Let's construct a triangle, given its perimeter, and the radii of two of its excircles.
I. Solution. Consider the problem as solved. The excircle of triangle $ABC$ opposite to side $BC$ touches the sides of the triangle at points $D, F, E$, and the excircle opposite to side $AC$ touches this side at point $G$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_4f43ef06636ad8220e47g-1.jpg?height=594&width=629&top_left_y=305&top_left_x=737)

We know that the two tangent segments from an external point to a circle are equal, so $BD = BE$ and $CD = CF$; therefore, the lengths of segments $AE = AB + BE$ and $AF = AC + CF$ are equal to the perimeter of triangle $ABC$. But $AE = AF$, so the length of each of these segments is equal to the semiperimeter of the triangle, $s$. Using this, we can complete the construction:

We take a circle with radius $r_a$ and arbitrarily choose point $F$ on its circumference. We construct the tangent to the circle at $F$ and measure the distance $FA = s$, which is given. Then, from $A$, we construct the other tangent to the circle (which is the line of the side $AB$ of the triangle). The other excircle touches the lines $AF$ and $AB$, and its radius is given, so we can construct this circle as well. (The center of the circle lies on the external angle bisector of $\angle FAB$ and on the line parallel to $AB$ at a distance $r_b$ from it.) Finally, we get the line $BC$ by constructing the common internal tangent of the two circles different from $AF$. (The construction of a common tangent of two circles can be found, for example, in the problem 633 of the Collection of Geometric Problems I.) From the above, it follows that the constructed triangle $ABC$ has one of its excircles with radius $r_a$, another excircle with radius $r_b$, and its perimeter is $2AF = 2s$, so the triangle meets the requirements.

The construction can be carried out precisely when point $G$ is an interior point of segment $AF$. In this case, there is 1 solution; otherwise, there is no solution. (In the second solution, we show that $G$ is precisely an interior point of segment $AF$ if $r_a \cdot r_b < s^2$. Since $\operatorname{tg} \frac{\alpha}{2} = \frac{r_a}{s}$, $\operatorname{tg} \frac{\beta}{2} = \frac{r_b}{s}$,

$$
\operatorname{tg}\left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \frac{\operatorname{tg} \frac{\alpha}{2} + \operatorname{tg} \frac{\beta}{2}}{1 - \operatorname{tg} \frac{\alpha}{2} \cdot \operatorname{tg} \frac{\beta}{2}} = \frac{\frac{r_a}{s} + \frac{r_b}{s}}{1 - \frac{r_a \cdot r_b}{s^2}}
$$

This fraction is positive if its denominator is positive, so the triangle to be constructed exists precisely when $s^2 > r_a \cdot r_b$. In this case, the problem has 1 solution; otherwise, it has no solution.

Based on the work of Imre Mikulás (Nyíregyháza, Krúdy Gyula Gymnasium, 2nd grade)