1. In triangle $A B C$ with sides $a, b$, and $c$, an inscribed circle is drawn. The points of tangency of the circle with the sides of the triangle determine triangle $E F G$. Calculate the area of $\triangle E F G$, if the area of $\triangle A B C$ is $P$.
Solution. Let $R$ and $r$ be the radii of the circumcircle and the incircle of $\triangle ABC$ respectively, and let $O$ be the center of the incircle. Let $P_{1}$ be the area of $\triangle EFG$. We have

![](https://cdn.mathpix.com/cropped/2024_06_05_6809eb833a150cf84798g-4.jpg?height=425&width=431&top_left_y=1047&top_left_x=1015)

$$
P_{1}=P_{EOF}+P_{FOG}+P_{GOE}=\frac{r^{2}}{2}\left(\sin \alpha_{1}+\sin \beta_{1}+\sin \gamma_{1}\right)
$$

On the other hand, $\alpha_{1}=180^{\circ}-\alpha, \beta_{1}=180^{\circ}-\beta$ and $\gamma_{1}=180^{\circ}-\gamma$, so

$$
P_{1}=\frac{r^{2}}{2}(\sin \alpha+\sin \beta+\sin \gamma)
$$

Using the sine theorem, we get

$$
P_{1}=\frac{r^{2}}{2}\left(\frac{a}{2 R}+\frac{b}{2 R}+\frac{c}{2 R}\right)=\frac{r^{2}}{4 R}(a+b+c)
$$

If we also use the known equalities $P=\frac{abc}{4 R}, P=rs, s=\frac{a+b+c}{2}$, we get $P_{1}=\frac{2 P^{3}}{\text{sabc}}$.