HMMT 二月 2002 · 冲刺赛 · 第 53 题

HMMT February 2002 — Guts Round — Problem 53

[10] ABC is a triangle with points E, F on sides AC, AB , respectively. Suppose that 2 BE, CF intersect at X . It is given that AF/F B = ( AE/EC ) and that X is the midpoint of BE . Find the ratio CX/XF .

解析

ABC is a triangle with points E, F on sides AC, AB , respectively. Suppose that 2 BE, CF intersect at X . It is given that AF/F B = ( AE/EC ) and that X is the midpoint of BE . Find the ratio CX/XF . Solution: Let x = AE/EC . By Menelaus’s theorem applied to triangle ABE and line CXF , 2 AF BX EC x 1 = · · = . F B XE CA x + 1 √ 2 Thus, x = x + 1, and x must be positive, so x = (1 + 5) / 2. Now apply Menelaus to triangle ACF and line BXE , obtaining AE CX F B CX x 1 = · · = · , 2 EC XF BA XF x + 1 √ 2 2 so CX/XF = ( x + 1) /x = (2 x − x ) /x = 2 x − 1 = 5 .