A triangle has two sides $a$ and $b$, and the median $k$ connecting vertex $C$ to the midpoint of side $AB$; calculate the other elements and construct the triangle.
From triangle $ABC$:

$$
a^{2}=b^{2}+c^{2}-2 b c \cos A
$$

From triangle $AEC$:

$$
k^{2}=b^{2}+\frac{c^{2}}{4}-b c \cos A
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_0e658494390c1b63d72cg-1.jpg?height=282&width=439&top_left_y=493&top_left_x=838)

These two equations can also be written as:

$$
2 a^{2}=2 b^{2}+2 c^{2}-4 b c \cos A
$$

$$
4 k^{2}=4 b^{2}+c^{2}-4 b c \cos A
$$

Subtracting (4) from (3):

$$
c^{2}=2 a^{2}+2 b^{2}-4 k^{2}
$$

and thus

$$
c=\sqrt{2\left(a^{2}+b^{2}-2 k^{2}\right)}
$$

Since we know the sides, the angles can be easily calculated.

Construction. From sides $A, b$, and $2k$, we construct triangle $ADC$; we bisect $DC$, connect $A$ to $E$, extend $AE$, and measure $AE$ again from point $E$, thereby obtaining point $B$. $ABC$ is the desired triangle.

Proof. Quadrilateral $ACBD$ is a parallelogram because its diagonals bisect each other; thus, $AD=BC=a$.

The problem is not solvable if a triangle cannot be constructed from sides $a, b$, and $2k$.

(Béla Grünhut, VII. grade, Pécs.)

The problem was also solved by: Bernát Friedmann, S.-A.-Ujhely; Zsigmond Goldstein, Nyíregyháza; Ervin Hofbauer and Nándor Kántor, Budapest; István Szabó, Debrecen; Aladár Visnya, Pécs; Győző Zemplén, Fiume.