HMMT 二月 2005 · 冲刺赛 · 第 33 题

英文解析

33. 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, byconsidering the homothety centered at A taking the excircle to the incircle, we notice 2
    that A is the intersection of ω and the tangent line parallel to BC . It follows that
    A B C is congruent to A B C by reflecting through the center of ω . We therefore 123
    1 1 1 3 3 3
    need only find [ A B C ] / [ ABC ]. Since
    1 1 1
    [ A BC ] A B · BC ((9 + 10 − 13) / 2) 12
    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