HMMT 二月 2016 · 几何 · 第 10 题

英文解析

10. The incircle of a triangle ABC is tangent to BC at D . Let H and Γ denote the orthocenter andcircumcircle of 4 ABC . The B -mixtilinear incircle , centered at O , is tangent to lines BA and BCand internally tangent to Γ. The C -mixtilinear incircle , centered at O , is defined similarly. Suppose b
    √Cthat DH ⊥ O O , AB = 3 and AC = 2. Find BC .
    B C
    Proposed by: Evan Chen
    √
    √
    Answer: (7 + 2 13)1
    Let the B -mixtilinear incircle ω touch Γ at T , BA at B and BC at B . Define T ∈ Γ, C ∈ CB ,3
    B B 1 2 C 1
    C ∈ CA , and ω similarly. Call I the incenter of triangle ABC , and γ the incircle.
    2 C
    We first identify two points on the radical axis of the B and C mixtilinear incircles:
    • The midpoint M of arc BC of the circumcircle of ABC . This follows from the fact that M , B ,
    T are collinear with 1
    2 2B
    M B = M C = M B · M T
    1 Band similarly for C .
    • The midpoint N of ID . To see this, first recall that I is the midpoint of segments B B and
    1 2
    C C . From this, we can see that the radical axis of ω and γ contains N (since it is the line
    1 2 Bthrough the midpoints of the common external tangents of ω , γ ). A similar argument for Cshows that the midpoint of ID is actually the radical center of the ω , ω , γ .B
    B C
    Now consider a homothety with ratio 2 at I . It sends line M N to the line through D and the A -excenter
    I (since M is the midpoint of II , by “Fact 5”). Since DH was supposed to be parallel to line M N ,
    A Ait follows that line DH passes through I ; however a homothety at D implies that this occurs only if
    H is the midpoint of the A -altitude.A
    √
    Let a = BC , b = CA = 2 and c = AB = 3. So, we have to just find the value of a such that theorthocenter of ABC lies on the midpoint of the A -altitude. This is a direct computation with the Lawof Cosines, but a more elegant solution is possible using the fact that H has barycentric coordinates
    2 2 21
    ( S S : S S : S S ), where S = ( b + c − a ) and so on. Indeed, as H is on the A -midline we
    B C C A A B Adeduce directly that 2
    1 1
    2 2 2 2 2
    S S = S ( S + S ) = a S = ⇒ ( a − 1)( a + 1) = a (7 − a ) .
    B C A B C A
    4 2
    Solving as a quadratic in a and taking the square roots gives 2
    √
    √
    4 21
    3 a − 14 a − 1 = 0 = ⇒ a = (7 + 2 13)
    as desired. 13