The distance from the orthocenter of triangle $ABC$ to vertex $C$ is equal to the radius of the circumscribed circle. Find the angle $ACB$.

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Let $H$ be the orthocenter of triangle $ABC$, $C_{1}$ the midpoint of $AB$, $O$ the center of the circumscribed circle, and $R$ its radius. Since $CH=2OC_{1}$, then

$$
OC_{1}=\frac{1}{2}CH=\frac{1}{2}R=\frac{1}{2}OA
$$

Therefore,

$$
\angle AOC_{1}=60^{\circ}, \angle AOB=2\angle AOC_{1}=120^{\circ}.
$$

If points $C$ and $O$ lie on the same side of line $AB$, then

$$
\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2} \cdot 120^{\circ}=60^{\circ}
$$

If points $C$ and $O$ lie on opposite sides of line $AB$, then

$$
\angle ACB=\frac{1}{2}(360^{\circ}-\angle AOB)=180^{\circ}-60^{\circ}=120^{\circ}.
$$

The distance from the orthocenter to a vertex of the triangle is twice the distance from the center of the circumscribed circle to the opposite side.

![](https://cdn.mathpix.com/cropped/2024_05_06_7b30f89ee421e63350f8g-13.jpg?height=429&width=460&top_left_y=1829&top_left_x=799)

Answer

$60^{\circ}$ or $120^{\circ}$.