8,9

The extensions of the altitudes of triangle $ABC$ divide the circumcircle of triangle $ABC$ into arcs whose lengths are in the ratio $p: q: r$. Find the angles of triangle $ABC$.
Let $\alpha, \beta$ and $\gamma$ be the angles at the vertices $A, B$ and $C$ of an acute triangle $ABC$ (Fig.1), $A_1, B_1$ and $C_1$ be the points of intersection of the extensions of the altitudes of triangle $ABC$ drawn from vertices $A, B$ and $C$ respectively, and $B_1 A C_1: A_1 B C_1: A_1 C B_1 = p: q: r$. Then

$$
\angle B_1 A_1 C_1 = \frac{1}{2} B_1 A C_1 = \frac{1}{2} \cdot \frac{2 \pi \varphi}{\rho + q + \varphi} = \frac{\pi \rho}{\rho + q + \varphi}
$$

On the other hand,

$$
\angle B_1 A_1 C_1 = \angle B_1 A_1 A + \angle C_1 A_1 A = \angle B_1 B A + \angle C_1 C A = \left(\frac{\pi}{2} - \alpha\right) + \left(\frac{\pi}{2} - \alpha\right) = \pi - 2 \alpha.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-12.jpg?height=80&width=1949&top_left_y=2052&top_left_x=0)
In triangle $ABC$, angle $BAC$ is obtuse (Fig. 2). Then

$$
\angle B_1 A_1 C_1 = \angle B_1 A_1 A + \angle C_1 A_1 A = \angle B_1 B A + \angle C_1 C A = \left(\alpha - \frac{\pi}{2}\right) + \left(\alpha - \frac{\pi}{2}\right) = 2 \alpha - \pi.
$$

From the equation $\frac{\pi p}{p + q + \varphi} = 2 \alpha - \pi$, we find that $\alpha = \frac{\pi}{2} \left(1 + \frac{p}{p + q + \varphi}\right)$. By the property of a cyclic quadrilateral,

$$
\angle A_1 B_1 C_1 = \pi - \frac{1}{2} A_1 C C_1 = \pi - \frac{\pi q}{-p + q + \varphi}.
$$

On the other hand,

$$
\angle A_1 B_1 C_1 = \pi - \angle A_1 C C_1 = \angle A_1 C B + \angle C_1 C B = \angle A_1 A B + \angle C_1 C B = \left(\frac{\pi}{2} - \beta\right) + \left(\frac{\pi}{2} - \beta\right) = \pi - 2 \beta.
$$

From the equation $\pi - \frac{\pi q}{p + q + \varphi} = \pi - 2 \beta$, we find that $\beta = \frac{\pi}{2} \cdot \frac{q}{p + q + \varphi}$. Similarly, $\gamma = \frac{\pi}{2} \cdot \frac{\varphi}{p + q + \varphi}$.

## Answer

$\frac{\pi}{2} \cdot \frac{q + \varphi}{p + q + \varphi}, \frac{\pi}{2} \cdot \frac{p + \varphi}{p + q + \varphi}, \frac{\pi}{2} \cdot \frac{p + q}{p + q + \varphi}$, if triangle $ABC$ is acute; $\frac{\pi}{2} \cdot \frac{q}{p + q + \varphi}, \frac{\pi}{2} \cdot \frac{\varphi}{p + q + \varphi}, \frac{\pi}{2} \left(1 + \frac{p}{p + q + \varphi}\right)$, if triangle $ABC$ is obtuse, with angle $BAC$ being obtuse.