Let's construct a right-angled triangle, given the hypotenuse $c$, and the median to one of the legs $s_{a}$.
I. Solution: Imagine the problem is solved. The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_4da15080389abbd32f2dg-1.jpg?height=363&width=572&top_left_y=244&top_left_x=755)

Figure 1

The vertex of the right angle $C$ lies on the Thales semicircle drawn over the hypotenuse $AB$. The geometric place of the endpoint $A_1$ of the given median $s_a = AA_1$, or the midpoint of side $CB$, as $C$ moves on the Thales semicircle, is the reduction of this Thales semicircle from point $B$ as the center of similarity in the ratio $1:2$. (This can also be understood as a Thales semicircle, since $C_1A_1$ is always perpendicular to $BA_1$.) Therefore, the geometric place of the points $A_1$ is the semicircle drawn over $C_1B$ as the diameter, and its intersection with the circle of radius $AA_1 = s_a$ centered at $A$ gives the point $A_1$. This implies that the solvability condition is: $AC_1 < s_a < AB$, or $\frac{c}{2} < s_a < c$.

There is at most one solution, because if a triangle is to be constructed not by position but only by shape and size, then congruent triangles count as one solution.

Similarly, we can directly construct the point $C$ if, instead of reducing the Thales circle passing through $C$ from point $B$ in the ratio $1:2$, we enlarge the circle of radius $s_a$ centered at $A$ in the ratio $2:1$.

Ferenc Beleznay (Bp. V., Piarist g. II. o. t.)

II. Solution: The centroid $S$ forms a triangle with points $A$ and $C_1$ where all three sides are known: $AC_1 = \frac{c}{2}, AS = \frac{2s_a}{3}$, and $C_1S = \frac{s_c}{3} = \frac{c}{6}$ (since $s_c = \frac{c}{2}$) (Figure 1). With the point $S$, the line $C_1S$ intersects the Thales semicircle at point $C$. (Verification: $SC = 2 \cdot \frac{c}{6}, SA_1 = \frac{s_a}{3}$). The solvability condition is: $\frac{c}{2} - \frac{c}{6} < \frac{2s_a}{3} < \frac{c}{2} + \frac{c}{6}$, or $\frac{c}{2} < s_a < c$.

Klara Huj (Bp., I. Fürst S. g. II o. t.)

III. Solution: Reflecting the triangle over $A_1$ (Figure 2), we obtain the parallelogram $AB A^* C$, whose diagonal $AA^* = 2s_a$, and side $A^*C$ is equal to the given $c$, and the angle between diagonal $BC$ and side $AC$ is $90^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4da15080389abbd32f2dg-1.jpg?height=450&width=749&top_left_y=1670&top_left_x=688)

Figure 2

Based on this, the construction process is: Take the segment $AA^* = 2s_a$. The midpoint of this segment is $A_1$. The intersection of the Thales semicircle drawn over $AA_1$ with the circle of radius $c$ centered at $A^*$ gives the point $C$. The reflection of $C$ over $A_1$ gives the point $B$. The solvability condition is: $s_a < c < 2s_a$, or $\frac{c}{2} < s_a < c$.