8,9 $[\quad$ The Law of Sines

In triangle $ABC$, altitudes $BM$ and $CN$ are drawn, and $O$ is the center of the inscribed circle. It is known that $BC=24$, $MN=12$.

Find the radius of the circumscribed circle of triangle $BOC$.

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Points $M$ and $N$ lie on the circle with diameter $B C$.

Let angle $A$ be acute (see the left figure). Triangle $A M N$ is similar to triangle $A B C$ with a similarity coefficient of $\cos \angle A = M N / B C = 1 / 2$. Therefore, $\angle A = 60^{\circ}$, and $\angle B O C = 120^{\circ}$ (see problem 55448).

By the Law of Sines, the radius of the circumcircle of triangle $B O C$ is

$\frac{B C}{2 \sin \angle B O C} = \frac{24}{2 \sin 120^{\circ}} = 8 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-30.jpg?height=454&width=400&top_left_y=2031&top_left_x=515)

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-30.jpg?height=454&width=415&top_left_y=2031&top_left_x=1119)

Now, let angle $A$ be obtuse (see the right figure). Then $\cos \angle C A N = 1 / 2$. Therefore, $\angle C A N = 60^{\circ}$, $\angle A = 120^{\circ}$, and $\angle B O C = 150^{\circ}$.

Thus, the radius of the circumcircle of triangle $B O C$ is $\frac{2 t}{2 \text{ sin } 150^{\circ}} = 24$.

## Answer

$8 \sqrt{3}$ or 24.