HMMT 二月 2016 · 冲刺赛 · 第 20 题

HMMT February 2016 — Guts Round — Problem 20

[ 11 ] Let ABC be a triangle with AB = 13, AC = 14, and BC = 15. Let G be the point on AC such that the reflection of BG over the angle bisector of ∠ B passes through the midpoint of AC . Let Y be AX the midpoint of GC and X be a point on segment AG such that = 3. Construct F and H on AB XG and BC , respectively, such that F X ‖ BG ‖ HY . If AH and CF concur at Z and W is on AC such that W Z ‖ BG , find W Z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2016, February 20, 2016 — GUTS ROUND Organization Team Team ID#

解析

[ 11 ] Let ABC be a triangle with AB = 13, AC = 14, and BC = 15. Let G be the point on AC such that the reflection of BG over the angle bisector of ∠ B passes through the midpoint of AC . Let Y be AX the midpoint of GC and X be a point on segment AG such that = 3. Construct F and H on AB XG and BC , respectively, such that F X ‖ BG ‖ HY . If AH and CF concur at Z and W is on AC such that W Z ‖ BG , find W Z . Proposed by: Ritesh Ragavender √ 1170 37 Answer: 1379 2 AG c Observe that BG is the B -symmedian, and thus = . Stewart’s theorem gives us 2 GC a √ √ 2 2 2 2 2 √ 2 a c b a b c ac 390 37 2 2 2 BG = − = 2( a + c ) − b = . 2 2 2 2 2 2 b ( a + c ) a + c a + c 197 Then by similar triangles, √ ZA Y C ZA 1 6 1170 37 ZW = HY = BG = BG = HA GC HA 2 7 1379 ZA where is found with mass points or Ceva. HA