Let $H$ be the point of intersection of the altitudes of triangle $ABC$, and $O$ be the center of the circumcircle. Then $\angle AHC = 180^{\circ} - \angle B$, $\angle AOC = 2 \angle B$.

## Solution

If $H$ is the point of intersection of the altitudes of triangle $ABC$, then $\angle AHC = 180^{\circ} - \angle ABC$. Therefore,

$$
\angle AOC = \cup AHC = 360^{\circ} - 2 \angle AHC = 2 \angle ABC
$$

where $O$ is the center of the circumcircle.

Since $\angle AOC + \angle ABC = 180^{\circ}$, then $3 \angle ABC = 180^{\circ}$. Therefore, $\angle ABC = 60^{\circ}$. Then

$$
AC = 2R \sin \angle ABC = 2 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-44.jpg?height=440&width=680&top_left_y=0&top_left_x=695)

## Answer
[ Law of Sines

The radius of the circumcircle of an acute triangle $ABC$ is 1. It is known that the center of another circle, passing through the vertices $A, C$ and the orthocenter of triangle $ABC$, lies on this circumcircle. Find $AC$.

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