Given one angle of a triangle: $\alpha=50^{\circ} 32^{\prime}$, the side opposite to this angle $a=18.5 \mathrm{dm}$, and the difference between the other two sides $\delta=8.5 \mathrm{dm}$. Determine the area and perimeter of the triangle.
According to the Carnot theorem:

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

and

$$
(b-c)^{2}=\delta^{2}=b^{2}+c^{2}-2 b c
$$

thus:

$$
a^{2}-\delta^{2}=2 b c(1-\cos \alpha)=4 b c \sin ^{2} \frac{\alpha}{2}
$$

and from this

$$
b c=\frac{a^{2}-\delta^{2}}{4 \sin ^{2} \frac{\alpha}{2}}
$$

and thus the area of the triangle is

$$
t=\frac{1}{2} b c \sin \alpha=\frac{2\left(a^{2}-\delta^{2}\right) \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2}}{8 \sin ^{2} \frac{\alpha}{2}}=\frac{1}{4}(a+\delta)(a-\delta) \operatorname{ctg} \frac{\alpha}{2}
$$

The perimeter $(k)$ can be calculated as follows:

$$
t=\frac{1}{4} \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
$$

thus

$$
t^{2}=\frac{1}{16}(a+\delta)^{2}(a-\delta)^{2} \operatorname{ctg}^{2} \frac{\alpha}{2}=\frac{1}{16} \cdot k \cdot(k-2 a)(a-\delta)(a+\delta)
$$

from which

$$
k^{2}-2 a k=(a+\delta)(a-\delta) \operatorname{ctg}^{2} \frac{\alpha}{2}
$$

In our case:

$$
\begin{gathered}
t=143.01 \mathrm{dm}^{2} \\
k=57.92 \mathrm{dm}
\end{gathered}
$$

(Riesz Marcel, Győr.)

The problem was also solved by: Ádámffy E., Bánó L., Bartók J., Dömény E., Dömény J., Eckhart F., Enyedi B., Fekete M., Frank K., Friedländer H., Füstös P., Glück J., Haar A., Hirschfeld Gy., Jánosy Gy., Kertész G., Kiss J., Kürti J., Liebner A., Neidenbach E., Pazsiczky G., Pám M., Pető L., Pichler S., Pivnyik J., Popoviciu M., Ragányi B., Rássy P., Riesz K., Rosenberg J., Schwarz Gy., Schwemmer J., Székely J., Söpkéz Gy., Szécsi J., Szűcs A., Winkler J.