2015 Thirteenth IMAR Mathematical Competition

Let the lines $B_2{C}_2$, $C_2{A}_2$, $A_2{B}_2$ meet the circle $ABC$ again at $A_3$, $B_3$, $C_3$, respectively. The angle $A{A}_3{A}_1$ is a right angle, so $A_1{A}_3$ is perpendicular to $B_2{C}_2$ and hence parallel to $XA$, and the parallel lines $XA$ and $A_1{A}_3$ are equidistant from the circumcentre $O$ of the triangle $ABC$. Hence $A_1{A}_3$ passes through the reflection $X^{\prime }$ of $X$ across $O$.

Similarly, the lines $B_1{B}_3$ and $C_1{C}_3$ pass through $X^{\prime }$, and therefore $X^{\prime }{A}_1\cdot X^{\prime }{A}_3=X^{\prime }{B}_1\cdot X^{\prime }{B}_3=X^{\prime }{C}_1\cdot X^{\prime }{C}_3$, a quantity denoted $r^2$.



It follows that $A_1$, $B_1$, $C_1$ are the poles of the lines $B_2{C}_2$, $C_2{A}_2$, $A_2{B}_2$, respectively, relative to the circle of radius $r$ centered at $X^{\prime }$, so the triangles $A_1{B}_1{C}_1$ and $A_2{B}_2{C}_2$ are in perspective, by the well-known lemma below.

Lemma. If $XYZ$ is a triangle, and $X^{\prime }$, $Y^{\prime }$, $Z^{\prime }$ are the poles of the lines $YZ$, $ZX$, $XY$, respectively, relative to a conic $\gamma$, then the triangles $XYZ$ and $X^{\prime }{Y}^{\prime }{Z}^{\prime }$ are in perspective.