Let's draw the external angle bisectors of a triangle. These angle bisectors form a triangle whose angles are $\alpha_{1}, \beta_{1}, \gamma_{1}$. What are the angles of the original triangle?
I. Solution: The notation is shown in the figure. In $\triangle O_{1} B C$, $\angle B = 90^{\circ} - \frac{\beta}{2}$ and $\angle C = 90^{\circ} - \frac{\gamma}{2}$, because the external angle bisectors are perpendicular to the internal angle bisectors.

Thus,

$$
\alpha_{1} + \left(90^{\circ} - \frac{\beta}{2}\right) + \left(90^{\circ} - \frac{\gamma}{2}\right) = 180^{\circ}
$$

from which

$$
\alpha_{1} = \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ} - \frac{\alpha}{2}
$$

and thus

$$
\alpha = 180^{\circ} - 2 \alpha_{1} = \beta_{1} + \gamma_{1} - a_{1}
$$

Similarly,

$$
\beta = 180^{\circ} - 2 \beta_{1} = \alpha_{1} + \gamma_{1} - \beta_{1}
$$

and

$$
\gamma = 180^{\circ} - 2 \gamma_{1} = \alpha_{1} + \beta_{1} - \gamma_{1}
$$

Balatoni Ferenc (Bp. VII., Rákóczi közg. k. i. II. o. t.)

II. Solution: $O B O_{1} C$ is a cyclic quadrilateral because $\angle B = \angle C = 90^{\circ}$.

From $\triangle B O C$, $\angle O = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2}$, and from the cyclic quadrilateral, $\angle O = 180^{\circ} - \alpha_{1}$,

Thus, $\alpha_{1} = \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ} - \frac{\alpha}{2}$, etc., as in the first solution.

Biczó Géza (Bp. II., Rákóczi g. I. o. t.)

Note: Since the internal angle bisectors of the original $\triangle A B C$ are the altitudes of the triangle $\triangle O_{1} O_{2} O_{3}$ formed by the external angle bisectors, the problem can also be stated as: Given the angles of a triangle, what are the angles of the pedal triangle?