In triangle $ABC$ , $AA_0,BB_0,CC_0$ are the angle bisectors , $A_0,B_0,C_0$are on sides $BC,CA,AB,$ draw $A_0A_1//BB_0,A_0A_2//CC_0$ ,$A_1$ lies on $AC$ ,$A_2$ lies on $AB$ , $A_1A_2$ intersects $BC$ at $A_3$.$B_3$ ,$C_3$ are constructed similarly.Prove that : $A_3,B_3,C_3$ are collinear.
Problem
Source: China South East Mathematical Olympiad 2011
Tags: geometry proposed, geometry
Luis González
18.08.2011 19:53
The result is still true for any P-cevian triangle $\triangle A_0B_0C_0.$ Let $A_0A_1$ cut $B_0C_0$ at $U.$ From the harmonic pencil $B_0(A_0,C_0,P,A),$ we deduce that $A_1$ is midpoint of $\overline{UA_0}.$ Similarly, if $A_0A_2$ cuts $B_0C_0$ at $V,$ then $A_2$ is midpoint of $\overline{VA_0}$ $\Longrightarrow$ $A_1A_2$ is the $A_0$ -midline of $\triangle A_0UV.$ If $X_0 \equiv B_0C_0 \cap BC,$ then $A_3$ is midpoint of $\overline{A_0X_0}.$ Defining $Y_0,Z_0$ on $CA,AB$ similarly as $X_0,$ we get that $B_3,C_3$ are midpoints of $\overline{B_0Y_0},$ $\overline{C_0Z_0}.$ $X_0,Y_0,Z_0$ are collinear on the trilinear polar $\tau$ of $P$ WRT $\triangle ABC,$ hence $A_3,B_3,C_3$ lie on the Newton's line of the complete quadrangle bounded by the sidelines of $\triangle A_0B_0C_0$ and $\tau,$ i.e. $\overline{A_3B_3C_3}$ is the trilinear polar of $P^2.$
erfan_Ashorion
26.10.2011 23:01
oh i have a solution but it is 11:25 in iran and i want to go sleep but i say a small of my solution and compleat it tomorrow! i proof that : $\frac{BA3}{A3C} = \frac{BC0.CB0}{(AC-BC0).(AB-CB0)}$ continuation of it tomorrow!