HMMT 十一月 2014 · 冲刺赛 · 第 29 题
HMMT November 2014 — Guts Round — Problem 29
题目详情
- [ 15 ] Let ω be a fixed circle with radius 1, and let BC be a fixed chord of ω such that BC = 1. The locus of the incenter of ABC as A varies along the circumference of ω bounds a region R in the plane. Find the area of R .
解析
- [ 15 ] Let ω be a fixed circle with radius 1, and let BC be a fixed chord of ω such that BC = 1. The locus of the incenter of ABC as A varies along the circumference of ω bounds a region R in the plane. Find the area of R . ( ) √ 3 − 3 Answer: π − 1 We will make use of the following lemmas. 3 A Lemma 1 : If ABC is a triangle with incenter I , then ∠ BIC = 90 + . 2 B Proof: Consider triangle BIC . Since I is the intersection of the angle bisectors, ∠ IBC = and 2 B C A C ∠ ICB = . It follows that ∠ BIC = 180 − − = 90 + . 2 2 2 2 Lemma 2: If A is on major arc BC , then the circumcenter of △ BIC is the midpoint of minor arc BC , and vice-versa. ◦ Proof: Let M be the midpoint of minor arc BC . It suffices to show that ∠ BM C + 2 ∠ BIC = 360 , since BM = M C . This follows from Lemma 1 and the fact that ∠ BM C = 180 − ∠ A . The other case is similar. Let O be the center of ω . Since BC has the same length as a radius, △ OBC is equilateral. We now break the problem into cases depending on the location of A. ◦ Case 1: If A is on major arc BC , then ∠ A = 30 by inscribed angles. If M is the midpoint of minor ◦ arc BC , then ∠ BM C = 150 . Therefore, if I is the incenter of △ ABC , then I traces out a circular ◦ segment bounded by BC with central angle 150 , on the same side of BC as A . Case 2: A similar analysis shows that I traces out a circular segment bounded by BC with central ◦ angle 30 , on the other side of BC . 1 1 2 2 The area of a circular segment of angle θ (in radians) is given by θR − R sin θ , where R is the radius 2 2 2 2 of the circular segment. By the Law of Cosines, since BC = 1, we also have that 2 R − 2 R cos θ = 1. Computation now gives the desired answer.