Download 2009 - OCTM Tournament

2009 OHMIO PRESSURE ROUND
QUESTIONS AND SOLUTIONS
1. Express in simplest form: (log2 8)(log2 4)
[log2 (1/4)][log2 (1/2)]
Solution:
__ (log3 9) + (log3 27)
log3 (1/3)
3 2
23 6 5

 
 35  8
 2   1 1 2 1
2. Given that xy = 1 and that x + y = 1, find the value of x3 + x2 + x + y3 + y2 + y + 1.
Solution:
 group like powers of x and y, then factor or rewrite  :We are given that xy  1. Factoring and
2
substituting given x 2  y 2   x  y   2 xy  12  2 1  1. Finally,  x3  y 3    x  y   x 2  xy  y 2  
1 1  1  2. Therefore the given expression simplifies to  2  1  1  1  1
Method 1:
Method 2:
 solve and substitute  :
The system x  y  1 and xy  1 leads to the equation x 
or equivalently, x 2  x  1  0. This equation has two complex solutions:
 exponential  form, e
i
 i
3

1
 1,
x

1
1  i 3 , or in polar
2
. One of these is x and one is y. It does not matter which we choose, so
 i
1
 e 3 . As points in the complex plane, we can see that x 2   y and y 2   x,
x
and x3  y 3  1, so x3  x 2  x  y 3  y 2  y  1  1
let x  e 3 and y 
3. Write a cubic equation in the form y = ax3 + bx2 + cx + d. The equation has integral
coefficients and roots of (- 1/3) and (2 + 2i).
Solution:
1
In order to have root  , the polynomial must have a factor of 3x  1. IF a polynomial real
3
coefficients and a complex root, then the complex conjugate of that root must also be a root, so the
other two factor of the polynomial are  x  2  2i  and  x  2  2i  . Therefore, the simplest
expression for the right side is the expansion of  3x  1 x  2  2i  x  2  2i  
2
  2i     3 x  1  x 2  4 x  8   3 x 3  11x 2  20 x  8.

3
2
y  3 x  11x  20 x  8 and multiples of the right side will also work.
 3x  1  x  2 
2
4. In a given arithmetic sequence, the first term is 2, the last term is 29, and the sum of
all the terms is 155. Find the common difference of the sequence.
Solution:
If d is the common difference and k is the number of terms, the ith term is 2   i  1 d . The
sum of the terms is 155   2  0d    2  1d   ...   2   k  1 d   2k   0  1  2  ...  k  1  d .
1
This simplifies to 155  2k  k  k  1 d . since the last term in the sequence is 2   k  1 d  29,
2
1
we know  k  1 d  27, so 155  2k  k  27  , and k  10, so d  3.
2
5. In how many non-congruent triangles of perimeter 16 do all the sides have an
integral length?
Solution:
We seek integers a  b  c with a  b  c  15 and a  b > c. If a  1, there are no solutions. If a  2,
the only solution is  2, 7, 7  . If a  3, the only solution is  3, 6, 7  . If a  4, the solutions are  4, 6, 6 
or  4, 5, 7  . If a  5, the solution is  5, 5, 6  . There are 5 integer solutions.
6. By adding the same constant to each of 20, 50, and 100, a geometric sequence results.
Find the common ratio of that geometric sequence.
Solution:
For the geometric sequence 20  x, 50  x, 100  x, the common ratio is
r
50  x 100  x
2

. cross-multiplying gives  50  x   100  x  20  x  , or
20  x 50  x
2500  100 x  x 2  2000  120 x  x 2 , so 500  20 x. Then x  25 and the ratio =
75 5

45 3
7. If 3x3 – kx2 + 15x – 9 is divisible by x – 3, evaluate 3k2 + k – 12.
Solution:
If the given polynomial is divisible by x  3, then it evaluates to 0 when x  3. Therefore,
3  3  k  3  15  3  9  81  9k  45  9  117  9k  0, so k  13. This means that
3
2
3k 2  k  12  3 13  13  12  508
2
8. A triangle is inscribed in a circle. The vertices of the triangle divide the circle into
three arcs of lengths 3, 4, and 5. What is the area of the triangle?
Solution:
12 6

2 
Let A, B, andC be the vertices of the triangle (with 3, 4, and 5 as the respective lengths of the arcs
Since 3  4  5  12  2 r , the radius of the circle is
AB, BC , and CA ), and let D be the triangle's circumcenter/ Then the area of the triangle is the
sum of the areas of the three isisceles triangles ADB, BDC, and CDA. The legs of these triangles have
length
6

(the radius of the circle), and the vertex engles are mADB 

2
, mBDC 
2
, and
3
5
1
. Using the area formula A= ab sin  , the area of the triangle is therefore
6
2
2
1 6   
2
5  18 
3 1 9
A=    sin  sin
 sin
 
3 3
  2 1 
2  
2
3
6   
2 2   2
mCDA 


9. Solve 2 cos2(θ) = 3 + 3sin(θ), 0° < θ < 360°.
Solution:
Let x  sin   , so that cos 2    1  x 2 , and the given equation is 2 1  x 2   3  3 x,
or 2 x 2  3 x  1  0. This factors into  2 x  1 x  1  0, so sin    
1
or sin    1.
2
In the range 0   < 360 , this means that   210 , 270 , or 330 .
10. Write a quadratic function f(x) = ax2 + bx + c if its graph passes through the points
(1, – .5), (5, 3.5), and (– 3, 11.5).
Solution:
Knowing that f 1  0.5, f  5   3.5, and f  3  11.5 gives us a system of three
equations and three unknowns:
 a  b  c  0.5

25a  5b  c  3.5
 9a  3b  c  11.5

Solving this system gives a  0.5, b  2, c  1, and the function is f  x   0.5 x 2  2 x  1