Let $A B C$ be an acute triangle with circumcenter $O$, incenter $I$, orthocenter $H$. If $O I=H I$, what are the possible values of the angles of triangle $A B C$ ?
Answer: this occurs if and only if some angle is 60 degrees.

One direction is immediate; if $\angle A=60^{\circ}$ then $B H O I C$ are cyclic since $\angle B H C=$ $\angle B I C=\angle B O C=120^{\circ}$.

For the other direction, note that we have an "SSA congruence" of triangles $A I H$ and $A I O$. Consequently, either $A$ lies on the circle $(O I H)$ or $\triangle A I H \cong \triangle A I O$. In latter case, $A H=A O$, but since it's known that $A H=2 A O \cos A$, it follows that $\cos A=\frac{1}{2} \Longrightarrow \angle A=60^{\circ}$.

Now it's impossible for $A, B, C \in(O I H)$ since the points $A, B, C, O$ are not concyclic. Thus some angle must be $60^{\circ}$.