The angles of a triangle are: $\alpha=74^{\circ} 15^{\prime} 46^{\prime \prime}, \beta=36^{\circ} 9^{\prime} 24^{\prime \prime}, \gamma=69^{\circ} 34^{\prime} 50^{\prime \prime}$, and the radius of the circle circumscribed around the triangle is $r=2$ dm; the question is, how large is the area of the triangle formed by the tangents drawn to the circle at the vertices?
In the triangle $ABC$, the radius of the circumscribed circle is $r=AO=BO=CO$. The area of the triangle $A_{1}B_{1}C_{1}$ circumscribed around the circle is:

$$
\begin{aligned}
T= & AOB C_{1}+BOC A_{1}+COA B_{1}= \\
& =AC_{1} \cdot r+A_{1}B \cdot r+B_{1}C \cdot r
\end{aligned}
$$

But

$$
AC_{1}=r \tan \gamma
$$

because

$$
\frac{\angle AOB}{2}=\angle AOC_{1}=\gamma
$$

Similarly,

$$
A_{1}B=r \tan \alpha \quad \text{and} \quad B_{1}C=r \tan \beta
$$

Thus,

$$
T=r^{2}(\tan \alpha+\tan \beta+\tan \gamma)
$$

or (see: K. M. L. II. 91. problem)

$$
T=r^{2} \tan \alpha \tan \beta \tan \gamma
$$

Substituting the given values:

$$
T=27.86 \mathrm{dm}^{2}
$$

(András Hrivzák, Selmeczbánya.)

The problem was also solved by: Dénes A., Freibauer E., Friedmann B., Goldziher K., and Kármánt T., Orlowszky F., Porde Gy., Porkoláb J., Schiffer H., Spitzer Ö., Szabó I., Szabó K., Weisz J.