NMO Selection Tests for the Balkan and International Mathematical Olympiads

The tangent at $A$ to the circumcircle $ABC$ meets the line $BC$ at the point $A^{\prime \prime }$; the points $B^{\prime \prime }$ and $C^{\prime \prime }$ are defined similarly. The points $A^{\prime \prime }$, $B^{\prime \prime }$ and $C^{\prime \prime }$ are collinear on Lemoine's line. We shall prove that the lines $A{A}^{\prime }$, $B{B}^{\prime }$ and $C{C}^{\prime }$ are the polars of the points $A^{\prime \prime }$, $B^{\prime \prime }$ and $C^{\prime \prime }$, respectively, relative to the nine-point circle $\gamma$, so they are indeed concurrent. Clearly, it is sufficient to prove that that $A{A}^{\prime }$ is the polar of $A^{\prime \prime }$ with respect to $\gamma$.

Let $A_1,B_1$ and $C_1$ be the perpendicular feet dropped from $A,B$ and $C$, respectively, on the lines $BC,CA$ and $AB$, respectively. Let further $A_2$ be the midpoint of the side $BC$, and let $A_3$ be the midpoint of the segment joining $A$ to the orthocenter of the triangle $ABC$. It is easily seen that the line $A_2{A}_3$ is the perpendicular bisector of the segment $B_1{C}_1$, so it is perpendicular to the tangent at $A$ to the circumcircle $ABC$. Consequently, $A_3$ is the orthocenter of the triangle $A{A}^{\prime \prime }{A}_2$, so the lines $A{A}_2$ and $A^{\prime \prime }{A}_3$ are perpendicular; it is easily seen that they meet at some point on $\gamma$, so $A^{\prime \prime }$ lies on the polar of $A$ with respect to $\gamma$. Finally, $A^{\prime \prime }$ lies on $BC$, which is the polar of $A^{\prime }$ with respect to $\gamma$, so $A^{\prime \prime }$ is the pole of $A{A}^{\prime }$ with respect to $\gamma$.