Selected Problems from the Final Round of National Olympiad

Let $\angle CAB=\alpha$ and $\angle CBA=\beta$ (see Fig. 13), and let $r$ and $s$ be the circumradii of the triangles $ACM$ and $BCM$, respectively. By the sine law in the triangle $ACM$ we obtain $\frac{|CM|}{\sin \alpha }=2r$, reducing to $r=\frac{|CM|}{2\sin \alpha }$. Analogously, $s=\frac{|CM|}{2\sin \beta }$. Let $U$ and $V$ be the feet of altitudes drawn from the point $M$ in triangles $ACM$ and $BCM$, respectively. Fig. 13 Then $|MU|=|AM|\sin \alpha =\frac{|AB|}{2}\sin \alpha$. Analogously, $|MV|=\frac{|AB|}{2}\sin \beta$. Thus $r\cdot |MU|=\frac{|CM|\cdot |AB|}{4}=s\cdot |MV|$.

Fig. 13

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Alternative solution.

Let $x$ and $y$ be the altitudes drawn from $M$ in the triangles $ACM$ and $BCM$, respectively. Let $r$ and $s$ be the circumradii of these triangles, respectively. The areas of triangles $ACM$ and $BCM$ are equal because of $|AM|=|BM|$ and the common altitude drawn from $C$. Therefore $|AC|\cdot x=|BC|\cdot y$. Denote $\angle AMC=\gamma$. The sine law gives $|AC|=2r\sin \gamma$ and $|BC|=2s\sin (180^{\circ }-\gamma )=2s\sin \gamma$. Hence $2r\sin \gamma \cdot x=2s\sin \gamma \cdot y$. As $\gamma \ne 0$, this implies $rx=sy$.

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Alternative solution.

The formulas $S=\frac{abc}{4R}$ and $S=\frac{ah}{2}$, where $a,b,c$ are the side lengths, $R$ is the circumradius and $h$ is the altitude corresponding to $a$, together give $Rh=\frac{bc}{2}$. The product of the circumradius of $ACM$ and the altitude drawn from $M$ is thus $\frac{|AM|\cdot |CM|}{2}$. Analogously, $\frac{|BM|\cdot |CM|}{2}$ for triangle $BCM$. These two products are equal, since $|AM|=|BM|$.