HMMT 二月 2017 · 团队赛 · 第 10 题
HMMT February 2017 — Team Round — Problem 10
题目详情
- [ 65 ] Let LBC be a fixed triangle with LB = LC , and let A be a variable point on arc LB of its circumcircle. Let I be the incenter of 4 ABC and AK the altitude from A . The circumcircle of 4 IKL intersects lines KA and BC again at U 6 = K and V 6 = K . Finally, let T be the projection of I onto line U V . Prove that the line through T and the midpoint of IK passes through a fixed point as A varies.
解析
- [ 65 ] Let LBC be a fixed triangle with LB = LC , and let A be a variable point on arc LB of its circumcircle. Let I be the incenter of 4 ABC and AK the altitude from A . The circumcircle of 4 IKL intersects lines KA and BC again at U 6 = K and V 6 = K . Finally, let T be the projection of I onto line U V . Prove that the line through T and the midpoint of IK passes through a fixed point as A varies. Proposed by: Sam Korsky Answer: Let M be the midpoint of arc BC not containing L and let D be the point where the incircle of triangle ABC touches BC . Also let N be the projection from I to AK . We claim that M is the desired fixed point. By Simson’s Theorem on triangle KU V and point I we have that points T, D, N are collinear and since quadrilateral N KDI is a rectangle we have that line DN passes through the midpoint of IK . Thus it suffices to show that M lies on line DN . Now, let I , I , I be the A, B, C -excenters of triangle ABC respectively. Then I is the orthocenter of a b c triangle I I I and ABC is the Cevian triangle of I with respect to triangle I I I . It’s also well-known a b c a b c that M is the midpoint of II . a ′ ′ Let D be the reflection of I over BC and let N be the reflection of I over AK . Clearly K is the ′ ′ ′ ′ midpoint of D N . If we could prove that I , D , K, N were collinear then by taking a homothety a 1 centered at I with ratio we would have that points M, D, N were collinear as desired. Thus it 2 ′ suffices to show that points I , D , K are collinear. a Let lines BC and I I intersect at R and let lines AI and BC intersect at S . Then it’s well-known that b c ( I , I ; A, R ) is harmonic and projecting from C we have that ( I , I ; S, A ) is harmonic. But KS ⊥ KA b c a ′ which means that KS bisects angle ∠ IKI . But it’s clear by the definition of D that KS bisects a ′ ′ angle ∠ IKD which implies that points I , K, D are collinear as desired. This completes the proof. a