36

3
28  6  3 28  6
x3  6 x  12  0 and therefore,  a, b, c    0, 6, 12 
Tie Breaker 2
If 2  4 x   6 x  9 x and x  log 2 a, find the numerical value of a.
3
Solution:
2  4 x  6 x  9 x , and therefore 2 2 x 1  2 x  3x  32 x  0. Factoring this equation gives
2
x 1
3
x
 2
so x  log 2
3
x
3
x
  0,
1
1
and a 
2
2
x
2  3 ,
x
x
2
x+1
=3
x
1
2
gives   
2
3
