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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 23 6 5 35 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 mADB 2 , mBDC 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 mCDA 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