Let's construct a triangle, given one side, $a$, and the radii of the excircles opposite to the other two sides, $\varrho_{b}$ and $\varrho_{c}$.
I. Solution: The problem can be solved with the help of calculations. Let the centers of the circles be denoted by $O_{b}$ and $O_{c}$, and their points of tangency on the line $B C$ by $T_{b}$ and $T_{c}$ (Figure 5).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9d5db5627628fae3ba5g-1.jpg?height=477&width=764&top_left_y=268&top_left_x=667)

Figure 5

It is known that $B T_{b} = C T_{c} = s$, where $s$ is half the perimeter of the triangle. By connecting $B$ to the centers of the circles, we know that $B O_{b}$ and $B O_{c}$ are the internal and external angle bisectors of the angle at $B$, and thus they are perpendicular to each other. This implies that

$$
\angle B O_{c} T_{c} = \angle O_{b} B T_{b},
$$

as right-angled angles are equal, hence the triangles $B O_{c} T_{c}$ and $O_{b} B T_{b}$ are similar. Writing the ratio of the legs,

$$
\frac{O_{c} T_{c}}{T_{c} B} = \frac{B T_{b}}{T_{b} O}, \quad \text{that is} \quad \frac{\varrho_{c}}{s-a} = \frac{s}{\varrho_{b}}, \quad s^{2} - a s - \varrho_{b} \varrho_{c} = 0
$$

From this, solving for $s$ (taking the positive root),

$$
s = \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^{2} + \varrho_{b} \varrho_{c}}
$$

This distance can be constructed; the second term, for example, can be constructed as follows: measure on a line from a common point $P$ in opposite directions the distances $P Q = \varrho_{b}$ and $P R = \varrho_{c}$ (Figure 6).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9d5db5627628fae3ba5g-1.jpg?height=374&width=547&top_left_y=1537&top_left_x=759)

Figure 6

The perpendicular to the line at $P$ intersects the semicircle erected over $Q R$ as diameter at $S$; then, as is known, $P S = \sqrt{\varrho_{b} \varrho_{c}}$. Measuring from $P$ on the line $Q R$ (in either direction) the distance $P T = \frac{a}{2}$,

$$
S T = \sqrt{P T^{2} + P S^{2}} = \sqrt{\left(\frac{a}{2}\right)^{2} + \varrho_{b} \varrho_{c}}
$$

Since $s = B T_{b} = C T_{c}$, the points $T_{b}$ and $T_{c}$ can be constructed on the line $B C$. The rest is trivial.

II. Solution: Using the notation from the previous solution, we saw that the segment $O_{b} O_{c}$ is seen from point $B$ - and similarly from point $C$ - at a right angle. This can be used much more simply for constructing the triangle. According to the derived relationship, the semicircle erected over $O_{b} O_{c}$ as diameter passes through $B$ and $C$ (Figure 7).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9d5db5627628fae3ba5g-2.jpg?height=675&width=986&top_left_y=129&top_left_x=537)

The perpendicular from the center $K$ of the circle to the line $B C$ bisects the segment $B C$ as a chord of the circle, and its length, as the midline of the trapezoid $O_{b} T_{b} T_{c} O_{c}$, is $\frac{\varrho_{b} + \varrho_{c}}{2}$.

Based on this, the construction can proceed as follows: At the midpoint $D$ of a segment $B C$ of length $a$, we draw a perpendicular of length $D K = \frac{\varrho_{b} + \varrho_{c}}{2}$ (Figure 8).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9d5db5627628fae3ba5g-2.jpg?height=578&width=878&top_left_y=1088&top_left_x=626)

Figure 8

We draw the circle centered at $K$ passing through $B$ and $C$ and intersect the semicircle on the side of point $C$ from the line $K D$ at a distance $\varrho_{b}$ from the line $B C$. The tangents from the point $O_{b}$ to the circle touching the line $B C$ at $B$ and $C$ will be the other two sides of the triangle.

We need to show that the triangle meets the conditions, i.e., that the circle inscribed to side $c$ has radius $\varrho_{c}$. We show that the center of this inscribed circle is the point $O_{c}$ on the line through $K$ and $O_{b}$, and this point is at a distance $\varrho_{c}$ from the line $B C$. The former follows from the fact that $O_{c}$ lies on the internal angle bisector of the angle at $C$ and the external angle bisector of the angle at $B$. To prove this, consider the circle centered at $O_{b}$ with radius $\varrho_{b}$. This is the excircle of the triangle, so $O_{b} C$ bisects the external angle at $C$, and $O_{b} B$ bisects the internal angle at $B$. Thus, the lines $C O_{c}$ and $B O_{c}$, which are perpendicular to the former, indeed bisect the internal angle at $C$ and the external angle at $B$. Furthermore, drawing perpendiculars from $O_{b}$ and $O_{c}$ to the line $B C$, we get a trapezoid whose side from $O_{b}$ is of length $\varrho_{b}$, and the midline from $K$ is of length $\frac{\varrho_{b} + \varrho_{c}}{2}$. Thus, the length of the side from $O_{c}$ is indeed $\varrho_{c}$.

III. Solution: Make a sketch (Figure 7). Let the circle inscribed to side $C A$ with radius $\varrho_{b}$ touch the line $B C$ at $D$, and the side $C A$ at $E$; let the circle inscribed to side $A B$ with radius $\varrho_{c}$ touch the side $C A$ at $F$; let the semiperimeter of the triangle be denoted by $s$.

Then,

$$
B D = C F = s, \quad C D = C E = s - a,
$$

so

$$
E F = C F - C E = s - (s - a) = a.
$$

Based on this, the following construction can be obtained: to a segment $E F$ of length $a$, draw a tangent circle of radius $\varrho_{b}$ at $E$ and a tangent circle of radius $\varrho_{c}$ at $F$ (Figure 9).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9d5db5627628fae3ba5g-3.jpg?height=661&width=810&top_left_y=130&top_left_x=641)

Figure 9

Construct the second internal common tangent and an external common tangent. The tangents enclose the desired triangle. Choosing the notation according to the figure, we need to show that $B C = a$, which follows similarly to the above calculation:

$$
a = E F = C F - C E = s - C E = D B - D C = C B
$$

Remark: Most of the correct solvers solved the problem with complex calculations.