Let's construct a triangle, given the side $c$, the corresponding height $m_{c}$, and the sum of the other two sides $d(=a+b)$.
I. Solution: Imagine the problem is solved and draw the inscribed and the excircle of the $ABC$ triangle to the side $c$. See Figure 1, which also shows the labeling.

![](https://cdn.mathpix.com/cropped/2024_05_02_c98be9c8834cb2e35f65g-1.jpg?height=710&width=854&top_left_y=260&top_left_x=619)

Figure 1

It is known that the distance from vertex $C$ to the points of tangency $T$ and $T_{c}$ is $s-c$ and $s$, respectively, where $s$ is the semiperimeter.

From the given data, $s=\frac{d+c}{2}$, and $s-c$ (assuming $d>c$) can be directly constructed.

We only need to construct $\varrho$.

It is known that the double area of the triangle is

$$
2t = c \cdot m_{c} = \varrho \cdot s, \text{ hence } s : c = m_{c} : \varrho
$$

and thus $\varrho$ is the fourth proportional. (Therefore, $m_{c} > \varrho$.)

The construction process is as follows: We draw a ray from point $C$, on which we measure the distances $CT = s - c$ and $CT_{c} = s$. On the perpendicular to $CT$ at $T$, we measure the constructed segment $\varrho = TO$. The intersection of the perpendicular to $CT$ at $T_{c}$ and the line $CO$ is $O_{c}$. Drawing the incircle and the excircle around $O$ and $O_{c}$, respectively, the other common external tangent of these two circles will be the line carrying side $CA$, and one of the common internal tangents will be the line carrying side $BA$. (The other common internal tangent leads to the symmetric solution with respect to the $CO$ axis.)

The solvability condition (besides $d > c$) is that the two constructed tangent circles do not intersect at two different points, because in the latter case there is no common internal tangent. If the two circles are tangent, we get an isosceles triangle $(a = b)$.

István Goldperger (Balassagyarmat, Balassa g. I. o. t.)

II. Solution: We can also reach the goal without using the excircle. After constructing the right triangle $CTO$, we construct the other tangent from $C$ to the incircle, then draw a circle around $C$ with radius $m_{c}$ (as we have seen, $m_{c} > \varrho$). Any common external tangent of the two circles will intersect the sides of $\angle C$ at points $A$ and $B$ (Figure 1).

The solvability condition is that $m_{c} < CO + \varrho$, or

$$
\left(m_{c} - \varrho\right)^{2} < CO^{2} = (s - c)^{2} + \varrho^{2}
$$

But

$$
\varrho = \frac{m_{c} c}{2s} = \frac{m_{c} c}{d + c}, \quad s - c = \frac{a + b + c}{2} - c = \frac{d - c}{2}
$$

Substituting these values into (1), we get

$$
m_{c}^{2} < \frac{(d - c)^{2}(d + c)}{4(d - c)} = \frac{d^{2} - c^{2}}{4}, \quad \text{ i.e., } \quad m_{c} < \frac{\sqrt{d^{2} - c^{2}}}{2}
$$

Tamás Endrödy (Bp., III., Árpád g. I. o. t.)

III. Solution: Start from the segment $AB = c$. Since $a + b = d$, the geometric locus of points $C$ is an ellipse with semi-major axis $OP = OQ = \frac{d}{2}$ and semi-minor axis

$$
OR = OS = \sqrt{\left(\frac{d}{2}\right)^{2} - \left(\frac{c}{2}\right)^{2}} = \frac{\sqrt{d^{2} - c^{2}}}{2}
$$

On the other hand, the geometric locus of points $C$ is a line $e$ parallel to $AB$ at a distance $m_{c}$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_c98be9c8834cb2e35f65g-2.jpg?height=580&width=829&top_left_y=125&top_left_x=629)

This line $e$ intersects the ellipse at the desired point $C$. The intersection of a line and an ellipse can be constructed most simply by transferring the line to a circle system that is affinely related to the ellipse, and then transferring the constructed intersection points back to the ellipse system.

In this case, the most convenient is to consider the circle $k'$ drawn over the minor axis as the affine counterpart of the ellipse. In this case, the minor axis is the $x$-axis of the affinity, the direction of the affinity is perpendicular to it, the point $P$ at the end of the major axis corresponds to the point $P'$ on the circle, and thus the affine counterpart of $e$ is $e'$, which coincides with $e$. The line $e'$ intersects the circle $k'$ at $C'$. The tangent $t'$ drawn at $C'$ intersects the tangent $p'$ at $T'$, and the $x$-axis at $X$. The affine counterpart of $T'$ in the ellipse system is the point $T$ on $p$. The line $TX$ intersects $e$ at the desired point $C$.

Péter Tatai (Bp., XIV., I. István g. I. o. t.)