In an acute triangle $A B C$ with angle $C$ equal to $30^{\circ}$, the altitudes intersect at point $M$. Find the area of triangle $A M B$, if the distances from the center of the circle circumscribed around triangle $A B C$ to sides $B C$ and $A C$ are $\sqrt{2}$ and $\frac{\sqrt{3}}{3}$, respectively.
Let $O$ be the center of the circle circumscribed around triangle $ABC$, and $P$ be the projection of point $O$ onto $BC$. Prove that $AM = 2OP$.

## Solution

Let $O$ be the center of the circle circumscribed around triangle $ABC$, and $P$ and $Q$ be the projections of point $O$ onto sides $BC$ and $AC$. It is known that the distance from the orthocenter of the triangle to its vertex is twice the distance from the center of the circumscribed circle to the opposite side of the triangle. Therefore,

$$
BM = 2OQ = \frac{2\sqrt{3}}{3}, \quad AM = 2OP = 2\sqrt{2}
$$

Since $\angle AMB = 180^\circ - 30^\circ = 150^\circ$, then

$$
S_{\triangle AMB} = \frac{1}{2} AM \cdot MB \sin 150^\circ = \frac{\sqrt{6}}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-22.jpg?height=466&width=531&top_left_y=1099&top_left_x=772)

## Answer

$\frac{\sqrt{6}}{3}$