HMMT 二月 2012 · 代数 · 第 9 题
HMMT February 2012 — Algebra — Problem 9
题目详情
- How many real triples ( a, b, c ) are there such that the polynomial p ( x ) = x + ax + bx + ax + c has exactly three distinct roots, which are equal to tan y , tan 2 y , and tan 3 y for some real y ?
解析
- How many real triples ( a, b, c ) are there such that the polynomial p ( x ) = x + ax + bx + ax + c has exactly three distinct roots, which are equal to tan y , tan 2 y , and tan 3 y for some real y ? Answer: 18 Let p have roots r, r, s, t . Using Vieta’s on the coefficient of the cubic and linear terms, 2 2 2 we see that 2 r + s + t = r s + r t + 2 rst . Rearranging gives 2 r (1 − st ) = ( r − 1)( s + t ). 2 If r − 1 = 0, then since r 6 = 0, we require that 1 − st = 0 for the equation to hold. Conversely, if 2 1 − st = 0, then since st = 1 , s + t = 0 cannot hold for real s, t , we require that r − 1 = 0 for the 2 equation to hold. So one valid case is where both these values are zero, so r = st = 1. If r = tan y π 3 π (here we stipulate that 0 ≤ y < π ), then either y = or y = . In either case, the value of tan 2 y 4 4 π 3 π 5 π 7 π is undefined. If r = tan 2 y , then we have the possible values y = , , , . In each of these cases, 8 8 8 8 π we must check if tan y tan 3 y = 1. But this is true if y + 3 y = 4 y is a odd integer multiple of , which 2 is the case for all such values. If r = tan 3 y , then we must have tan y tan 2 y = 1, so that 3 y is an odd π integer multiple of . But then tan 3 y would be undefined, so none of these values can work. 2 2 2 Now, we may assume that r − 1 and 1 − st are both nonzero. Dividing both sides by ( r − 1)(1 − st ) 2 r s + t and rearranging yields 0 = + , the tangent addition formula along with the tangent double 2 1 − r 1 − st angle formula. By setting r to be one of tan y , tan 2 y , or tan 3 y , we have one of the following: (a) 0 = tan 2 y + tan 5 y (b) 0 = tan 4 y + tan 4 y (c) 0 = tan 6 y + tan 3 y . π We will find the number of solutions y in the interval [0 , π ). Case 1 yields six multiples of . Case 2 7 π yields tan 4 y = 0, which we can readily check has no solutions. Case 3 yields eight multiples of . In 9 total, we have 4 + 6 + 8 = 18 possible values of y .