We draw a circle around a triangle whose angles are $\alpha, \beta, \gamma$. The tangents drawn at the vertices of the triangle form a new triangle. How does the area of this triangle compare to the area of the original triangle? $\left(\alpha=50^{\circ}, \beta=60^{\circ}, \gamma=70^{\circ}\right)$. $)$
Auxiliary theorems. If the radius of the circumcircle of triangle $ABC$ is $R$, the radius of the incircle is $r$, and the angles of the triangle are $\alpha, \beta, \gamma$, then the area of the triangle is

$$
T=2 R^{2} \sin \alpha \cdot \sin \beta \cdot \sin \gamma
$$

or

$$
T=r^{2} \cdot \operatorname{ctg} \frac{\alpha}{2} \cdot \operatorname{ctg} \frac{\beta}{2} \cdot \operatorname{ctg} \frac{\gamma}{2}
$$

Proof. (a) If the lengths of the sides of the triangle are $a, b, c$, then

$$
T=\frac{a b c}{4 R}
$$

and since

$$
R=\frac{a}{2 \sin \alpha}=\frac{b}{2 \sin \beta}=\frac{c}{2 \sin \gamma}
$$

thus

$$
T=\frac{2 R \sin \alpha \cdot 2 R \sin \beta \cdot 2 R \sin \gamma}{4 R}=2 R^{2} \sin \alpha \cdot \sin \beta \cdot \sin \gamma
$$

(b) On the other hand

$$
T=\frac{r}{2}(a+b+c)
$$

but

$$
r=\frac{a}{\operatorname{ctg} \frac{\beta}{2}+\operatorname{ctg} \frac{\gamma}{2}}=\frac{b}{\operatorname{ctg} \frac{\gamma}{2}+\operatorname{ctg} \frac{\alpha}{2}}=\frac{c}{\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\gamma}{2}}
$$

thus

$$
T=r^{2}\left(\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}+\operatorname{ctg} \frac{\gamma}{2}\right)
$$

However,

$$
\begin{gathered}
\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}+\operatorname{ctg} \frac{\gamma}{2}=\frac{\sin \frac{\alpha+\beta}{2}}{\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}}+\frac{\cos \frac{\gamma}{2}}{\sin \frac{\gamma}{2}}= \\
=\frac{\cos \frac{\gamma}{2}}{\sin \frac{\gamma}{2}}\left[\frac{\sin \frac{\gamma}{2}+\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}}{\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}}\right]=\operatorname{ctg} \frac{\gamma}{2} \cdot \frac{\cos \frac{\alpha+\beta}{2}+\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}}{\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}}= \\
=\operatorname{ctg} \frac{\gamma}{2} \cdot \frac{\cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}+\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}}{\sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2}}= \\
=\operatorname{ctg} \frac{\alpha}{2} \cdot \operatorname{ctg} \frac{\beta}{2} \cdot \operatorname{ctg} \frac{\gamma}{2}
\end{gathered}
$$

thus indeed

$$
T=r^{2} \operatorname{ctg} \frac{\alpha}{2} \cdot \operatorname{ctg} \frac{\beta}{2} \cdot \operatorname{ctg} \frac{\gamma}{2}
$$

After this, we can proceed to the solution of our problem. Let the center of the circumcircle of the given $ABC$ triangle be $O$ and its radius $r$. The vertices of the tangential triangle are denoted by $A_{1}, B_{1}, C_{1}$ and its angles by $\alpha_{1}, \beta_{1}, \gamma_{1}$. Let the area of the $ABC$ triangle be $t$, and that of the $A_{1}, B_{1}, C_{1}$ triangle be $T$, then

$$
T=r^{2} \operatorname{ctg} \frac{\alpha_{1}}{2} \cdot \operatorname{ctg} \frac{\beta_{1}}{2} \cdot \operatorname{ctg} \frac{\gamma_{1}}{2}
$$

and

$$
t=2 r^{2} \cdot \sin \alpha \cdot \sin \beta \cdot \sin \gamma
$$

We only need to calculate the angles $\alpha_{1}, \beta_{1}, \gamma_{1}$. From the quadrilateral $A_{1} B O C$ (since the angles $A_{1} B O \varangle$ and $A_{1} C O$ are right angles):

$$
\alpha_{1}+B O C \varangle=\alpha_{1}+2 \alpha=180^{\circ}
$$

thus

$$
\alpha_{1}=2\left(90^{\circ}-\alpha\right)
$$

and thus

$$
T=r^{2} \cdot \operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma
$$

Now

$$
T: t=r^{2} \operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma: 2 r^{2} \sin \alpha \sin \beta \sin \gamma=1: 2 \cos \alpha \cdot \cos \beta \cdot \cos \gamma
$$

In this case

$$
T: t=1: 0.219865
$$

(József Kiss, Pápa.)

The problem was also solved by: I. Bartók, I. Braun, I. Dömény, E. Dömény, B. Enyedi, I. Glück, A. Haar, Gy. Jánosy, G. Kertész, I. Kürti, A. Liebner, P. Messer, E. Neidenbach, M. Pám, S. Pichler, I. Pivnyik, M. Popoviciu, P. Rássy, K. Riesz, I. Rosenberg, Gy. Schwarz, I. Sonnenfeld, A. Szúcs, I. Schwemmer.