HMMT 十一月 2014 · 冲刺赛 · 第 29 题

英文解析

29. [ 15 ] Let ω be a fixed circle with radius 1, and let BC be a fixed chord of ω such that BC = 1. Thelocus 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.
    A3
    Lemma 1 : If ABC is a triangle with incenter I , then ∠ BIC = 90 + .
    B2
    Proof: Consider triangle BIC . Since I is the intersection of the angle bisectors, ∠ IBC = and
    B C A2
    ∠ ICB = . It follows that ∠ BIC = 180 − − = 90 + .C
    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 caseis similar.
    Let O be the center of ω . Since BC has the same length as a radius, △ OBC is equilateral. We nowbreak 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.