Baltic Way 2024 · Problem 11

Solution: Denote the circumradius of $A B C D$ by $r$ and the circumcircle of triangle $B O D$ by $\omega$. Let $T=A C \cap B D$, let $O T$ meet $\omega$ again at $S$, and let $O E$ be a diameter of $\omega$ (Fig. 11). We see that $A C \| O E$ as $A C \perp B D$ and $B D \perp O E$. Furthermore, note that

$$\measuredangle D S T=\measuredangle D S O=\measuredangle D B O=\measuredangle O D B=\measuredangle O D T$$

so $O D$ is tangent to the circumcircle of the triangle $D S T$ and thus $O T \cdot O S=O D^{2}=r^{2}$.

Figure 11

We find $\angle A X O=90^{\circ}=\angle O X E$ so $A, X, E$ are collinear. Since also $\angle A M O=90^{\circ}$, points $A$, $O, X, M$ are concyclic. Next, we can see that points $A, X, T, S$ are concyclic since $\measuredangle X A T=$ $\measuredangle E A C=\measuredangle A E O=\measuredangle X E O=\measuredangle X S O=\measuredangle X S T$. Moreover, $A O$ is tangent to this circle as $O T \cdot O S=r^{2}=O A^{2}$. Hence $\measuredangle M T X=\measuredangle A T X=\measuredangle O A X=\measuredangle O M X$, so $O M$ is tangent to the circumcircle of triangle $X M T$ at $M$. By interchanging the roles of $A$ and $C$ and the roles of $X$ and $Y$, we can similarly prove that $O M$ is also tangent to the circumcircle of triangle $Y M T$ at $M$. But then these two circles must coincide. Now $O M \perp A C$ implies that $M T$ is a diameter of this one circle, and $A C \perp B D$ implies that $B D$ is tangent to it. The desired result follows.