HMMT 二月 2008 · TEAM1 赛 · 第 14 题
HMMT February 2008 — TEAM1 Round — Problem 14
题目详情
- [ 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. 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 they pass 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. The incenter is the center of the incircle , which is the unique circle inscribed in ABC , tangent to all three sides. 2 • 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 if sin ∠ BAX sin ∠ CBY sin ∠ ACZ · · = 1 . sin ∠ XAC sin ∠ Y BA sin ∠ ZCB 3
解析
- [ 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 P E ′ E ′ F B C D Using sine law, we have ′ ′ 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 . ′ AD Do this for all the factors in ( † ), and we get ( ) 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 on triangle DEF . The result follows. Remark: This result is known as Steinbart Theorem . Beware that its converse is not com- pletely true. For more information and discussion, see Darij Grinberg’s paper “Variations of the 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 they pass through a common point. A set of points are concyclic if they lie on a common circle. 6 • Given ABC a triangle, the three angle bisectors are concurrent at the incenter of the triangle. The incenter is the center of the incircle , which is the unique circle inscribed in ABC , tangent to 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 if sin ∠ BAX sin ∠ CBY sin ∠ ACZ · · = 1 . sin ∠ XAC sin ∠ Y BA sin ∠ ZCB 7