HMMT 十一月 2021 · 冲刺赛 · 第 21 题

HMMT November 2021 — Guts Round — Problem 21

[11] Circle ω is inscribed in rhombus HM M T so that ω is tangent to HM at A , M M at I , M T at 1 2 1 1 2 2 M , and T H at E . Given that the area of HM M T is 1440 and the area of EM T is 405, find the area 1 2 of AIM E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2021, November 13, 2021 — GUTS ROUND Organization Team Team ID#

解析

[11] Circle ω is inscribed in rhombus HM M T so that ω is tangent to HM at A , M M at I , M T 1 2 1 1 2 2 at M , and T H at E . Given that the area of HM M T is 1440 and the area of EM T is 405, find the 1 2 area of AIM E . Proposed by: Joseph Heerens Answer: 540 Solution: First, from equal tangents, we know that T E = T M. As the sides of a rhombus are also equal, this gives from SAS similarity that EM T ∼ T HM . Further, the ratio of their areas is 2 405 9 3 = . This means that T E = T M = HT. Then, we get that M M = M I, so M M I ∼ M T M , 2 2 2 1 1440 / 2 16 4 1 1 720 and since M M = M T, we get that [ M M I ] = [ M T M ] = = 45 . From here, 2 2 2 2 1 4 16 16 [ AIM E ] = [ HM M T ] − [ EHA ] − [ AM I ] − [ IM M ] − [ M T E ] = 1440 − 2(405 + 45) = 540 . 1 2 1 2