The equation of the base of a certain triangle is $55 y+18 x-256=0$. The x-coordinates of its endpoints are $x_{1}=2, x_{2}=7.5$. The coordinates of the centroid of the triangle are $\alpha=5, \beta=6$. What are the lengths of the sides and the angles?
The equation of line $BC$, substituting the abscissas of points $B$ and $C$, gives the coordinates of the two points: $y_{1}=4$ and $y_{2}=2.2$. The coordinates of the midpoint $D$ of the base $BC$ are:

$$
x_{4}=\frac{x_{1}+x_{2}}{2}=4.75, y_{4}=\frac{y_{1}+y_{2}}{2}=3.1
$$

The coordinates of vertex $A$, $x_{3}$ and $y_{3}$, can be determined from the condition that the centroid divides the distance $AD$ in the ratio $2:1$, which is why:

$$
\alpha=\frac{2 x_{4}+x_{3}}{3} \text { and } \beta=\frac{2 y_{4}+y_{3}}{3}
$$

from which

$$
x_{3}=3 \alpha-2 x_{4}=5.5, y_{3}=3 \beta-2 y_{4}=11.8
$$

Knowing the coordinates of the vertices of the triangle, the sides are:

$$
\begin{gathered}
a=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}, b=\sqrt{\left(x_{2}-x_{3}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}} \\
c=\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}}
\end{gathered}
$$

Substituting the given values: $a=5.79, b=9.81, c=8.55$; the angles calculated using the tangents of the angles are: $\alpha=35^{\circ} 56^{\prime} 26^{\prime \prime}, \beta=83^{\circ} 58^{\prime} 34^{\prime \prime}, \gamma=60^{\circ} 5^{\prime}$.

(Béla Grünhut, Pécs.)

The problem was also solved by Frigyes Riesz and István Szabó.