HMMT 二月 2020 · 冲刺赛 · 第 24 题
HMMT February 2020 — Guts Round — Problem 24
题目详情
- [12] In 4 ABC , ω is the circumcircle, I is the incenter and I is the A -excenter. Let M be the midpoint A ̂ of arc BAC on ω , and suppose that X , Y are the projections of I onto M I and I onto M I , respectively. A A If 4 XY I is an equilateral triangle with side length 1, compute the area of 4 ABC . A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2020, February 15, 2020 — GUTS ROUND Organization Team Team ID# 4
解析
- [12] In 4 ABC , ω is the circumcircle, I is the incenter and I is the A -excenter. Let M be the A ̂ midpoint of arc BAC on ω , and suppose that X , Y are the projections of I onto M I and I onto A A M I , respectively. If 4 XY I is an equilateral triangle with side length 1, compute the area of 4 ABC . A Proposed by: Michael Diao √ 6 Answer: 7 Solution 1: Using Fact 5, we know that II intersects the circle ( ABC ) at M , which is the center A A 1 √ of ( II BCXY ). Let R be the radius of the latter circle. We have R = . A 3 π We have ∠ AIM = ∠ Y II = ∠ Y IX = . Also, ∠ II M = ∠ IM I by calculating the angles from the A A A 3 equilateral triangle. Using 90-60-30 triangles, we have: 1 1 AI = M I = II = R A 2 2 √ √ 3 AM = M I = 3 R 2 2 2 2 2 M M = AM + AM = 7 R A A Now, let J and N be the feet of the altitudes from A and B respectively on M M . Note that as M is A an arc midpoint of BC , N is actually the midpoint of BC . 2 AM 4 A √ M J = = R A M M 7 A 2 BM 1 A M N = = √ R A M M 7 A 3 √ Thus JN = R . Also, we have, 7 6 2 2 BN = M N · M N = R A 7 √ √ 1 3 6 6 2 Now, [ ABC ] = JN · BC = JN · BN = R = . 2 7 7 Solution 2: By Fact 5, we construct the diagram first with 4 XY I as the reference triangle. A M A X I B C D M A Y I A 1 √ Let M be the circumcenter of 4 XY I and let Ω be the circumcircle, which has circumradius R = . A A 3 ̂ Then by Fact 5, M is the midpoint of minor arc BC , and B, C ∈ Ω. Now we show the following A result: Claim. b + c = 2 a . 1 Proof. Letting D be the intersection of II with BC , we have M D = R . Then by the Shooting A A 2 Lemma, 2 M A · M D = R = ⇒ M A = 2 R. A A A On the other hand, by Ptolemy, M B · AC + M C · AB = M A · BC = ⇒ R · ( b + c ) = 2 R · a, A A A whence the result follows. By the triangle area formula, we have 1 3 [ ABC ] = sr = ( a + b + c ) r = ar. 2 2 Therefore, we are left to compute a and r . Since M is the antipode of M on ( ABC ) we get M B, M C are tangent to Ω; in particular, M BM C A A is a kite with M B = M C and M B = M C . Then by Power of a Point, we get A A 2 2 M B = M C = M I · M Y = 2 R · 3 R = 2 , where M I = II = 2 R and IY = R . A Since ID = DM , we know that r is the length of the projection from M to BC . Finally, given our A A above information, we can compute 2 2 BM BM 1 A A √ √ r = = = 2 2 M M 21 A BM + M B A √ BM 2 2 A a = 2 r = . M B 7 The area of ABC is then ( ) √ √ ( ) 3 2 2 1 6 √ √ = . 2 7 7 21 4