HMMT 二月 2020 · 冲刺赛 · 第 28 题

HMMT February 2020 — Guts Round — Problem 28

[15] Let 4 ABC be a triangle inscribed in a unit circle with center O . Let I be the incenter of 4 ABC , and let D be the intersection of BC and the angle bisector of ∠ BAC . Suppose that the circumcircle of 12 4 ADO intersects BC again at a point E such that E lies on IO . If cos A = , find the area of 4 ABC . 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2020, February 15, 2020 — GUTS ROUND Organization Team Team ID#

解析

[15] Let 4 ABC be a triangle inscribed in a unit circle with center O . Let I be the incenter of 4 ABC , and let D be the intersection of BC and the angle bisector of ∠ BAC . Suppose that the circumcircle 12 of 4 ADO intersects BC again at a point E such that E lies on IO . If cos A = , find the area of 13 4 ABC . Proposed by: Michael Diao 15 Answer: 169 Solution: Consider the following lemma: Lemma. AD ⊥ EO . Proof. By the Shooting Lemma, the reflection of the midpoint M of arc BC not containing A over BC lies on ( ADO ). Hence ′ ′ ◦ ] ADE + ] DEO = ] M DC + ] DM O = ] M DC + ] M M D = 90 . This is enough to imply AD ⊥ EO . A O I ′ M E D B C M Thus I is the foot from O onto AD . Now 2 2 2 AI + IO = AO . By Euler’s formula, ( ) 2 r 2 2

R − 2 Rr = R . A sin 2 Hence A 2 r = 2 R sin . 2 Then r s = a + = a + R sin A = 3 R sin A A tan 2 and ( ) A 2 [ ABC ] = rs = 2 R sin (3 R sin A ) . 2 Since R = 1, we get [ ABC ] = 3 (1 − cos A ) sin A. 5 12 Plugging in sin A = and cos A = , we get 13 13 1 5 15 [ ABC ] = 3 · · = . 13 13 169 5 Remark. On the contest, this problem stated that 4 ABC is an acute triangle and that sin A = , 13 12 rather than cos A = . This is erroneous because there is no acute triangle satisfying the conditions 13 of the problem statement (the given diagram is not accurate). We apologize for the mistake.