72nd Czech and Slovak Mathematical Olympiad

Let us denote $M$ the midpoint of $AB$, $N$ the midpoint of $AC$, $K$ and $L$ the intersections of the angle $CAB$ bisector with perpendicular bisectors of $AB$ and $AC$, respectively. The intersection of perpendicular bisectors of $AB$ and $AC$ is denoted by $O$. The triangle $KLO$ is therefore the triangle from the problem statement. Its orthocenter we denote by $H$. All these points are marked in the figure 1 for the case $|AB|<|AC|$. (The case $|AB|>|AC|$ looks analogously, the case $|AB|=|AC|$ is excluded by the specification—then points $K,L,O$ merge into one point.)



We have to prove that $H$ lies on the median from the vertex $A$ of triangle $ABC$. It is sufficient to show that triangles $ABH$ and $ACH$ have the same area.

Since $HL\perp OK\perp AB$ we have $HL\parallel AB$. Therefore $H$ and $L$ have the same distance from the line $AB$. However, it is equal to the length of the segment $LN$, since the point $L$ lies on the angle bisector of $CAB$ and $N$ is the perpendicular projection of $L$ onto $AC$. Hence, we get that the area of $ABH$ is equal to $\frac{1}{2}|AB|\cdot |LN|$. Analogously, we deduce that area of $ACH$ is equal to $\frac{1}{2}|AC|\cdot |KM|$. It remains to prove $|AB|\cdot |LN|=|AC|\cdot |KM|$.

For the points $K$ and $L$ lying on the angle bisector of $CAB$ we have $|\angle MAK|=|\angle NAL|$. The rectangular triangles $AKM$ and $ALN$ are therefore similar, and therefore $|KM|:|AM|=|LN|:|AN|$. Hence with respect to $|AM|=\frac{1}{2}|AB|$ and $|AN|=\frac{1}{2}|AC|$ we get $|KM|:|AB|=|LN|:|AC|$, i.e. $|AB|\cdot |LN|=|AC|\cdot |KM|$ as we needed to prove.

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Alternative solution.

In addition to the points from the first solution, we also consider the intersection $E$ of lines $AB$ and $KH$ and intersection $F$ of lines $AC$ and $LH$. Again we observe that $KH\parallel AC$ and $LH\parallel AB$, so the quadrilateral $AEHF$ is a parallelogram (see fig. 2).



We use again similarity of triangles $AKM$ and $ANL$ and deduce $|KM|:|LN|=|AM|:|AN|=|AB|:|AC|$. Equality of exterior angles at vertices $E$, $F$ of the parallelogram $AEHF$ yields $|\angle KEM|=|\angle LFN|$. Thus, the rectangular triangles $EKM$ and $FLN$ are also similar, whence it follows $|EM|:|FN|=|KM|:|LN|=|AB|:|AC|$. Hence, $$\frac{|AE|}{|AF|}=\frac{|AM|-|EM|}{|AN|-|FN|}=\frac{\frac{|AB|}{|AC|}\cdot |AN|-\frac{|AB|}{|AC|}\cdot |FN|}{|AN|-|FN|}=\frac{|AB|}{|AC|},$$ and triangles $AEF$ and $ABC$ are similar by the SAS condition (they coincide in the angle at the vertex $A$ and in the ratio of the adjacent sides). We thus obtain the decisive relation $EF\parallel BC$.

Since in the parallelogram $AEHF$ the line $AH$ bisects the diagonal $EF$, this line also bisects the segment $BC$, which is homothetic with $EF$ with center $A$. In other words, $H$ lies on the median from $A$ of the triangle $ABC$, as we had to prove.

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Alternative solution.

We consider points $K,L,M,N,O,H,E,F$ from the second solution. As there, we conclude that $AEHF$ is a parallelogram, and therefore the midpoint of the segment $EF$ lies on the ray $AH$. If we prove $EF\parallel BC$ then the midpoint $P$ of $BC$ also lies on the ray $AH$.



Due to the construction from the problem statement and the parallelogram $AEHF$, the six angles $KAE,KAF,KBA,LCA,AKE$ and $ALF$ marked in fig. 3 are congruent. According to the theorem $AA$, we have $△ABK\sim △ACL$ and $△AKE\sim △ALF$. According to the first similarity, $|AB|:|AC|=|AK|:|AL|$, and by the second similarity $|AK|:|AL|=|AE|:|AF|$. Together we get $|AB|:|AC|=|AE|:|AF|$, so by the SAS theorem, $△AEF\sim △ABC$ holds, and hence $EF\parallel BC$, as we promised to prove.