In triangle $ABC$, we draw the altitude $CC_{1}$, as well as the angle bisectors of angles $A$ and $B$. These angle bisectors intersect each other at $O$, and the altitude at points $E$ and $F$. What is the area of triangle $EFO$, if we know the altitude $CC_{1}=m$ and the angles of the original triangle.
Given that $\angle OEF = 90^\circ - \frac{\angle A}{2}$ and $\angle OFE = 90^\circ - \frac{\angle B}{2}$, the area in question is

$$
t = \frac{\overline{EF}^2 \cos \frac{A}{2} \cos \frac{B}{2}}{2 \sin \frac{A+B}{2}} = \frac{\overline{EF}^2 \cos \frac{A}{2} \cos \frac{B}{2}}{2 \cos \frac{C}{2}}
$$

From the right triangles $BC_1F$ and $AC_1E$,

$$
FC_1 = BC_1 \tan \frac{B}{2}
$$

and

$$
EC_1 = AC_1 \tan \frac{A}{2}
$$

thus

$$
EF = FC_1 - EC_1 = BC_1 \tan \frac{B}{2} - AC_1 \tan \frac{A}{2}
$$

But

$$
BC_1 = m \cot B \text{ and } AC_1 = m \cot A
$$

so

$$
EF = m \left( \cot B \tan \frac{B}{2} - \cot A \tan \frac{A}{2} \right) = \frac{m}{2} \left( \tan^2 \frac{A}{2} - \tan^2 \frac{B}{2} \right)
$$

Therefore,

$$
\begin{aligned}
& t = \frac{m^2}{8} \left( \tan^2 \frac{A}{2} - \tan^2 \frac{B}{2} \right) \frac{\cos \frac{A}{2} \cos \frac{B}{2}}{\cos \frac{C}{2}} \\
& = \frac{m^2}{8} \left( \tan^2 \frac{A}{2} - \tan^2 \frac{B}{2} \right) \left( \tan \frac{A}{2} - \tan \frac{B}{2} \right)
\end{aligned}
$$

or

$$
t = \frac{m^2}{8} \cdot \frac{\sin \frac{A+B}{2} \sin^2 \frac{A-B}{2}}{\cos^3 \frac{A}{2} \cos^3 \frac{B}{2}}
$$

(Jánosy Gyula, Budapest.)

The problem was also solved by: Ádámffy E., Blum J., Dömény I., Epstein K., Erdős V., Fekete M., Fodor H., Földes R., Glasel G., Glück J., Haar A., Heimlich P., Krampera Gy., Kürti I.; Merse P., Pető L., Rássy P., Rosenberg J., Schwarz Gy., Székely J., Sümegi Gy., Tandlich E.