HMMT 二月 2011 · ALGGEO 赛 · 第 10 题

HMMT February 2011 — ALGGEO Round — Problem 10

Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD , respectively, so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X and BX = 6, determine DF . 2

解析

Let ABCD be a square of side length 13. Let E and F be points on rays AB and AD , respectively, so that the area of square ABCD equals the area of triangle AEF . If EF intersects BC at X and BX = 6, determine DF . √ Answer: 13 F x Y C D y 7 X 6 A B E Algebra & Geometry Individual Test First Solution Let Y be the point of intersection of lines EF and CD . Note that [ ABCD ] = [ AEF ] implies that [ BEX ] + [ DY F ] = [ CY X ]. Since 4 BEX ∼ 4 CY X ∼ 4 DY F , there exists some constant r such 2 2 2 2 2 2 that [ BEX ] = r · BX , [ Y DF ] = r · CX , and [ CY X ] = r · DF . Hence BX + DF = CX , so √ √ √ 2 2 DF = CX − BX = 49 − 36 = 13. Second Solution Let x = DF and y = Y D . Since 4 BXE ∼ 4 CXY ∼ 4 DF Y , we have BE CY DY y = = = . BX CX DF x 6 y 13 − y y Using BX = 6, XC = 7 and CY = 13 − y we get BE = and = . Solving this last equation x 7 x 13 x for y gives y = . Now [ ABCD ] = [ AEF ] gives x +7 ( ) 1 1 6 y 169 = AE · AF = 13 + (13 + x ) . 2 2 x 78 y 169 = 6 y + 13 x + x 6 x 78 13 = + x + x + 7 x + 7 2 0 = x − 13 . √ Thus x = 13. 2