In a certain triangle, the angles formed by the angle bisectors are in the ratio $37: 41: 42$. What are the angles of the triangle?
The angles enclosed by the angle bisectors are given in the ratio $111^{\circ}, 123^{\circ}, 126^{\circ}$.

Thus,

$$
\frac{\alpha}{2}+\frac{\beta}{2}=69^{\circ}, \frac{\alpha}{2}+\frac{\gamma}{2}=57^{\circ}, \frac{\beta}{2}+\frac{\gamma}{2}=54^{\circ}
$$

from which

$$
\alpha=69^{\circ}, \beta=66^{\circ}, \gamma=42^{\circ}
$$

(József Kiss, Pápa.)

The problem was also solved by: L. Bánó, E. Freund, P. Füstös, Z. Harsányi, P. Heimlich, Gy. Jánosy, E. Makó, A. Paunz, Z. Reich, E. Sárközy, Gy. Schilhanek, Gy. Schlesinger, Gy. Schuster, J. Steiger, J. Szabó, B. Tóth.