2016 OHMIO Individual Competition
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1 06 OHMIO Individual Competition. Taylor thought of three positive integers a, b, c, all between and 0 inclusive. The three integers form a geometric sequence. Taylor then found the number of positive integer divisors of each number: d(a), d(b), and d(c). She was astonished, not only because these values of d were all odd, but they formed an increasing arithmetic sequence! What is the sum a + b+ c?. All the positive integers with an initial digit of 6 are written down in succession: 6, 60, 6, 6, 6, and so on. In this process, what is the 06 th digit that will be written?. In parallelogram OCTM, the bisector of ÐOCT intersects OM at X. If XM = 5, CX = 6, and TX = 6, what is the value of OC? 4. When connected in the complex plane, the solutions to (z +) 6 = 64 form a regular polygon. Find the exact area of this polygon in simplest radical form. 5. Find the value of ê 06! + 0! ú ê ë05! + 04! ú. û 6. For the function f (x) = x 4 - x - x + ax +, determine the real number a where f (a) = a is the relative minimum. 7. In DABC, AB =, BC = 6, and mða = 60. What is the exact area of the triangle s circumcircle? 8. Suppose log + log 4 5 = log b n where b and n are positive integers and b is as small as possible. Find the value of b + n. 9. A committee of 5 to grade the OHMIO is to be chosen from a group of 9 people. How many ways can this be done if Jon and Duane must serve together or not at all, and Chris and Brian refuse to serve with each other? 0. For a monic fourth-degree polynomial function p(x) with integer coefficients, the sum of two of the roots is and the product of the other two roots is 6-5. If two of the four roots are integers, find p(0). [Note: monic means the leading coefficient is ]. Find the product of the real solutions of x log x = 8x.. The product can be written as a b Find the value of a + b. in simplest terms.
2 . A square is cut by two parallel segments a perpendicular distance of units apart. If these lines partition the square into three regions of equal area, what is the area of the square? 4. Two distinct integers are selected at random from {,,,, 00} and are multiplied together. The probability that the product of these selected integers is divisible by can be written in simplest form as a. Find a + b. b 5. The three angles of a triangle form an arithmetic sequence. The shortest side of the triangle has length and the area of the triangle is. The length of the largest side can be written in simplified form as a b. Find a + b. 6. In a cube, a regular octahedron is inscribed by connecting the midpoints of adjacent faces of the cube. What is the ratio of the volume of the cube to the volume of the octahedron? 7. Neil the astronaut discovered the equation x - 5x + 66 = 0 carved into a moon rock on the planet Kardashyan. Under the equation was written x = 4 or x = 9. What base number system is used by aliens on planet Kardashyan? 8. Suppose x and y are defined parametrically as x = tant, y = sin t. What is lim x y? 9. Consider the sequence of numbers: a =,a =,a = 5,a 4 = 57,a 5 = 579,a 6 = 579, etc. Here a n is formed by writing the next odd number at the end of a n-. Now make a new sequence b n of those terms in a n divisible by, taken in order. What is the units digit of b 00? 0. Find the area of the region enclosed by êë x ú û + ê ë y ú û 6 TB. Find the exact value of cot 75 + tan 75 cot75 + tan75 TB. Determine the value of 6 ò 0 ê éê xù ú - x ú ê ë údx û
3 06 OHMIO Individual Competition ANSWERS or 6: 7. π TB. TB. 08
4 06 OHMIO Individual Competition SOLUTIONS. It is easy to show that the perfect squares are exactly those integers with an odd number of divisors. Thus, a, b, c must come from the set {, 4, 9, 6}. We see that {, 4, 6} is the only such geometric sequence and that the number of divisors forms the sequence {,, 5} which is arithmetic. Thus, the sum a + b + c = =.. There is one-digit number, 0 two-digit numbers, and 00 three-digit numbers for a total of digits accounted for leaving 06 = 695 remaining digits. The numbers 6000, 600, 600, etc. each use 4 digits and = 4. So 64 will be the last full four-digit 4 number written, resulting in 64. The 06 th digit is.. Consider the diagram shown. We see that DXOC ~ DCXT since base angles are congruent. Thus a 6 = 6 b so b = 6 a. Thus, a + 5 = 6 a or (a + 9)(a - 4) = 0 so a = Let u = z +. The solutions to u 6 = 64 form a regular hexagon whose distance from the center to a vertex of the hexagon is. The total area is 6 ( sin(60 )) = = 6. Note the shift of z + only translates this hexagon to the left not affecting its area. 06! + 0! 04!( Note 05! + 04! = 04 ) = 05+. From this, we see that 04!(05 +) the floor of this value is f '(x) = 4x - x - x + a and f '(a) = 4a - a - a = a(4a +)(a -) = 0 when a = 0,- 4,. Checking, we see a = is the only value satisfying f(a) = a. 7. By the extended Law of Sines, circumcircle s area is p sin(60 ) 6 = Rwhere R is the circumradius. Thus, the sin(60 ) = 9p 4 = p. 8. log + log 4 5 = log log 4 5 = log So b = 4, n = 45 and b + n = 49.
