HMMT 二月 2008 · TEAM1 赛 · 第 13 题

HMMT February 2008 — TEAM1 Round — Problem 13

[ 40 ] Let M be the midpoint of BC , and T diametrically opposite to D on the incircle of ABC . Show that DT, AM, EF are concurrent.

解析

[ 40 ] Let M be the midpoint of BC , and T diametrically opposite to D on the incircle of ABC . Show that DT, AM, EF are concurrent. Solution: If AB = AC , then the result is clear as AM and DT coincide. So, assume that AB 6 = AC . A T F X Z Y E I B C M D Let lines DT and EF meet at Z . Construct a line through Z parallel to BC , and let it ◦ ◦ meet AB and AC at X and Y , respectively. We have ∠ XZI = 90 , and ∠ XF I = 90 . Therefore, F, Z, I, X are concyclic, and thus ∠ IXZ = ∠ IF Z . By similar arguments, we also have ∠ IY Z = ∠ IEZ . Thus, triangles IF E and IXY are similar. Since IE = IF , we must also have IX = IY . Since IZ is an altitude of the isosceles triangle IXY , Z is the midpoint of XY . 5 Since XY and BC are parallel, there is a dilation centered at A that sends XY to BC . So it must send the midpoint Z to the midpoint M . Therefore, A, Z, M are collinear. It follows that DT, AM, EF are concurrent.