HMMT 二月 2025 · 冲刺赛 · 第 31 题

HMMT February 2025 — Guts Round — Problem 31

[16] There exists a unique circle that is both tangent to the parabola y = x at two points and tangent q 3 y to the curve x = . Compute the radius of this circle. 1 − y

解析

[16] There exists a unique circle that is both tangent to the parabola y = x at two points and tangent q 3 y to the curve x = . Compute the radius of this circle. 1 − y Proposed by: Karthik Venkata Vedula √ 5 Answer: 2 Solution: y x 3 y 2 We can square both sides of the second curve to get x = , which further rearranges to 1 − y 2 x y = . 2 2 2 2 2 ( x + y ) x + y 3 y 2 2 This relation implies that curves y = x and x = map to each other under inversion about the 1 − y 2 2 unit circle x + y = 1. Therefore, the unique circle we seek must be invariant under inversion about 2 2 x + y = 1. 2 Since the circle is tangent to y = x , we know that the circle is of the form 2 2 2 x + ( y − y ) = r . 0 We know that the length of the tangent from (0 , 0) to this circle is 1. Since the distance from (0 , 0) to 2 2 the center of the circle is y , using the Pythagorean Theorem gives r + 1 = y . Because the parabola 0 0 2 2 2 2 2 2 y = x is tangent to this circle at two distinct points, the equation x + ( x − y ) = r = y − 1 must 0 0 have two double roots. Therefore, 2 2 2 2 4 2 x + ( x − y ) − ( y − 1) = x − (2 y − 1) x + 1 0 0 0 3 5 2 2 must be a perfect square, so 2 y − 1 = 2 = ⇒ y = . This means r = y − 1 = , so the radius of 0 0 0 2 4 √ 5 the circle is . 2