Medians $A M$ and $C N$ of triangle $A B C$ intersect at point $O$. It is known that $\angle B A C=\alpha, \angle B C A=\beta, A C=b$. Find the distance from point $O$ to line $A C$.
The desired distance is one third of the height of triangle $ABC$, drawn from vertex $B$.

## Solution

Let $L$ and $H$ be the projections of points $O$ and $B$ respectively onto the line $AC$, and $K$ be the midpoint of $AC$.

From the right triangles $ABH$ and $CBH$, we find that $AH = BH \cot \alpha$ and $CH = BH \cot \beta$. Since

$$
b = AC = AH + CH = BH \cot \alpha + BH \cot \beta = BH (\cot \alpha + \cot \beta),
$$

then $BH = \frac{b}{\cot \alpha + \cot \beta}$.

Since $BK$ is also the median of triangle $ABC$, point $O$ lies on segment $BK$ and divides it in the ratio 2:1, counting from point $B$.

From the similarity of triangles $OLK$ and $BHK$, it follows that

$$
OL = BH \cdot \frac{OK}{BK} = \frac{1}{3} BH = \frac{1}{3} \cdot \frac{b}{\cot \alpha + \cot \beta} = \frac{b \sin \alpha \sin \beta}{3 \sin (\alpha + \beta)}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-34.jpg?height=403&width=652&top_left_y=1565&top_left_x=703)

Answer

$b \sin \alpha \sin \beta$

$\overline{3 \sin (\alpha+\beta)}$.

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