5 9. If Jon and Duane serve together, there are must subtract 5 not serve, there are 5 7 ways to select the others. However, we ways where Chris and Brian both serve together. If Jon and Duane do 7 5 ways to choose the remaining members. Yet, we must subtract ways that Chris and Brian can both serve together. This results in = = 4ways. 0. Let a ± b,c,d denote the four roots. Then (a + b) + c = for which a + c = 6 and b = 5. Also (a - b)d = 6-5for which ad = 6, d =, b = 5. We can then deduce a = and c = 4. This yields the polynomial p(x) = (x - ( + 5))(x - ( - 5))(x - 4)(x - ) and p(0) = (-( + 5))(-( - 5))(-4)(-) = -. x. Take a log of both sides, log log x = log (8x). x By a power property, log x log = log (8x). Using expanding properties, log x (log x -) = + log x which simplifies to (log x) - log x - = 0. This factors as (log x - )(log x +) = 0 so x = 8, whose product is = Looking at the odd-positioned terms, = 6. Looking at the other terms in the product, =. Thus, the 7 product is and a + b = Since one-third the area of the square equals the area of the triangle, s = s(s - x). Solving for x yields x = s. The hypotenuse of the right triangle (and the base of the parallelogram) has length ( s) + s = s. Calculating the area of the parallelogram two ways, s s = s so we find s = and the area of the square is s = 5
6 4. If divides ab then divides a or divides b. There are 00 ways to select two distinct 67 integers from the set. Note that of the numbers are divisible by so there are ways to select a pair of integers neither of which is divisible by. The desired probability is then - = - = = 8 50 and a + b =. 5. Let the three angles be y d, y, and y + d. Then the sum y = 80 and y = 60. This medium-sized angle is included between the smallest and largest sides. Letting the largest side length be x, ()(x)sin60 = and thus x = = 4. So a + b = 4 + = Consider a cube of side length whose volume is 8. Connecting two midpoints of adjacent faces, we find that the edge of the octahedron is. An octahedron s volume is the sum of two volumes of congruent square pyramids. The volume of the octahedron is then ( ) = 4. The volume ratio is 8 = Let the base of the planet be K. Note that the base is at least 0 since 4 and 9 are digits. Since (x - 4)(x - 9) = x - 9x +08in our base 0 system, we know that 9 = a +5for which a = 7. Checking, 08 = 6(7) + 6 as well. 8. Since y = sin t = (- cos t)then cos t = - y and sec t = y -. Since sec t -= tan tthen y - -= x and thus + = y. We see lim y=. x + x 9. Looking at the remainders of a n when we divide by, we get the sequence,, 0,,, 0,,, 0, etc. Thus, b n = a n. Here are the first few: b = 5 b = 579 b = b 4 = b 5 = b 6 = From this we see that the units digit cycles every 5 terms. The units digit of the 00 th term will be the same as the units digit of the 5 th term, which is 9.
7 0. Since either square is non-zero, values on -4 x 4,-4 y 4. When êë y ú û = 0, we see that ê ëx ú û = -4,-,-,-,0,,,,4. When êë y ú û =, we see that ê ëx ú û = -,-,-,0,,,. When êë y ú û =, we see that ê ëx ú û = -,-,-,0,,,. When êë y ú û =, we see that ê ëx ú û = -,-,0,,. When êë y ú û = 4, we see that ê ëx ú û = 0. By symmetry we will have similar behavior for when êë y ú û is negative. The total area of unit squares is then ( ) + 9 = 49. A graph of this region is below. This could be said to be a Taxi Cab circle of radius 4. TB. Let u = cot75,v = tan75. Then u + v u + v = (u + v)(u - uv + v ) = u - uv + v. (u + v) Using Pythagorean identities, this equals ( csc 75 -) -+ ( sec 75 -). Convert to sines and cosines to find sin 75 + cos 75 - = cos 75 + sin 75 sin 75 cos This equals sin 75 cos 75-4 or (sin75 cos75 ) - = 4 (sin50 ) - = 4 (sin75 cos75 ) - = 4 - = 6 - =. (sin50 ) TB. 6 ò 0 éê xù ú dx represents the sum of areas of rectangles of base and heights of,,,,. The total area of this is ( ) = = 8. 6 ê x ú ò ê ë údx represents the sum of areas of rectangles of base and heights of,,, 7. The 0 û total area of this is ( ) = 7 8 = 8 7. The difference is the desired integral which is 8(- 7) = 8(6) = 08.
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