HMMT 二月 2026 · 几何 · 第 9 题
HMMT February 2026 — Geometry — Problem 9
题目详情
- Let ABC be triangle with incenter I and incircle ω . The circumcircle of triangle BIC intersects ω at points E and F . Suppose that Γ ̸ = ω is a circle passing through E and F and tangent to lines AB and AC . If AB = 5 , AC = 7 , and Γ has twice the radius of ω , compute BC .
解析
- Let ABC be triangle with incenter I and incircle ω . The circumcircle of triangle BIC intersects ω at points E and F . Suppose that Γ ̸ = ω is a circle passing through E and F and tangent to lines AB and AC . If AB = 5 , AC = 7 , and Γ has twice the radius of ω , compute BC . Proposed by: Aprameya Tripathy √ Answer: 3 + 11 Solution: ©2026 HMMT A Q R E I ′ Q F C B P ′ R Let P , Q , and R be the points the incircle touches BC , CA , and AB respectively. Note that by the radical axis theorem on ( CP IQ ) , ( BIC ) , and ω , we have that CI , EF and P Q are concurrent. Hence, the midpoint of P Q lies on the EF . Similarly, by the radical axis theorem on ( BP IR ) , ( BIC ) , and ω , the midpoint of P R lies on EF ′ ′ Let Q and R be the points where Γ touches AB and AC respectively. Since Γ has twice the radius ′ ′ ′ of ω , we have Q and R are midpoints of AQ and AR respectively. However, since midpoints of QQ ′ and RR lie on the radical axes on ω and Γ , which is EF , we get that EF is equidistant to lines QR ′ ′ ′ ′ and Q R . Note that EF is also the P -midline of △ P QR , so P lies on Q R . x x We finish with simple length chasing. Let BC = x . Then BP = − 1 , CP = + 1 , and 2 2 ( ) x ′ ′ AQ = AR = 2 · AQ = 2 6 − = 12 − x. 2 Thus, by Menelaus’s Theorem, ′ ′ AR BC CQ 12 − x x − 2 x − 5 1 = · · = · · . ′ ′ R B P C Q A 7 − x x + 2 12 − x √ 2 Simplifying, we x − 6 x − 2 = 0 , which gives us BC = 3 + 11