In the interior of triangle $A B C$, point $P$ is taken such that the lines connecting it to the vertices form equal angles with each other. If these connecting lines are sequentially $d_{1}=4 \mathrm{~cm}, d_{2}=6 \mathrm{~cm}, d_{3}=10 \mathrm{~cm}$, what are the lengths of the sides and the angles of the triangle?
For the triangle $A B P$:

$$
A B=\sqrt{d_{1}^{2}+d_{2}^{2}-2 d_{1} d_{2} \cos 120^{\circ}}=\sqrt{d_{1}^{2}+d_{2}^{2}+d_{1} d_{2}}=8.717 \text{ cm}
$$

Similarly,

$$
A C=\sqrt{d_{1}^{2}+d_{2}^{2}+d_{1} d_{3}}=12.49 \text{ cm}
$$

and

$$
B C=\sqrt{d_{2}^{2}+d_{3}^{2}+d_{2} d_{3}}=14 \text{ cm}
$$

Knowing the sides of the triangle, we can easily calculate the angles using the cosine theorem. The result of the calculation is:

$$
\alpha=80^{\circ} 29^{\prime} 4^{\prime \prime}, \quad \beta=61^{\circ} 37^{\prime} 36^{\prime \prime} \quad \text { and } \quad \gamma=37^{\circ} 53^{\prime} 20^{\prime \prime}
$$

(György Schuster, Budapest.)

Number of solutions: 14.