HMMT 二月 2008 · TEAM1 赛 · 第 14 题

英文解析

14. [ 40 ] Let P be a point inside the incircle of ABC . Let lines DP, EP, F P meet the incircle
    ′ ′ ′ ′ ′ ′
    again at D , E , F . Show that AD , BE , CF are concurrent.
    Solution: Using the trigonometric version of Ceva’s theorem, it suffices to prove that
    ′ ′ ′
    sin ∠ BAD sin ∠ CBE sin ∠ ACF
    · · = 1 . ( † )
    ′ ′ ′
    sin ∠ D AC sin ∠ E BA sin ∠ F CB
    ′A
    F D
    ′EP
    ′E
    B CF
    Using sine law, we have d
    ′ ′
    F D F D
    ′ ′ ′
    sin ∠ BAD = · sin ∠ AF D = · sin ∠ F DD
    ′ ′
    AD AD
    ′ ′
    Let r be the inradius of ABC . Using the extended sine law, we have F D = 2 r sin ∠ F DD .
    Therefore,
    2 r
    ′ 2 ′
    sin ∠ BAD = · sin ∠ F DD .
    ′
    Do this for all the factors in ( † ), and we get aD
    ( )
    ′ ′ ′ ′ ′ ′2
    sin ∠ BAD sin ∠ CBE sin ∠ ACF sin ∠ F DD sin ∠ DEE sin ∠ EF F
    · · = · ·
    ′ ′ ′ ′ ′ ′
    sin ∠ D AC sin ∠ E BA sin ∠ F CB sin ∠ D DE sin ∠ E EF sin ∠ F F D
    ′ ′ ′
    Since DD , EE , F F are concurrent, the above expression equals to 1 by using trig Ceva ontriangle DEF . The result follows.
    Remark: This result is known as Steinbart Theorem . Beware that its converse is not completely true. For more information and discussion, see Darij Grinberg’s paper “Variations ofthe Steinbart Theorem” at http://de.geocities.com/darij_grinberg/ .
    Glossary and some possibly useful facts
    • A set of points is collinear if they lie on a common line. A set of lines is concurrent if theypass through a common point. A set of points are concyclic if they lie on a common circle.
    • Given ABC a triangle, the three angle bisectors are concurrent at the incenter of the triangle.6
    The incenter is the center of the incircle , which is the unique circle inscribed in ABC , tangentto all three sides.
    • Ceva’s theorem states that given ABC a triangle, and points X, Y, Z on sides BC, CA, AB ,
    respectively, the lines AX, BY, CZ are concurrent if and only if
    BX CY AZ
    · · = 1 .
    XB Y A ZB
    • “Trig” Ceva states that given ABC a triangle, and points X, Y, Z inside the triangle, the lines AX, BY, CZ are concurrent if and only ifsin ∠ BAX sin ∠ CBY sin ∠ ACZ
    · · = 1 .
    sin ∠ XAC sin ∠ Y BA sin ∠ ZCB 7