[ Law of Cosines ]

In triangle $ABC$, it is known that $AB=2, AC=5, BC=6$. Find the distance from vertex $B$ to the point of intersection of the altitudes.

#
The distance $B H$ from vertex $B$ to point $H$, the intersection of the altitudes of triangle $A B C$, is equal to twice the distance from the center $O$ of the circumscribed circle to side $A C$.

## Solution

By the cosine theorem, we find that

$$
\cos \angle B=\frac{A B^{2}+B C^{2}-A C^{2}}{2 A B \cdot B C}=\frac{4+36-25}{2 \cdot 2 \cdot 6}=\frac{5}{8}
$$

$$
\sin \angle B=\sqrt{1-\cos ^{2} \angle B}=\sqrt{1-\left(\frac{5}{8}\right)^{2}}=\frac{\sqrt{39}}{8}
$$

Let $O$ be the center of the circumscribed circle of triangle $A B C$, $R$ be its radius, and $M$ be the projection of the center onto side $A C$. Then $B H=2 O M$. By the sine theorem,

$$
A O=R=\frac{A C}{2 \sin \angle B}=\frac{5}{2 \cdot \frac{\sqrt{39}}{8}}=\frac{20}{\sqrt{39}}
$$

Therefore,

$$
\begin{aligned}
A H=2 O M= & \sqrt{A O^{2}-A M^{2}}=2 \sqrt{A O^{2}-\frac{A C^{2}}{4}}= \\
& =2 \sqrt{\frac{400}{39}-\frac{25}{4}}=\frac{25}{\sqrt{39}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-09.jpg?height=343&width=604&top_left_y=1009&top_left_x=344)

Fig. 2.

Answer

$\frac{25}{\sqrt{39}}$.