Let's construct a triangle, given one side $a$, the angle $\alpha$ opposite to it, and the sum of the other two sides $b+c=s$.
$\frac{\alpha}{2}$ is the angle from one arm of the angle, from the vertex $A$, we measure $a+b$; from the point $B$ thus obtained, we draw a circular arc with radius $a$, which intersects the other arm of the angle at points $M$ and $M^{\prime}$. At these points, we draw $\frac{\alpha}{2}$ angles to $MA$ and $M^{\prime}A$, respectively, whose other sides intersect the line $AB$ at $K$ and $K^{\prime}$, respectively. $KBM$ and $K^{\prime}BM^{\prime}$ are the sought triangles.

Proof. $KMA$ is an isosceles triangle, so $MK = AK$ and thus $BK + KM = BK + AK = BA = s$; the angle $BKM$, as an exterior angle, $=\frac{\alpha}{2} + \frac{\alpha}{2} = \alpha$.

The problem is only solvable if $as \sin \frac{\alpha}{2}$. If this condition is met, we get two solutions.

(Mihály Devecis, real gymnasium VI. class, Budapest, VIII. district.)

Number of solutions: 41.