HMMT 二月 2022 · 冲刺赛 · 第 30 题

HMMT February 2022 — Guts Round — Problem 30

[16] Let ( x , y ) , . . . , ( x , y ) be the distinct real solutions to the equation 1 1 k k 2 2 6 2 2 4 3 2 3 ( x + y ) = ( x − y ) = (2 x − 6 xy ) . P k a Then ( x + y ) can be expressed as , where a and b are relatively prime positive integers. Compute i i i =1 b 100 a + b . 2 2 2

解析

[16] Let ( x , y ) , . . . , ( x , y ) be the distinct real solutions to the equation 1 1 k k 2 2 6 2 2 4 3 2 3 ( x + y ) = ( x − y ) = (2 x − 6 xy ) . P k a Then ( x + y ) can be expressed as , where a and b are relatively prime positive integers. Compute i i i =1 b 100 a + b . Proposed by: Akash Das Answer: 516 Solution: Using polar coordinates, we can transform the problem to finding the intersections between r = cos 2 θ and r = 2 cos 3 θ . Drawing this out gives us a four-leaf clover and a large 3-leaf clover, which intersect at 7 points (one point being the origin). Note that since this graph is symmetric about the x - 3 axis, we are only interested in finding the x -coordinates, which is r cos θ = cos 2 θ cos θ = 2 cos θ − cos θ . Now note that all points of intersection satisfy 3 2 cos 2 θ = 2 cos 3 θ ⇐⇒ 8 cos θ − 2 cos θ − 6 cos θ + 1 = 0 . 3 Now, we want to compute the sum of 2 cos θ − cos θ over all values of cos θ that satisfy the above cubic. P 3 In other words, if the solutions for cos θ to the above cubic are a, b, and c , we want 2 2 a − a , cyc since each value for cos θ generates two solutions (symmetric about the x -axis). This is X X 1 3 2 4 a − 2 a = a + a − , 2 cyc cyc 1 1 3 2 where we have used the fact that a = a + 3 a − . By Vieta’s formulas, a + b + c = , while 2 4 2 1 3 25 2 2 2 a + b + c = + 2 · = . 4 4 16 5 Thus the final answer is . 16 2 2 2