Problem 7. In an isosceles triangle with a perimeter of 60 cm, the point of intersection of the medians lies on the inscribed circle. Find the sides of the triangle.
Answer: 25, 25, 10.

Solution. Let in the original triangle $ABC$, $AB=BC$, $BM$ be the median, and $O$ be the point of intersection of the medians. Note that $OM$ is the diameter of the circle, so $BM=3OM=6r$ (where $r$ is the radius of the inscribed circle).

The area of the triangle, on one hand, is $\frac{1}{2} AC \cdot BM = 3r \cdot AC$, and on the other hand, $\frac{1}{2}(AB+BC+AC)r=30r$. Therefore, $AC=\frac{30r}{3r}=10$, and $AB=BC=\frac{60-10}{2}=25$.