HMMT 二月 2013 · 冲刺赛 · 第 23 题

HMMT February 2013 — Guts Round — Problem 23

[ 14 ] Let ABCD be a parallelogram with AB = 8, AD = 11, and ∠ BAD = 60 . Let X be on segment CD with CX/XD = 1 / 3 and Y be on segment AD with AY /Y D = 1 / 2. Let Z be on segment AB such that AX , BY , and DZ are concurrent. Determine the area of triangle XY Z .

解析

[ 14 ] Let ABCD be a parallelogram with AB = 8, AD = 11, and ∠ BAD = 60 . Let X be on segment CD with CX/XD = 1 / 3 and Y be on segment AD with AY /Y D = 1 / 2. Let Z be on segment AB such that AX , BY , and DZ are concurrent. Determine the area of triangle XY Z . √ 19 3 Answer: Let AX and BD meet at P . We have DP/P B = DX/AB = 3 / 4. Now, applying 2 Ceva’s Theorem in triangle ABD , we see that AZ DP AY 3 1 3 = · = · = . ZB P B Y D 4 2 8 Now, [ AY Z ] [ AY Z ] 1 1 3 1 = = · · = , [ ABCD ] 2[ ABD ] 2 3 11 22 and similarly [ DY X ] 1 2 3 1 = · · = . [ ABCD ] 2 3 4 4 Also, ( ) [ XCBZ ] 1 1 8 43 = + = . [ ABCD ] 2 4 11 88 The area of XY Z is the rest of the fraction of the area of ABCD not covered by the three above polygons, which by a straightforward calculation 19 / 88 the area of ABCD , so our answer is √ 19 19 3 ◦ 8 · 11 · sin 60 · = . 88 2