Can a triangle be constructed if the radius of the incircle corresponding to two given sides and the height to the third side are known?
I. solution. Let the given $\varrho_{a}$ and $\varrho_{b}$ be the radii of the circles that touch the side segments $AB$ and $CA$ of $\triangle ABC$, and the extensions of the other two sides, respectively, and let the given height to side $AB$ be $m_{c}$. Let the centers of the two circles be $O_{a}$ and $O_{b}$, and their points of tangency on the line of side $BC$ be $A'$ and $A''$.

![](https://cdn.mathpix.com/cropped/2024_05_02_642ad368786bd7b8fb2dg-1.jpg?height=471&width=746&top_left_y=304&top_left_x=684)

$CO_{a}$ and $CO_{b}$ bisect the exterior angles at $C$ of the triangle, so they lie on the extensions of each other. Thus, $\triangle CO_{a}A'$ and $\triangle CO_{b}A''$ are similar right triangles, so

$$
\frac{CO_{a}}{CO_{b}} = \frac{O_{a}A'}{O_{b}A''} = \frac{\varrho_{a}}{\varrho_{b}}
$$

Let the line through $O_{b}$ parallel to $AB$ intersect the perpendiculars from $C$ and $O_{a}$ to $AB$ at points $D$ and $E$, respectively. If $\varrho_{a} > \varrho_{b}$, then $D$ lies on the height segment $CC_{0}$. Using the similarity of triangles $O_{b}CD$ and $O_{b}O_{a}E$ and (1),

$$
\begin{aligned}
& CD = \frac{CO_{b}}{O_{a}O_{b}} \cdot O_{a}E = \frac{CO_{b}}{O_{a}C + CO_{b}} \cdot O_{a}E = \frac{1}{\frac{CO_{a}}{CO_{b}} + 1} \cdot O_{a}E \\
& m_{c} - \varrho_{b} = \frac{1}{1 + \varrho_{a} / \varrho_{b}} \cdot (\varrho_{a} - \varrho_{b}), \text{ hence } \\
& m_{c} = \frac{2 \varrho_{a} \varrho_{b}}{\varrho_{a} + \varrho_{b}}
\end{aligned}
$$

If $\varrho_{a} = \varrho_{b}$, then the line $O_{a}CO_{b}$ is parallel to $AB$, and thus $m_{c} = \varrho_{a} = \varrho_{b}$, so the result (2) is also valid for this special case.

Accordingly, $m_{c}$ is not independent of the given radii in either case, and the triangle cannot be constructed uniquely from the given data. If the data triplet satisfies (2) (which is as if there were only two data points), infinitely many suitable triangles can be constructed (the two tangent circles are placed on one side of the line $AB$ so that they do not intersect, and the other two sides of the triangle are given by the common internal tangents of the two circles). If (2) is not satisfied, no triangle meets the requirements.

Tóth Teréz (Makó, József A. g. III. o. t.)

II. solution. The given radii and height can be expressed in a known manner by the area $t$ of the triangle, its sides $a, b, c$, and the semiperimeter $s = \frac{a+b+c}{2}$:

$$
\varrho_{a} = \frac{t}{s-a}, \quad \varrho_{b} = \frac{t}{s-b}, \quad m_{c} = \frac{2t}{c}
$$

From this, on the one hand,

$$
c = 2s - (a+b) = (s-a) + (s-b) = \frac{t}{\varrho_{a}} + \frac{t}{\varrho_{b}}
$$

and on the other hand, $c = \frac{2t}{m_{c}}$, and from the equality of the two expressions,

$$
\frac{1}{m_{c}} = \frac{1}{2} \left( \frac{1}{\varrho_{a}} + \frac{1}{\varrho_{b}} \right)
$$

(regardless of the size relationship between the two radii).

From this, the same conclusion follows as in the first solution.