The distance from the vertex of a triangle to the point of intersection of the altitudes is twice the distance from the center of the circumscribed circle to the side opposite the given vertex.

## Solution

From the right triangles $A H M$ and $A H N$, we find that

$$
\begin{gathered}
H M^{2}=A M^{2}-A H^{2}=m^{2}-h^{2}, H N=\frac{1}{2} H M=\frac{\sqrt{m^{2}-h^{2}}}{2} \\
A N^{2}=A H^{2}+H N^{2}=h^{2}+\frac{m^{2}-h^{2}}{4}=\frac{m^{2}+3 h^{2}}{4}
\end{gathered}
$$

Let $O$ be the center of the circle circumscribed around triangle $A B C$, and $K$ be the point of intersection of line $A N$ with the circumscribed circle, different from $A$. Since $A K$ is the angle bisector of $\angle B A C$, then $\cup B K=\cup K C$. Therefore, $O K$ is the angle bisector of $\angle B O C$ of the isosceles triangle $B O C$. Consequently, point $M$ lies on $O K$. Therefore, $K M \perp B C$, and since $N H=N M$, it follows from the equality of the right triangles $A H N$ and $K M N$ that $A N=N K$. Therefore, $O N$ is perpendicular to the chord $A K$.

In the right triangle $O N K$,

$$
N K^{2}=O K \cdot K M, \text{ or } \frac{m^{2}+3 h^{2}}{4}=(h+O M) h .
$$

From this, we find that

$$
O M=\frac{m^{2}-h^{2}}{4 h}
$$

Let $F$ be the point of intersection of the altitudes of triangle $A B C$. Then

$$
A F=2 O M=\frac{m^{2}-h^{2}}{2 h}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-27.jpg?height=426&width=504&top_left_y=702&top_left_x=777)

Answer

$\frac{m^{2}-h^{2}}{2 h}$.
[ Auxiliary circle

In triangle $A B C$, the altitude $A H$, equal to $h$, the median $A M$, equal to $m$, and the angle bisector $A N$ are drawn. Point $N$ is the midpoint of segment $M H$. Find the distance from vertex $A$ to the orthocenter of triangle $A B C$.

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