HMMT 二月 2009 · 几何 · 第 7 题

HMMT February 2009 — Geometry — Problem 7

[ 5 ] In triangle ABC , D is the midpoint of BC , E is the foot of the perpendicular from A to BC , and F is the foot of the perpendicular from D to AC . Given that BE = 5, EC = 9, and the area of triangle ABC is 84, compute | EF | .

解析

[ 5 ] In 4 ABC , D is the midpoint of BC , E is the foot of the perpendicular from A to BC , and F is the foot of the perpendicular from D to AC . Given that BE = 5, EC = 9, and the area of triangle ABC is 84, compute | EF | . √ √ 6 37 21 Answer: , 7585 5 205 Solution: There are two possibilities for the triangle ABC based on whether E is between B and C or not. We first consider the former case. We find from the area and the Pythagorean theorem that AE = 12, AB = 13, and AC = 15. We can √ then use Stewart’s theorem to obtain AD = 2 37. 1 Since the area of 4 ADC is half that of ABC , we have AC · DF = 42, so DF = 14 / 5. Also, 2 DC = 14 / 2 = 7 so ED = 9 − 7 = 2. √ Notice that AEDF is a cyclic quadrilateral. By Ptolemy’s theorem, we have EF · 2 37 = (28 / 5) · 12 + √ 6 37 2 · (54 / 5). Thus EF = as desired. 5 The latter case is similar.