HMMT 二月 2025 · 团队赛 · 第 5 题
HMMT February 2025 — Team Round — Problem 5
题目详情
- [35] Let △ ABC be an acute triangle with orthocenter H . Points E and F are on segments AC and ◦ AB , respectively, such that ∠ EHF = 90 . Let X be the foot of the altitude from H to EF . Prove ◦ that ∠ BXC = 90 .
解析
- [35] Let △ ABC be an acute triangle with orthocenter H . Points E and F are on segments AC and ◦ AB , respectively, such that ∠ EHF = 90 . Let X be the foot of the altitude from H to EF . Prove ◦ that ∠ BXC = 90 . Proposed by: Pitchayut Saengrungkongka Solution 1: A E Y X F Z H B C We use ∡ to denote directed angles. Let Y and Z be the feet of the altitudes from B and C to AC ◦ and AB , respectively. Then ∡ HZF = ∡ HXF = 90 , so HZF X is cyclic. Similarly, HY EX is cyclic. Therefore, ∡ BY X = ∡ HY X = ∡ HEX = ∡ F HX = ∡ F ZX = ∡ BZX. Hence, BZXY is cyclic. A symmetric argument shows C lies on this circle as well. It follows that ◦ ∠ BXC = ∠ BY C = 90 , as desired. ′ Solution 2: Let T be the foot of altitude from A to BC . For any point X , let X denote the image √ ′ ′ of X under the negative inversion at H with radius HA · HT . Then B and C are the feet of the altitudes from B and C to sides AC and AB , respectively. ′ ◦ Claim 1. ∠ BX C = 90 . ′ ′ ′ Proof. Because HX ⊥ EF and HE ⊥ HF , the quadrilateral HE X F is a rectangle. Note that ′ ′ ◦ ′ ′ ◦ ′ ′ ∠ BE H = ∠ EB H = 90 and ∠ X E H = 90 . Consequently, X , B , and E are collinear. Similarly, ′ ′ ′ ′ ′ ′ ◦ X , C , and F are collinear. Then, ∠ BX C = ∠ E X F = 90 , as desired. ′ ′ ′ From the claim, X lies on the circle with diameter BC (which B and C also lie on). Since this circle ◦ is invariant under the inversion, X lies on the circle with diameter BC as well, and ∠ BXC = 90 . A E X F H B C ′ E ′ F ′ X Solution 3: We begin by proving the following lemma. ◦ Lemma 2. Let ABCD be a quadrilateral and P be a point such that ∡ AP B + ∡ CP D = 180 . Then, the feet of the altitudes from P to each side of ABCD are concyclic. Proof. Let P , P , P , P the feet of the altitudes from P to AB , BC , CD , and DA respectively. A B C D Note that quadrilateral P P P B is cyclic. By angle chasing, A B ∡ P P P + ∡ P P P = ∡ P P P + ∡ P P P + ∡ P P P + ∡ P P P D A B B C D D A A B B C C D = ∡ P AP + ∡ P BP + ∡ P CP + ∡ P DP D B B D ◦ ◦ = (180 − ∡ AP D ) + (180 − ∡ BP C ) = ∡ BP A + ∡ DP C ◦ = 180 . Therefore, P P P P is cyclic as desired. A B C D A E X F H C B Let P be the point at infinity on line AC . Let Y and Z be the feet of the altitudes from H to AB ∞ ◦ ◦ ◦ and AC , respectively. Note that ∠ EHF + ∠ BHP = 90 + 90 = 180 . Thus, the feet of the altitudes ∞ from H to EF , EB , BP , and CP are concyclic. In other words, XY ZB is cyclic. Since BCY Z is ∞ ∞ ◦ a cyclic quadrilateral, we conclude X lies on this circle, giving us that ∠ BXC = 90 as desired. Remark. Here’s another way to prove the lemma. ′ It is well known that, with the provided condition, there is a point P that is the isogonal conjugate of P with respect to quadrilateral ABCD . Let P , P , P , and P be the feet of the altitudes from P A B C D ′ to AB , BC , CD , and DA , respectively, and let Q be the foot of the altitude from P to AB . Because ′ P and P are isogonal conjugates with respect to the triangle formed by lines AB , BC , and CD , we ′ have P P P Q is cyclic. Similarly, because P and P are also isogonal conjugate with respect to the A B C triangle formed by lines DA , AB , and BC , we have P P P Q is cyclic. Consequently, P P P P D A B A B C C is cyclic as desired.