58th Ukrainian National Mathematical Olympiad

Let $A{H}_1$, $B{H}_2$, $C{H}_3$ be the altitudes of $△ABC$ (Fig. 29), $M$ is the midpoint of $BC$, and let $O_1$ be the circumcenter of $△YHX$. We want to show that $△YHX\sim △ABC$.

Since $\angle B{H}_{3}C=\angle YDC=90^{\circ }$, $\angle ABC=\angle HYX$. Similarly, $\angle ACB=\angle HXY$. Therefore, $△YHX\sim △ABC$.

Let $H{H}^{\prime }$ be the altitude of $△YHX$, and let $M^{\prime }$ be the midpoint of $XY$.

Clearly, $O_1{M}^{\prime }\perp XY$, $OM\perp BC$. Thus, $H{H}^{\prime }$ and $A{H}_1$ are the corresponding elements in similar triangles. Similarly, $O_1{M}^{\prime }$ and $OM$ are the corresponding elements in similar triangles. Thus, $$\frac{H{H}^{\prime }}{A{H}_1}=\frac{O_1{M}^{\prime }}{OM}.$$ From the other side, it is clear that $△OMD\sim △A{H}_{1}D$. Hence, $$\frac{H_{1}D}{A{H}_1}=\frac{MD}{OM}.$$ These two equalities, together with $H{H}^{\prime }=H_{1}D$, since $H{H}_{1}D{H}^{\prime }$ is a rectangle, give $O_1{M}^{\prime }=MD$. Thus, $O_1{M}^{\prime }DM$ is a rectangle. Therefore, $O_{1}M\perp BC$, so $O_1$ belongs to the perpendicular bisector of $BC$. Hence, $O_{1}B=O_{1}C$.