A triangle's two vertices, the center of its inscribed circle, and its orthocenter lie on a circle. Calculate the angle at the third vertex of the triangle!
Let the center of the inscribed circle of the triangle $ABC$ be $O$, its orthocenter be $M$, and let the circle $k_0$ in question pass through the vertices $A$ and $B$. (The points $O, M, A, B$ are, of course, considered distinct, for otherwise we would be relying on the trivial assumption that three points lie on a circle.)

It is known that for any triangle $ABC$, the circle determined by the points $A, B, O$ - in our case, this is $k_0$ - has its center $D$ on the arc of the circumcircle $k$ of the triangle that does not contain $C$ (and, of course, bisects this arc), see Gy. 1762 (1978, No. 8-9, p. 143).

Furthermore, it is known that for any triangle, the circumcircle $k$ passes through the reflection of $M$ over the line $AB$, denoted as $M_c$. Therefore, in our case, $k_0$ is the reflection of $k$ over the line $AB$. Thus, their radii are equal, so $k_0$ passes through the center $K$ of $k$, and $K$ is on the same side of $AB$ as $C$, and thus as $O$. Consequently, the triangle $DKA$ is equilateral, and $\angle AKB = 120^\circ$. From this, it follows that the angle $\angle ACB$ is half of that, i.e., $60^\circ$.

The points $A, B, O$ are distinct in any case, but $M$ can exceptionally coincide with one of them. If $M$ coincides with $O$, we reach our goal more quickly: the line $CO$ plays a double role: it is both an angle bisector and an altitude, so $CA = CB$, and this holds for $AO$ and $BO$ as well, making $ABC$ an equilateral triangle, $\gamma = 60^\circ$.

If, however, $M$ coincides with $A$ or $B$, then there is a right angle at that vertex of the triangle, and $A, B, M$ do not determine a circle, and the angle at the third vertex can be any acute angle. However, as we have noted, the assumption only makes sense if it states that four distinct points lie on a circle.

Remarks. 1. Some examined the question in two cases, depending on whether $M$ and $O$ are on the same side of the line $AB$ or not. This caution, which is otherwise correct, would be unnecessary in the above reasoning; it would be appropriate if (instead of reflecting $A$ over $M$) the angle $\angle AMB$ - which is indeed either equal to $\angle AOB$ or its supplement to $180^\circ$ - were calculated using the altitudes.

2. Some mistakenly did not consider the first sentence of the problem as an assumption but as a theorem valid for all triangles. How, however, could it be calculated? The use of "any" in the sense of "all" or "every" - as if it were an exclamation - may occur in literary texts, but it is not customary in mathematical texts. There are several other precise mathematical formulations in which the assumption is not stated in a subordinate clause beginning with "if".