What are the conditions for a triangle's two vertices, its orthocenter, and the center of its inscribed circle to lie on a circle?
If the triangle is right-angled and the right angle is between the two selected vertices, then the orthocenter also falls there, and the question is meaningless, as it only concerns 3 different points, not 4; if, however, the selected vertices are the endpoints of the hypotenuse, then the desired property does not hold, since the center of the inscribed circle is inside the circle determined by the first 3 points. Therefore, we exclude right-angled triangles from our investigation.

We can rephrase the question as follows: what is the condition for the circle $k_{m}$ determined by the vertices $A$ and $B$ of triangle $ABC$ and its orthocenter $M$ to pass through the center $O$ of the inscribed circle of the triangle, or, in other words, for $k_{m}$ to be identical to the circumcircle $k_{0}$ of triangle $ABO$.

One aspect that facilitates the answer is the known fact that the reflection of $M$ over each side of the triangle lies on the circumcircle $k$ of the triangle. This can also be stated as follows: with the vertices $A$ and $B$ of the inscribed triangle $ABC$ in $k$ fixed, and $C$ moving along one of the arcs $AB$ of $k$, $M$ moves along the reflection $k_{m}$ of $k$ over the line $AB$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_dac247744297b9b71f20g-1.jpg?height=534&width=569&top_left_y=595&top_left_x=759)

Figure 1

We will show, on the other hand, that with fixed $A$ and $B$, and $C$ moving as described above, $O$ moves on a circle $k_{0}$ (on the side of the line $AB$ containing $C$), whose center $F$ bisects the arc $AB$ of $k$ not containing $C$, and whose radius is $FA=FB$. Indeed, with the usual notation, the angle subtended by $AB$ from $O$ is

$$
\delta=180^{\circ}-\angle OAB-\angle OBA=180^{\circ}-\frac{\alpha+\beta}{2}=90^{\circ}+\frac{\gamma}{2},
$$

an obtuse angle, so $F$ is on the opposite side of the line $AB$, on the perpendicular bisector of $AB$, and the angle subtended by $AB$ from $F$ is twice the complement of $\delta$, which is $180^{\circ}-\gamma$. This is only true for the position of $F$ described above. The vertices $A$ and $B$ themselves, of course, cannot be $O$.

For the purpose of our problem, $k_{m}$ and $k_{0}$ have a (different from $A$ and $B$) common point only if $k_{0}$ is identical to $k_{m}$, their radii are $FA=FK=AK$ (where $K$ is the center of $k$), $FAK$ is an equilateral triangle, $\angle AKB=120^{\circ}$, $\angle ACB=60^{\circ}$.

Conversely, it is immediately clear that this condition is also sufficient for the 4 points under consideration to lie on a circle (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_dac247744297b9b71f20g-1.jpg?height=642&width=678&top_left_y=1820&top_left_x=710)

Figure 2

Juhász Júlia (Balassagyarmat, Balassi B. Gymnasium, 4th grade)