HMMT 十一月 2016 · 团队赛 · 第 6 题

HMMT November 2016 — Team Round — Problem 6

[ 5 ] Let ABC be a triangle with AB = 5, BC = 6, and AC = 7. Let its orthocenter be H and the feet of the altitudes from A, B, C to the opposite sides be D, E, F respectively. Let the line DF intersect the circumcircle of AHF again at X . Find the length of EX .

解析

[ 5 ] Let ABC be a triangle with AB = 5, BC = 6, and AC = 7. Let its orthocenter be H and the feet of the altitudes from A, B, C to the opposite sides be D, E, F respectively. Let the line DF intersect the circumcircle of AHF again at X . Find the length of EX . Proposed by: Allen Liu 190 Answer: 49 ◦ Since ∠ AF H = ∠ AEH = 90 , E is on the circumcircle of AHF . So ∠ XEH = ∠ HF D = ∠ HBD , EX EY which implies that XE ‖ BD . Hence = . Let DF and BE intersect at Y . Note that BD Y B ◦ ◦ ∠ EDY = 180 − ∠ BDF − ∠ CDE = 180 − 2 ∠ A , and ∠ BDY = ∠ A . Applying the sine rule to EY D and BY D , we get EY ED sin ∠ EDY ED sin 2 ∠ A ED = · = · = · 2 cos ∠ A Y B BD sin ∠ BDY BD sin ∠ A BD Next, letting x = CD and y = AE , by Pythagoras we have 2 2 2 2 2 AB − (6 − x ) = AD = AC − x 2 2 2 2 2 BC − (7 − y ) = BE = BA − y 19 Solving, we get x = 5 , y = . Drop the perpendicular from E to DC at Z . Then ED cos ∠ A = 7 AE 95 ED cos ∠ EDZ = DZ . But AD ‖ EZ , so DZ = · DC = . Therefore AC 49 EY 190 EX = · BD = 2 ED cos ∠ A = 2 DZ = Y B 49