The points of tangency of the incircle of a given triangle determine a new triangle. Express the area of this triangle in terms of the sides of the original triangle.
I. solution. Let the area of the triangle $ABC$ be $T$, the semi-perimeter of its perimeter be $s$, and the area of the triangle determined by the points of tangency $A', B', C'$ of the inscribed circle be $t$. The other notations can be seen in the figure.

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It is known that $AC' = s - a$, and thus from the figure, $t_1 = \frac{1}{2}(s-a)^2 \sin \alpha; T = \frac{1}{2} bc \sin \alpha$, from which $t_1 = T \cdot \frac{(s-a)^2}{bc}$. We get similar relationships for $t_2$ and $t_3$.

The sought area is then $t = T - t_1 - t_2 - t_3$, which, using the above, is:

$$
t = T - T \cdot \frac{(s-a)^2}{bc} - T \cdot \frac{(s-b)^2}{ac} - T \cdot \frac{(s-c)^2}{ab}
$$

If we now express $T$ using Heron's formula, then after factoring out, we arrive at the relationship required by the problem:

$$
t = \sqrt{s(s-a)(s-b)(s-c)} \left(1 - \frac{(s-a)^2}{bc} - \frac{(s-b)^2}{ac} - \frac{(s-c)^2}{ab}\right)
$$

II. solution. In the cyclic quadrilateral $AC'O B'$, $\angle B'OC' = 180^\circ - \alpha$, and thus $T_1 = \frac{1}{2} r^2 \sin \alpha$. Similarly for $T_2$ and $T_3$ (where $r$ is the radius of the inscribed circle):

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

Since, for example, $T = \frac{1}{2} bc \sin \alpha$, it follows that $\sin \alpha = \frac{2T}{bc}$, and similarly for $\sin \beta$ and $\sin \gamma$. Therefore, (1) can be written as:

$$
t = \frac{r^2}{2} \left(\frac{2T}{bc} + \frac{2T}{ac} + \frac{2T}{ab}\right)
$$

Factoring out $T$ and using the relationship $T = rs$, we get $t = \frac{T^3}{s^2} \cdot \frac{a+b+c}{abc}$, and substituting $T$ again using Heron's formula:

$$
t = \frac{2 \cdot \sqrt{s \cdot (s-a)^3 (s-b)^3 (s-c)^3}}{abc}
$$