shortlistBMO 2011

We must show that $AN$ is symmedian in the triangle $ABC$. Let $R$ denote the radius of the circle $ABC$ and notice that the cross-ratio $(EDBC)=-1$ to deduce that $L{A}^2=L{D}^2=LB\cdot LC=L{O}^2-R^2=L{O}^2-A{O}^2$, so $L$ and $O$ are antipodal in the circle $ALO$. It then follows that $N$ is the reflection of $A$ in the diameter $LO$ and lies on the Apollonius circle $ADE$. Let $AN$ and $BC$ meet at $K$. We shall prove that $AK$ is the symmedian through $A$ in the triangle $ABC$. Notice that $K$ has equal powers relative to the circles $ABC$ and $ADE$, $KB\cdot KC=KA\cdot LK=KD\cdot KE$, to deduce that $$KB=\frac{BD\cdot BE}{BE+CD}\text{and}KC=\frac{CD\cdot CE}{BD+CE}.$$ Finally, recall that $$BD=\frac{AB\cdot BC}{AB+AC},CD=\frac{AC\cdot BC}{AB+AC},$$ $$BE=\frac{AB\cdot BC}{|AB-AC|}\text{and}CE=\frac{AC\cdot BC}{|AB-AC|},$$ from the bisectrix theorem, to get $$KB=\frac{A{B}^2\cdot BC}{A{B}^2+A{C}^2}\text{and}KC=\frac{A{C}^2\cdot BC}{A{B}^2+A{C}^2},$$ so $KB/KC=A{B}^2/A{C}^2$, and conclude by Steiner's theorem that $AK$ is indeed the symmedian from $A$ in the triangle $ABC$.



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Alternative solution.

As in the previous solution, $LA$ is tangent to the circle $ABC$, and $N$ is the reflection of $A$ in the diameter $LO$ and lies on the Apollonius circle $ADE$. It then follows that $LN$ is the tangent at $N$ to the circle $ABC$, so $\angle BNL=\angle BAN$. On the other hand, $\angle LNE=\angle LEN=\angle DEN=\angle DAN$, so $\angle BNE=\angle BNL+\angle LNE=\angle BAN+\angle DAN=\angle BAD$. Extend $AD$ to meet again the circle $ABC$ at the midpoint $J$ of the arc $BC$ that does not contain $A$. Notice that $\angle BAD=\angle BAJ$ is supplementary to $\angle BNJ$ to deduce, by the preceding, that $J$ lies on the diameter $EJ$ of the circle $AEJ$. The latter passes through $M$, for the lines $EM$ and $JM$ are perpendicular. Consequently, $\angle MAD=\angle MAJ=\angle MEJ=\angle LEN=\angle NAD$ and the conclusion follows.

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Alternative solution.

We must show that $AN$ is symmedian in the triangle $ABC$. Notice first that $L$ and $O$ are antipodal points on the circle $ALO$, so $N$ is the reflection of $A$ about the diameter $LO$; also, $M$ lies on the circle $ALO$, for the lines $LM$ and $MO$ are perpendicular.



Now set the pole at $A$ and invert the configuration; write $X^{\ast }$ for the image of a point $X$ different from $A$. The points $B^{\ast },C^{\ast },D^{\ast },E^{\ast },L^{\ast }$ and $M^{\ast }$ all lie on a circle $\gamma$ through $A$; $D^{\ast }$ and $E^{\ast }$ are antipodal, for $AD$ and $AE$ are perpendicular; $B^{\ast }$ and $C^{\ast }$ are reflections of one another in the diameter $D^{\ast }{E}^{\ast }$, for $AD$ and $AE$ are the two bisectrices of the angle $BAC$; and the lines $A{M}^{\ast }$ and $A{L}^{\ast }$ are symmedians in triangles $A{B}^{\ast }{C}^{\ast }$ and $A{D}^{\ast }{E}^{\ast }$, respectively, for $AM$ and $AL$ are medians in triangles $ABC$ and $ADE$, respectively. Since $D^{\ast }$ and $E^{\ast }$ are antipodal points on $\gamma$, the lines $A{D}^{\ast }$ and $A{E}^{\ast }$ are perpendicular, so $L^{\ast }$ is the reflection of $A$ in the diameter $D^{\ast }{E}^{\ast }$. Let $N^{\prime }$ be the midpoint of the chord $B^{\ast }{C}^{\ast }$, extend $A{N}^{\prime }$ to meet $\gamma$ at $M^{\prime }$ and recall that $A{M}^{\ast }$ is symmedian in triangle $A{B}^{\ast }{C}^{\ast }$ to deduce that $M^{\prime }$ is the reflection of $M^{\ast }$ in the diameter $D^{\ast }{E}^{\ast }$. Consequently, the segments $A{M}^{\prime }$ and $L^{\ast }{M}^{\ast }$ are reflections of one another in the diameter $D^{\ast }{E}^{\ast }$, so $N^{\ast }=N^{\prime }$ lies on $L^{\ast }{M}^{\ast }$; that is, $A{N}^{\ast }$ is median in triangle $A{B}^{\ast }{C}^{\ast }$. Back to the original configuration, we conclude that $AN$ is indeed symmedian in the triangle $ABC$.