Thứ Sáu, 12 tháng 9, 2014

8-Fermat point and concurrent point

Problem: Let ABC be a triangle, Let F is the first or the secon Fermat point. Construc two line through F and angle of two lie is 60 degrees. Two lines meet three sideline at A_b,A_c,B_c,B_a, C_a,C_b (show in the figure attachment). Prove that A_aA_bB_aB_cC_aC_b are concurrent.


Solutionby Luis González:


We also show that the concurrency point D of the referred lines is on the trilinear polar of F WRT ABC.

Let \triangle F_aF_bF_c be the cevian triangle of F WRT \triangle ABC. X \equiv F_bF_c \cap BCand Y \equiv F_cF_a \cap CA \Longrightarrow XY \equiv f is trilinear polar of F WRT \triangle ABC, i.e. perspectrix of \triangle ABC and \triangle F_aF_bF_c. 



We use polarity WRT an abitrary circle (F) centered at F. Polars of A,B,Cbound an equilateral triangle since they are perpendicular to FA,FB,FC, i.e. poles P,Q,R of BC,CA,AB are vertices of equilateral triangle. Polars of F_b,F_c are then the parallels through Q,R to PR,PQ, respectively, meeting at the pole U of F_bF_c. If V,W are the poles of F_cF_a, F_aF_b, then \triangle UVW is also equilateral, being the antimedial of \triangle PQR. PU and QV are polars of X,Ymeeting at the pole O \equiv PU \cap QV of f \equiv XY; the center of \triangle PQR. Polars of A_b,B_a are two parallels through Q,R and polars of B_c,A_c are two parallels through P,Q (these directions forming 120^{\circ}, due to \angle (A_bB_a,A_cB_c)=60^{\circ}\Longrightarrow poles M,L,K of A_bB_c, B_cB_a,A_cA_b form parallelogram RKMLcircumscribed to \triangle PQR.

Since \angle PLR=\angle POR=\angle ROQ=\angle RKQ=60^{\circ} \pmod\pi, then clearly the circles \odot(PLR) and \odot(RKQ) meet at R and O, which is then midpoint of their arcs PR,RQ. If LO cuts \odot(RKQ) again at K', we have \angle QK'O=30^{\circ}=\angle RLO \Longrightarrow QK' \parallel RL \Longrightarrow K \equiv K' \Longrightarrow O,K,L are collinear \Longrightarrow their polars f, A_bA_c, B_aB_c concur at D. By similar reasoning C_aC_b goes through D \equiv f \cap A_bA_c \cap B_aB_c.

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