Express the side $c$ of the triangle using the sides $a$ and $b$, if we know that the radius of the circle tangent to the side $c$ is the geometric mean of the radii of the circles tangent to the sides $a$ and $b$.
Let the vertices of the triangle be denoted by $A, B, C$, the radii of the excircles by $r_{a}, r_{b}, r_{c}$, and the center of the excircle opposite to side $a$ by $O_{A}$.

The area of triangle $ABC$ can be obtained by subtracting the area of triangle $O_{A}BC$ from the sum of the areas of triangles $O_{A}AB$ and $O_{A}AC$. In each of these triangles, the altitude corresponding to $O_{A}$ is $r_{a}$, so

$$
T_{ABC}=\frac{r_{a} c}{2}+\frac{r_{a} b}{2}-\frac{r_{a} a}{2}
$$

Similarly, we get

$$
T_{ABC}=\frac{r_{b} c}{2}+\frac{r_{b} a}{2}-\frac{r_{b} b}{2} \quad \text{and} \quad (2) T_{ABC}=\frac{r_{c} a}{2}+\frac{r_{c} b}{2}-\frac{r_{c} c}{2} \cdot(3)
$$

The product of the right-hand sides of equations (1) and (2) equals the square of the right-hand side of equation (3), that is,

$$
\left[\frac{r_{a}}{2}(c+b-a)\right]\left[\frac{r_{b}}{2}(c+a-b)\right]=\left[\frac{r_{c}}{2}(a+b-c)\right]^{2}
$$

Using the fact that $r_{a}$ and $r_{b}$ are the geometric mean of $r_{c}$, i.e., $r_{a} r_{b}=r_{c}^{2}$:

$$
(c+b-a)(c+a-b)=(a+b-c)^{2}
$$

From this, the solution to the problem follows: $c=\frac{a^{2}+b^{2}}{a+b}$.

Based on the work of Viktor Csendes (Kaposvár, Táncsics M. Gymnasium, 11th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_5fed1a2349145cd28384g-1.jpg?height=410&width=743&top_left_y=1161&top_left_x=669)