HMMT 二月 2014 · 冲刺赛 · 第 19 题

HMMT February 2014 — Guts Round — Problem 19

[ 11 ] Let ABCD be a trapezoid with AB k CD . The bisectors of \ CDA and \ DAB meet at E , the bisectors of \ ABC and \ BCD meet at F , the bisectors of \ BCD and \ CDA meet at G , and the bisectors of \ DAB and \ ABC meet at H . Quadrilaterals EABF and EDCF have areas 24 and 36, respectively, and triangle ABH has area 25. Find the area of triangle CDG .

解析

[ 11 ] Let ABCD be a trapezoid with AB ‖ CD . The bisectors of ∠ CDA and ∠ DAB meet at E , the bisectors of ∠ ABC and ∠ BCD meet at F , the bisectors of ∠ BCD and ∠ CDA meet at G , and the bisectors of ∠ DAB and ∠ ABC meet at H . Quadrilaterals EABF and EDCF have areas 24 and 36, respectively, and triangle ABH has area 25. Find the area of triangle CDG . 256 Answer: Let M, N be the midpoints of AD, BC respectively. Since AE and DE are bisectors 7 of supplementary angles, the triangle AED is right with right angle E . Then EM is the median of a right triangle from the right angle, so triangles EM A and EM D are isosceles with vertex M . But then ∠ M EA = ∠ EAM = ∠ EAB , so EM ‖ AB . Similarly, F N ‖ BA . Thus, both E and F are on the midline of this trapezoid. Let the length of EF be x . Triangle EF H has area 1 and is similar to triangle ABH , which has area 25, so AB = 5 x . Then, letting the heights of trapezoids EABF and 6 xh EDCF be h (they are equal since EF is on the midline), the area of trapezoid EABF is = 24. So 2 9 xh the area of trapezoid EDCF is 36 = . Thus DC = 8 x . Then, triangle GEF is similar to and has 2 1 64 times the area of triangle CDG . So the area of triangle CDG is times the area of quadrilateral 64 63 256 EDCF , or . 7