HMMT 二月 2020 · 团队赛 · 第 8 题
HMMT February 2020 — Team Round — Problem 8
题目详情
- [50] Let ABC be a scalene triangle with angle bisectors AD , BE , and CF so that D , E , and F lie on segments BC , CA , and AB respectively. Let M and N be the midpoints of BC and EF respectively. Prove that line AN and the line through M parallel to AD intersect on the circumcircle of ABC if and only if DE = DF .
解析
- [50] Let ABC be a scalene triangle with angle bisectors AD , BE , and CF so that D , E , and F lie on segments BC , CA , and AB respectively. Let M and N be the midpoints of BC and EF respectively. Prove that line AN and the line through M parallel to AD intersect on the circumcircle of ABC if and only if DE = DF . Proposed by: Michael Ren Solution 1: L A A E N F M B C D Y Z P X Let X , Y be on AB , AC such that CX ‖ BE and BY ‖ CF . Then BX = BC = CY . Let Z be the − − → − − → − − → 1 midpoint of XY . Then M Z = ( BX + CY ), which bisects the angle between BX and CY because 2 they have the same length. Therefore M Z ‖ AD . Furthermore, by similar triangles we have AE · AX = AB · AC = AF · AY. AE AY This rearranges to = , so EF ‖ XY . Therefore Z is the intersection of the lines in the problem AF AX statement. Then sin ∠ ZBX BX sin ∠ BZX XZ = = 1 sin ∠ ZCY sin ∠ CZY CY Y Z iff Z ∈ ( ABC ), so XY is the external angle bisector of ∠ BZC iff Z ∈ ( ABC ). Thus if P = AD ∩ XY , P ∈ ( ABC ) if Z ∈ ( ABC ). Additionally the spiral similarity from BX to CY gives L Z ⊥ XY where A ◦ L is the midpoint of arc BAC , so if P ∈ ( ABC ) then Z must be on ( ABC ) because ∠ L ZP = 90 . A A Therefore Z ∈ ( ABC ) iff P ∈ ( ABC ). √ From the previous length computation, we know that an inversion at A with radius AB · AC com- posed with reflection about AD will send X and Y to E and F . We have P ∈ ( ABC ) iff its image under the inversion is D , but since P was defined as AD ∩ XY this is true iff ( AEDF ) is cyclic. Since ABC is scalene and AE 6 = AF, this is true iff DE = DF. Solution 2: L ′ A Y Y A E F N C T B D M ′ X X M A Let L be the midpoint of arc BAC and let M be diametrically opposite L . Let EF, AL , and BC A A A A ◦ meet at T so ∠ DAT = 90 ; note that DE = DF iff DN ⊥ EF, which is equivalent to ( T AN D ) being ′ cyclic. Let AN ∩ ( ABC ) = X and XM ∩ ( ABC ) = Y, and let Y be the reflection of Y over L M A A ′ with similarly X the reflection of X over L M . We wish to show N ∈ ( T AD ) iff XY ‖ AM . A A A X ′ ′ We claim AY ‖ EF. By projecting − 1 = ( B, C ; M, ∞ ) = ( B, C ; Y, X ) and reflecting over L M , we A A ′ ′ find ( X, Y ; B, C ) = − 1 . Then projecting through A gives ( N, AY ∩ EF ; F, E ) = − 1 , and since N is ′ the midpoint of EF we find AY ‖ EF. ′ Now ( T AN D ) cyclic iff ] DAN = ] DT N, and ] DT N = ] Y Y A by the parallel lines. But we have ] DAN = ] M AX, so arcs M X and Y A are equal iff ( T AN D ) is cyclic. Thus XY ‖ AM iff A A A DE = DF as desired.