HMMT 十一月 2014 · 冲刺赛 · 第 24 题

HMMT November 2014 — Guts Round — Problem 24

[ 12 ] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. We construct isosceles right ◦ triangle ACD with ∠ ADC = 90 , where D, B are on the same side of line AC , and let lines AD and ◦ CB meet at F . Similarly, we construct isosceles right triangle BCE with ∠ BEC = 90 , where E, A are on the same side of line BC , and let lines BE and CA meet at G . Find cos ∠ AGF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2014, 15 NOVEMBER 2014 — GUTS ROUND Organization Team Team ID#

解析

[ 12 ] Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. We construct isosceles right ◦ triangle ACD with ∠ ADC = 90 , where D, B are on the same side of line AC , and let lines AD and ◦ CB meet at F . Similarly, we construct isosceles right triangle BCE with ∠ BEC = 90 , where E, A are on the same side of line BC , and let lines BE and CA meet at G . Find cos ∠ AGF . 5 ◦ Answer: − We see that ∠ GAF = ∠ GBF = 45 , hence quadrilateral GF BA is cyclic. 13 ◦ Consequently ∠ AGF + ∠ F BA = 180 . So cos ∠ AGF = − cos ∠ F BA . One can check directly that 5 cos ∠ CBA = (say, by the Law of Cosines). 13