HMMT 二月 2016 · 团队赛 · 第 10 题

英文解析

10. [ 50 ] Let ABC be a triangle with incenter I whose incircle is tangent to BC , CA , AB at D , E , F .
    Point P lies on EF such that DP ⊥ EF . Ray BP meets AC at Y and ray CP meets AB at Z . Point
    Q is selected on the circumcircle of 4 AY Z so that AQ ⊥ BC .
    Prove that P , I , Q are collinear.
    Proposed by: Evan Chen
    EYXA
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    B CI
    The proof proceeds through a series of seven lemmas.D
    Lemma 1. Lines DP and EF are the internal and external angle bisectors of ∠ BP C .
    Proof. Since DEF the cevian triangle of ABC with respect to its Gregonne point, we have that
    ( ) − 1 = EF ∩ BC, D ; B, C .
    °
    Then since ∠ DP F = 90 we see P is on the Apollonian circle of BC through D . So the conclusionfollows.
    Lemma 2. Triangles BP F and CEP are similar.
    Proof. Invoking the angle bisector theorem with the previous lemma gives
    BP BP CP CP = = = .
    BF BD CD CE
    But ∠ BF P = ∠ CEP , so 4 BF P ∼ 4 CEP .
    Lemma 3. Quadrilateral BZY C is cyclic; in particular, line Y Z is the antiparallel of line BC through
    ∠ BAC .
    Proof. Remark that ∠ Y BZ = ∠ P BF = ∠ ECP = ∠ Y CZ .
    Lemma 4. The circumcircles of triangles AY Z , AEF , ABC are concurrent at a point X such that
    4 XBF ∼ 4 XCE .
    Proof. Note that line EF is the angle bisector of ∠ BP Z = ∠ CP Y . Thus
    ZF ZP Y P Y E = = = .
    F B P B P C EC
    Then, if we let X be the Miquel point of quadrilateral ZY CB , it follows that the spiral similaritymapping segment BZ to segment CY maps E to F ; therefore the circumcircle of 4 AEF must passthrough X too.
    Lemma 5. Ray XP bisects ∠ F XE .
    Proof. The assertion amounts to
    XF BF F P = = .
    XE EC P E
    The first equality follows from the spiral similarity 4 BF X ∼ 4 CEX , while the second is from
    4 BF P ∼ 4 CEP . So the proof is complete by the converse of angle bisector theorem.
    Lemma 6. Points X , P , I are collinear.
    1 1 1
    Proof. On one hand, ∠ F XI = ∠ F AI = ∠ A . On the other hand, ∠ F XP = ∠ F XE = ∠ A . Hence,
    2 2 2
    X , Y , I collinear.
    Lemma 7. Points X , Q , I are collinear.
    °
    Proof. On one hand, ∠ AXQ = 90 , because we established earlier that line Y Z was antiparallel to
    °
    line BC through ∠ A , hence AQ ⊥ BC means exactly that ∠ AZQ = AY Q = 90 . On the other hand,
    °
    ∠ AXI = 90 according to the fact that X lies on the circle with diameter AI . This completes theproof of the lemma.
    Finally, combining the final two lemmas solves the problem.