A circle is divided into three arcs by the points $X, Y$, and $Z$, with the central angles corresponding to these arcs being $60^{\circ}, 100^{\circ}$, and $200^{\circ}$.

If $A, B$, and $C$ are the vertices of a triangle, let $M_{A}$ and $M_{B}$ denote the intersection points of the altitudes from $A$ and $B$ with the circumcircle of the triangle, and $F_{C}$ denote the intersection point of the angle bisector at $C$ with the circumcircle.

Determine all acute triangles $ABC$ for which the points $M_{A}, M_{B}$, and $F_{C}$ coincide with the points $X, Y$, and $Z$ in some order.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_2d987390b62ca0db3bf1g-1.jpg?height=578&width=596&top_left_y=218&top_left_x=754)

Solution. The triangle $ABC$ is acute-angled, so the orthocenter and the circumcenter $O$ of the circumcircle are inside the triangle. We apply the theorem of inscribed and central angles:

$$
A O F_{C} \angle=2 \cdot A C F_{C} \angle=2 \cdot \frac{\gamma}{2}=\gamma
$$

Since $M_{B} B \perp A C$, we have $M_{B} B A \angle=90^{\circ}-\alpha$. Again, by the theorem of inscribed and central angles:

$$
\begin{aligned}
& M_{B} O A \angle=2 \cdot M_{B} B A \angle= \\
& =2 \cdot\left(90^{\circ}-\alpha\right)=180^{\circ}-2 \alpha
\end{aligned}
$$

Thus, $M_{B} O F_{C} \angle=\gamma+180^{\circ}-2 \alpha$. Similarly, we find that $M_{A} O F_{C} \angle=\gamma+180^{\circ}-2 \beta$. From these,

$$
\begin{gathered}
M_{A} O M_{B} \angle=360^{\circ}-M_{B} O F_{C} \angle-M_{A} O F_{C} \angle= \\
=360^{\circ}-\left(\gamma+180^{\circ}-2 \alpha\right)-\left(\gamma+180^{\circ}-2 \beta\right)=2 \alpha+2 \beta-2 \gamma=360^{\circ}-4 \gamma
\end{gathered}
$$

(using that $\alpha+\beta+\gamma=180^{\circ}$).

Thus, the three given angles expressed in terms of the angles of the triangle are: $360^{\circ}-4 \gamma, 180^{\circ}+\gamma-2 \alpha, 180^{\circ}+\gamma-2 \beta$. Since in the problem $\alpha$ and $\beta$ can be interchanged, we can distinguish three main cases:

1. $360^{\circ}-4 \gamma=60^{\circ}$, hence $\gamma=75^{\circ}$,

$180^{\circ}+75^{\circ}-2 \alpha=100^{\circ} \Rightarrow \alpha=77.5^{\circ}$

$180^{\circ}+75^{\circ}-2 \beta=200^{\circ} \Rightarrow \beta=27.5^{\circ}$.
2. $360^{\circ}-4 \gamma=100^{\circ}$, hence $\gamma=65^{\circ}$,

$180^{\circ}+65^{\circ}-2 \alpha=60^{\circ} \quad \Rightarrow \alpha=92.5^{\circ}>90^{\circ}$: the triangle is not acute-angled

$\left(180^{\circ}+65^{\circ}-2 \beta=200^{\circ} \Rightarrow \beta=22.5^{\circ}\right)$.
3. $360^{\circ}-4 \gamma=200^{\circ}$, hence $\gamma=40^{\circ}$,

$180^{\circ}+40^{\circ}-2 \alpha=60^{\circ} \Rightarrow \alpha=80^{\circ}$

$180^{\circ}+40^{\circ}-2 \beta=100^{\circ} \Rightarrow \beta=60^{\circ}$.

Therefore, the problem has two solutions among acute-angled triangles: $\gamma=75^{\circ}$, with $\alpha$ and $\beta$ being $77.5^{\circ}$ and $27.5^{\circ}$ respectively; or $\gamma=40^{\circ}$, with $\alpha$ and $\beta$ being $60^{\circ}$ and $80^{\circ}$ respectively.