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Find the cosine of the angle at the base of an isosceles triangle if the point of intersection of its altitudes lies on the inscribed circle of the triangle.

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Let $O$ be the center of the circle inscribed in triangle $ABC (AC = BC)$, $H$ be the point of intersection of the altitudes, $\angle A = \angle B = 2\alpha$, $K$ be the midpoint of $AB$. Then $OK = AK \tan \alpha$, $HK = AK \tan(90^\circ - 2\alpha) = AK \cot 2\alpha$.

Since $HK = 2OK$, then $2 \tan \alpha = \cot 2\alpha$. Let $\tan \alpha = t$. Then $2t = 1 - t^2 / 2t$. From this, $t^2 = 1/5$. Therefore, $\cos 2\alpha = 1 - t^2 / 1 + t^2 = 2/3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7099908082d8b9f44e68g-50.jpg?height=297&width=464&top_left_y=1898&top_left_x=797)

## Answer