HMMT 二月 2006 · 冲刺赛 · 第 38 题

英文解析

38. Suppose ABC is a triangle with incircle ω , and ω is tangent to BC and CA at D and
    E respectively. The bisectors of ∠ A and ∠ B intersect line DE at F and G respectively,
    such that BF = 1 and F G = GA = 6. Compute the radius of ω .
    √
    2 5
    Answer:
    1 1 15
    Solution: Let α, β, γ denote the measures of ∠ A, ∠ B, ∠ C, respectively. We have
    2 2 2
    ° ° ° °
    m ∠ CEF = 90 − γ, m ∠ F EA = 90 + γ, m ∠ AF G = m ∠ AF E = 180 − α − (90 + γ ) =
    β = m ∠ ABG , so ABF G is cyclic. Now AG = GF implies that BG bisects ∠ ABF .
    Since BG by definition bisects ∠ ABC , we see that F must lie on BC . Hence, F = D .
    If I denotes the incenter of triangle ABC , then ID is perpendicular to BC , but since
    A, I, F are collinear, we have that AD ⊥ BC . Hence, ABC is isoceles with AB = AC .
    Furthermore, BC = 2 BF = 2. Moreover, since ABF G is cyclic, ∠ BGA is a right angle.
    ′ ′ ′
    Construct F on minor arc GF such that BF = 6 and F G = 1, and let AB = x . By
    √13
    ′ ′
    the Pythagorean theorem, AF = BG = x − 36, so that Ptolemy applied to ABF G2
    yields x − 36 = x + 36. We have ( x − 9)( x + 8) = 0. Since x is a length we find 2
    x = 9. Now we have AB = AC = 9. Pythagoras applied to triangle ABD now yields
    √ √ √ √
    2 21
    AD = 9 − 1 = 4 5, which enables us to compute [ ABC ] = · 2 · 4 5 = 4 5.
    Since the area of a triangle is also equal to its semiperimeter times its inradius, we 2
    √
    √
    2 5
    have 4 5 = 10 r or r = .
    REMARK. In fact, ABF G is always a cyclic quadrilateral for which AB plays a di-5
    ameter. That is, we could have proven this fact without using F G = GA .