In a triangle, the sum of two sides $(a+c=k)$, the angle $\beta$ enclosed by the sides $a$ and $c$, and the height $m$ corresponding to the side $b$ are given. Solve the triangle.
According to Carnot's theorem,

$$
b^{2}=a^{2}+c^{2}-2 a c \cos \beta=(a+c)^{2}-2 a c(1+\cos \beta)
$$

But

$$
2 t=a c \sin \beta=m b
$$

thus

$$
b^{2}=k^{2}-4 \frac{m b}{\sin \beta} \cos ^{2} \frac{\beta}{2}=k^{2}-2 b m \operatorname{ctg} \frac{\beta}{2}
$$

or

$$
b^{2}+2 b m \operatorname{ctg} \frac{\beta}{2}-k^{2}=0
$$

from which

$$
b=-m \operatorname{ctg} \frac{\beta}{2} \pm \sqrt{m^{2} \operatorname{ctg}^{2} \frac{\beta}{2}+k^{2}}
$$

Given the nature of the problem, the root can only be taken with a positive sign. Furthermore, from (1)

$$
a c=\frac{m b}{\sin \beta}
$$

and

$$
a+c=k
$$

therefore $a$ and $c$ can be easily calculated.

(Henrik Fodor, Beregszász.)

Number of solutions: 33.