HMMT 二月 2005 · 冲刺赛 · 第 33 题

HMMT February 2005 — Guts Round — Problem 33

[10] Triangle ABC has incircle ω which touches AB at C , BC at A , and CA at B . 1 1 1 Let A be the reflection of A over the midpoint of BC , and define B and C similarly. 2 1 2 2 Let A be the intersection of AA with ω that is closer to A , and define B and C 3 2 3 3 similarly. If AB = 9, BC = 10, and CA = 13, find [ A B C ] / [ ABC ]. (Here [ XY Z ] 3 3 3 denotes the area of triangle XY Z .) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

解析

Triangle ABC has incircle ω which touches AB at C , BC at A , and CA at B . Let 1 1 1 A be the reflection of A over the midpoint of BC , and define B and C similarly. 2 1 2 2 Let A be the intersection of AA with ω that is closer to A , and define B and C 3 2 3 3 similarly. If AB = 9, BC = 10, and CA = 13, find [ A B C ] / [ ABC ]. (Here [ XY Z ] 3 3 3 denotes the area of triangle XY Z .) Solution: 14/65 Notice that A is the point of tangency of the excircle opposite A to BC . Therefore, by 2 considering the homothety centered at A taking the excircle to the incircle, we notice that A is the intersection of ω and the tangent line parallel to BC . It follows that 3 12 A B C is congruent to A B C by reflecting through the center of ω . We therefore 1 1 1 3 3 3 need only find [ A B C ] / [ ABC ]. Since 1 1 1 2 [ A BC ] A B · BC ((9 + 10 − 13) / 2) 1 1 1 1 1 = = = , [ ABC ] AB · BC 9 · 10 10 and likewise [ A B C ] / [ ABC ] = 49 / 130 and [ AB C ] / [ ABC ] = 4 / 13, we get that 1 1 1 1 [ A B C ] 1 49 4 14 3 3 3 = 1 − − − = . [ ABC ] 10 130 13 65