The point symmetric to the orthocenter of a triangle with respect to a side lies on the circumcircle of the triangle. Triangle $B M N$ is similar to triangle $B A C$, and the similarity coefficient is $\cos \angle A B C$.

## Solution

Let $H$ be the intersection point of the altitudes. It is known that the point symmetric to the orthocenter of a triangle with respect to a side lies on the circumcircle of the triangle. Therefore, $H M = M P$ and $H N = N Q$. Thus, $M N$ is the midline of triangle $H Q P$. Consequently,

$$
M N = \frac{1}{2} P Q = \frac{3 a}{5}
$$

Triangle $B M N$ is similar to triangle $B A C$, and the similarity coefficient is $\cos \angle A B C$. Therefore,

$$
\cos \angle A B C = \frac{M N}{A C} = \frac{\frac{3}{5} a}{a} = \frac{3}{5}, \sin \angle A B C = \frac{4}{5}
$$

If $R$ is the desired radius, then

$$
R = \frac{A C}{2 \sin \angle A B C} = \frac{a}{2 \cdot \frac{4}{5}} = \frac{5 a}{8}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-01.jpg?height=446&width=489&top_left_y=1646&top_left_x=790)

## Answer

$\frac{5 a}{8}$

Problem
[ $\left[\begin{array}{l}\text { Orthocenter and orthic triangle } \\ {[\text { Law of Sines }}\end{array}\right]$

The extensions of the altitudes $A M$ and $C N$ of an acute triangle $A B C$ intersect the circumcircle of the triangle at points $P$ and $Q$. Find the radius of the circumcircle if $A C=a, P Q=\frac{6 a}{5}$.