HMMT 十一月 2025 · THM 赛 · 第 10 题
HMMT November 2025 — THM Round — Problem 10
题目详情
- The orbits of Pluto and Charon are given by the ellipses 2 2 2 2 x + xy + y = 20 and 2 x − xy + y = 25 , respectively. These orbits intersect at four points that form a parallelogram. Compute the largest of the slopes of the four sides of this parallelogram. © 2025 HMMT
解析
- The orbits of Pluto and Charon are given by the ellipses 2 2 2 2 x + xy + y = 20 and 2 x − xy + y = 25 , respectively. These orbits intersect at four points that form a parallelogram. Compute the largest of the slopes of the four sides of this parallelogram. Proposed by: Pitchayut Saengrungkongka √ 7+1 Answer: 2 Solution 1: For any real number t , adding t times the first orbit equation and 1 − t times the second orbit equation shows that all four intersection points must satisfy the equation 2 2 t + 2(1 − t ) x + t − (1 − t ) xy + t + (1 − t ) y = 20 t + 25(1 − t ) which simplifies to 2 2 (2 − t ) x + (2 t − 1) xy + y = 25 − 5 t. We want to choose t so that the left hand side is a perfect square. This choice can be found by equating its discriminant to 0: 2 (2 t − 1) − 4(2 − t ) = 0 . √ 2 This simplifies to 4 t − 7 = 0, so t = ± 7 / 2. Fixing a choice of sign for t , we see all four intersection points must satisfy the equation ! √ √ 7 2 2 2 ∓ x + ( ± 7 − 1) xy + y = (some constant) , 2 which simplifies to ! √ 2 − 1 ± 7 y + x = (some constant) . 2 √ 1 ± 7 The graph of this equation consists of two parallel lines of slope , so these lines must be opposite 2 √ 1+ 7 sides of the parallelogram. Hence, the maximum slope is . 2 © 2025 HMMT Solution 2: Let ( x , y ) and ( x , y ) be two distinct intersection points which are not reflections across 1 1 2 2 y − y 1 2 the origin; the other two are ( − x , − y ) and ( − x , − y ). Thus, the two slopes are m = and 1 1 2 2 1 x − x 1 2 y + y 1 2 2 m = . Since subtracting the two orbit equations yields x − 2 xy = 5, 2 x + x 1 2 2 2 2 x y − 2 x y ( x − 5) − ( x − 5) 1 1 2 2 1 2 m + m = = = 1 . 1 2 2 2 2 2 x − x x − x 1 2 1 2 2 2 Likewise, adding the two equations yields 3 x + 2 y = 45, so 2 2 2 2 y − y 1 (45 − 3 x ) − (45 − 3 x ) 3 1 2 1 2 m m = = · = − . 1 2 2 2 2 2 x − x 2 x − x 2 1 2 1 2 √ 2 1+ 7 Thus, the two slopes are roots of m − m − 3 / 2 = 0, and the larger root of this polynomial is . 2 © 2025 HMMT