IMAR Mathematical Competition

Clearly, it is sufficient to prove that $AX$ is the $A$-symmedian of the triangle $ABC$. To this end, we show that the triangles $XAB$ and $XCA$ are similar. It then follows that the line $AX$ bisects the angle $BXC$, and $XB/XC=A{B}^2/A{C}^2$, so $AX$ is indeed the $A$-symmedian of the triangle $ABC$. To prove similarity, we first show that the quadrangle $ANXP$ is cyclic. By the preceding, $\angle DXP=180^{\circ }-\angle PBD$, and $\angle NXD=180^{\circ }-\angle DCN$, so $$\begin{array}{rl}\angle PXN& =360^{\circ }-\angle DXP-\angle NXD=360^{\circ }-(180^{\circ }-\angle PBD)-\\ & (180^{\circ }-\angle DCN)\\ & =\angle PBD+\angle DCN=\angle ABC+\angle BCA=180^{\circ }-\angle CAB\\ & =180^{\circ }-\angle NAP,\end{array}$$ showing that the quadrangle $ANXP$ is indeed cyclic.

Alternative Solution. As in the previous solution, we show that $AX$ is the A-symmedian of the triangle $ABC$. To this end, invert from $A$ with power $AB\cdot AC$. The images of $B,C,D,N,P$ and $X$ under this inversion are located as follows: (1) The image of $B$ is the point $B^{\prime }$ on the ray $AB$ emanating from $A$ such that $A{B}^{\prime }=AC$; similarly, the image of $C$ is the point $C^{\prime }$ on the ray $AC$ emanating from $A$ such that $A{C}^{\prime }=AB$, so the triangles $ABC$ and $A{C}^{\prime }{B}^{\prime }$ are reflexions of one another in their common internal A-bisectrix; (2) The image of $D$ is the antipode $D^{\prime }$ of $A$ in the circle $A{B}^{\prime }{C}^{\prime }$, since $$A{D}^{\prime }=\frac{AB\cdot AC}{AD}=\frac{AB\cdot AC\cdot BC}{2\cdot \text{area }ABC}=\frac{A{C}^{\prime }\cdot A{B}^{\prime }\cdot B^{\prime }{C}^{\prime }}{2\cdot \text{area }A{C}^{\prime }{B}^{\prime }}$$ which is the diameter of the circle $A{C}^{\prime }{B}^{\prime }$; (3) The images $N^{\prime }$ and $P^{\prime }$ of $N$ and $P$, respectively, are the reflexions of $A$ across $C^{\prime }$ and $B^{\prime }$, respectively; and (4) Since the angles $A{B}^{\prime }{D}^{\prime }$ and $A{C}^{\prime }{D}^{\prime }$ are both right, by (2), and $B^{\prime }$ and $C^{\prime }$ are the midpoints of the segments $A{P}^{\prime }$ and $A{N}^{\prime }$, respectively, by (3), $D^{\prime }$ is the centre of the circle $A{N}^{\prime }{P}^{\prime }$, so the image of $X$ is the midpoint $X^{\prime }$ of the segment $N^{\prime }{P}^{\prime }$. Finally, since $B^{\prime }{C}^{\prime }$ and $N^{\prime }{P}^{\prime }$ are parallel, the line $AX{X}^{\prime }$ also bisects the segment $B^{\prime }{C}^{\prime }$, and since the triangles $ABC$ and $A{C}^{\prime }{B}^{\prime }$ are reflexions of one another in their common internal A-bisectrix, it is indeed the A-symmedian of the triangle $ABC$.