HMMT 二月 2018 · 团队赛 · 第 4 题

HMMT February 2018 — Team Round — Problem 4

[ 30 ] In acute triangle ABC , let D, E, and F be the feet of the altitudes from A, B, and C respectively, and let L, M, and N be the midpoints of BC, CA, and AB , respectively. Lines DE and N L intersect ZB at X , lines DF and LM intersect at Y , and lines XY and BC intersect at Z . Find in terms of ZC AB , AC , and BC .

解析

[ 30 ] In acute triangle ABC , let D, E, and F be the feet of the altitudes from A, B, and C respectively, and let L, M, and N be the midpoints of BC, CA, and AB , respectively. Lines DE and N L intersect ZB at X , lines DF and LM intersect at Y , and lines XY and BC intersect at Z . Find in terms of ZC AB , AC , and BC . Proposed by: Faraz Masroor Because N L || AC we have triangles DXL and DEC are similar. From angle chasing, we also have ◦ ◦ that triangle DEC is similar to triangle ABC . We have ∠ XN A = 180 − ∠ XN B = 180 − ∠ LN B = N X XD · XE AB XE N M AB ED BC ED AB 180 − CAB = ∠ LM A . In addition, we have = = = = = = N A XL · N A BC LC N A BC DC AB DC AC M L . These two statements mean that triangles AN X and AM L are similar, and ∠ XAB = ∠ XAN = M A ∠ LAM = ∠ LAC . Similarly, ∠ XAY = ∠ LAC , making A, X , and Y collinear, with ∠ Y AB = ∠ XAB = ∠ LAC ; ie. line AXY is a symmedian of triangle ABC . ZB AB sin ∠ ZAB AB sin ∠ LAC LB Then = = , by the ratio lemma. But using the ratio lemma, 1 = = ZC AC sin ∠ ZAC AC sin ∠ LAB LC 2 AB sin ∠ LAB sin ∠ LAC AB ZB AB , so = , so = . 2 AC sin ∠ LAC sin ∠ LAB AC ZC AC