63rd Czech and Slovak Mathematical Olympiad

Assume first that $|\angle ABC|+|\angle ACM|=90^{\circ }$. Using the notation $\varphi =|\angle ACM|$ and $\psi =|\angle BCM|$ (Fig. 1), we conclude from our assumption that $|\angle ABC|=90^{\circ }-\varphi$, and hence $|\angle BAC|=90^{\circ }-\psi$ as well, because of an easy angle computation in $△ABC$: $$\begin{array}{rl}|\angle BAC|& =180^{\circ }-|\angle ABC|-|\angle ACB|\\ & =180^{\circ }-(90^{\circ }-\varphi )-(\varphi +\psi )=90^{\circ }-\psi .\end{array}$$ Fig. 1

Applying Law of Sines to $△ACM$ and $△BCM$, we get $$\frac{\sin (90^{\circ }-\psi )}{\sin \varphi }=\frac{|CM|}{|AM|}=\frac{|CM|}{|BM|}=\frac{\sin (90^{\circ }-\varphi )}{\sin \psi }.$$ Comparing the two ratios of sines and using the formula $\sin (90^{\circ }-\omega )=\cos \omega$, we obtain an equality $\sin \varphi \cos \varphi =\sin \psi \cos \psi$ or $\sin 2\varphi =\sin 2\psi$. Since the angles $\varphi$ and $\psi$ are acute, both $2\varphi$ and $2\psi$ are between $0^{\circ }$ and $180^{\circ }$. Thus by a well known sine property, the equality $\sin 2\varphi =\sin 2\psi$ means that either $2\varphi =2\psi$ or $2\varphi +2\psi =180^{\circ }$. In the first case (when $\varphi =\psi$), the interior angle of $△ABC$ at the vertices $A$ and $B$ are equal, in the second case (when $\varphi +\psi =90^{\circ }$) the interior angle at the vertex $C$ is right. This completes the proof of one of the two implications stated in the problem.

Fig. 1

To prove the second (converse) implication, let us assume that (i) $|AC|=|BC|$ or (ii) $|\angle ACB|=90^{\circ }$.

Case (i). It follows from $|AC|=|BC|$ that the triangles $ACM$ and $BCM$ are congruent (by SSS theorem), with right interior angles at the vertex $M$. Consequently,

Applying Law of Sines to $△ACM$ and $△BCM$, we get $$\frac{\sin (90^{\circ }-\psi )}{\sin \varphi }=\frac{|CM|}{|AM|}=\frac{|CM|}{|BM|}=\frac{\sin (90^{\circ }-\varphi )}{\sin \psi }.$$ The converse implication is proven.

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Alternative solution.

Let $k$ be the circumcircle of the given triangle $ABC$. Its median $CM$ can be extended to the chord $C{C}^{\prime }$ of the circle $k$ (Fig. 2). Since the inscribed angles $AB{C}^{\prime }$ and $AC{C}^{\prime }$ (or $ACM$) are congruent, the considered sum of angles $ABC$ and $ACM$ is equal to the angle $CB{C}^{\prime }$. By Thales' theorem the last angle $CB{C}^{\prime }$ is right if and only if the chord $C{C}^{\prime }$ is a diameter of the circle $k$. This happens if and only if the centre $S$ of $k$ lies on the ray $CM$. For such a situation, we distinguish two cases: $S=M$ and $S\ne M$. Note that $S=M$ holds if and only if the angle $ACB$ is right (by Thales' theorem again). Thus let us analyse the second case $S\ne M$: The three distinct points $C,M$ and $S$ are obviously collinear if and only if the line $MS$, a perpendicular bisector of a segment $AB$, passes through the point $C$. However, the last condition is equivalent to the desired equality $|AC|=|BC|$. This completes the proof (common for the both composing implications).

Fig. 2

Remark. Instead of the chord $C{C}^{\prime }$ of the circumcircle $k$, it is possible to consider the tangent line $t$ to the circle $k$ at its point $C$ (Fig. 3). Since the inscribed angle $ABC$ is always congruent to the marked angle between $AC$ and $t$, the sum of the angles $ABC$ and $ACM$ equals $90^{\circ }$ if and only if the tangent $t$ is perpendicular to the ray $CM$. The last is equivalent to the condition from the above solution, namely that the ray $CM$ passes through the centre $S$ of $k$.