The DANUBE Mathematical Competition

Let $A{A}^{\prime }$, $D{D}^{\prime }$ be altitudes in the triangle $△APD$ and let $B{B}^{\prime }$, $C{C}^{\prime }$ be altitudes in the triangle $△BPC$. We have $\angle XAC=\angle XBD$ and $\angle XCA=\angle XDB$, and thus $△XAC\sim △XBD$. We prove that the line $H_1{H}_2$ passes through the point $X$ if and only if the above triangles are congruent.



We consider the circles ${\omega }_1,{\omega }_2$ of diameter $AC$ and $BD$ respectively. The points $H_1,H_2$ belong to the radical axis of these two circles, because $A,A^{\prime }\in {\omega }_1,D,D^{\prime }\in {\omega }_2$

şi $P_{{\omega }_1}(H_1)=-A{H}_1\cdot A^{\prime }{H}_1=-D{H}_1\cdot D^{\prime }{H}_1=P_{{\omega }_2}(H_1)$; and similar for $H_2$. Thus, $X\in H_1{H}_2$ if and only if $P_{{\omega }_1}(X)=P_{{\omega }_2}(X)$. We have $P_{{\omega }_1}(X)=X{M}^2-M{C}^2$ and $P_{{\omega }_2}(X)=X{N}^2-N{D}^2$, were $M,N$ are the centers of the circles ${\omega }_1$ and ${\omega }_2$ respectively. We notice that $(X{M}^2-M{C}^2)/(X{N}^2-N{D}^2)$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$, or $X{M}^2-M{C}^2=X{N}^2-N{D}^2=0$. In the former case, the triangles $AXC$ and $BXD$ should be similar and right angles in $X$. But now the line $CD$ corresponds to the line $AB$ by a similarity of center $X$ and angle $90^{\circ }$, so $AB\perp CD$ – contradiction. Thus $(X{M}^2-M{C}^2)/(X{N}^2-N{D}^2)$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$. We obtain that $X\in H_1{H}_2$ if and only if the similarity ratio of the triangles $AXC$ and $BXD$ is 1, that is $\Delta AXC\equiv \Delta BXD$. But this is equivalent to $AC=BD$. In the same manner, $Y\in H_1{H}_2$ if and only if $AC=BD$, and we get the conclusion of the problem.