Let's draw the external angle bisectors of a triangle. These angle bisectors form a triangle whose angles are: $52^{\circ}, 61^{\circ}$, and $67^{\circ}$. What are the angles of the original triangle?
The external angle bisectors of triangle $ABC$ determine the triangle $A_{1} B_{1} C_{1}$. If the angles of triangle $ABC$ are $\alpha, \beta$, and $\gamma$, then

$$
A_{1} B C \varangle=90^{\circ}-\frac{\beta}{2}, \quad A_{1} C B \varangle=90^{\circ}-\frac{\gamma}{2},
$$

thus in the quadrilateral $A B A_{1} C$

$$
\alpha+\left(\beta+90^{\circ}-\frac{\beta}{2}\right)+52^{\circ}+\left(90^{\circ}-\frac{\gamma}{2}+\gamma\right)=360^{\circ}
$$

or

$$
\begin{gathered}
\alpha+\frac{\beta}{2}+\frac{\gamma}{2}=\alpha+90^{\circ}-\frac{\alpha}{2}=128^{\circ} \\
\alpha=76^{\circ}
\end{gathered}
$$

Similarly, we find that

$$
\beta=58^{\circ} \quad \text { and } \quad \gamma=46^{\circ} \text {. }
$$

(Bauer Elemér, Budapest.)

The problem was also solved by: Dénes M., Döri V., Fried E., Engler J., Erdős V., Koffler B., Köhler I., Lendvai D., Lengyel P., Neumann L., Rosenthal M., Schnabel L., Sichermann F., Spitzer L., Steiner L., Stolzer I., Szántó L., Szilárd V., Vilcsek A., Viola R.