Berkeley Math Circle: Monthly Contest 7

Solution:

Answer: this occurs if and only if some angle is $60$ degrees.

One direction is immediate; if $\angle A = 60^{\circ}$ then $BHOIC$ are cyclic since $\angle BHC = \angle BIC = \angle BOC = 120^{\circ}$.

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

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