In triangle $ABC$, point $S$ is the center of the inscribed circle. The area of quadrilateral $ABCS$ is equal to four-fifths of the area of triangle $ABC$. The lengths of the sides of triangle $ABC$, expressed in centimeters, are all integers, and the perimeter of triangle $ABC$ is $15 \, \text{cm}$.

Determine the lengths of the sides of triangle $ABC$. Find all possibilities.

(E. Semerádová)

Hint. Can you determine the area of a triangle using its perimeter and the radius of the inscribed circle?
Triangle $ABC$ can be decomposed into triangles $ABS$, $BCS$, and $ACS$. The height of each of these triangles from vertex $S$ is equal to the radius of the inscribed circle. Therefore, the ratios of their areas are the same as the ratios of the lengths of the sides opposite vertex $S$.

Similarly, these partial triangles can be compared with the entire triangle $ABC$ (whose area is equal to the product of the perimeter and the radius of the inscribed circle).

![](https://cdn.mathpix.com/cropped/2024_04_17_fc767241fb97c8703187g-3.jpg?height=477&width=851&top_left_y=341&top_left_x=608)

Since the area of quadrilateral $ABCS$ is equal to four-fifths of the area of triangle $ABC$, the remaining area for triangle $ACS$ is one-fifth of the area of triangle $ABC$. Therefore, the length of side $AC$ is one-fifth of the perimeter of triangle $ABC$, which in our case is $15: 5 = 3(\mathrm{~cm})$. The sum of the lengths of the remaining two sides is therefore $12 \mathrm{~cm}$; the following pairs of side lengths (in cm, regardless of order) are possible:

$$
1,11 ; \quad 2,10 ; \quad 3,9 ; \quad 4,8 ; \quad 5,7 ; \quad 6,6 .
$$

For the segments to form the sides of a triangle, the triangle inequalities must be satisfied. Only the following triples meet these requirements - possible side lengths of triangle $ABC$ in cm:

$$
3,5,7 ; \quad 3,6,6 \text {. }
$$