HMMT 十一月 2016 · 团队赛 · 第 6 题

英文解析

6. [ 5 ] Let ABC be a triangle with AB = 5, BC = 6, and AC = 7. Let its orthocenter be H and the feetof the altitudes from A, B, C to the opposite sides be D, E, F respectively. Let the line DF intersectthe circumcircle of AHF again at X . Find the length of EX .
   Proposed by: Allen Liu
   Answer:190
   °49
   Since ∠ AF H = ∠ AEH = 90 , E is on the circumcircle of AHF . So ∠ XEH = ∠ HF D = ∠ HBD ,
   EX EYwhich 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 Dand 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
   Solving, we get x = 5 , y = . Drop the perpendicular from E to DC at Z . Then ED cos ∠ A =19
   AE 957
   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