To determine the volume of a rectangular parallelepiped, whose edges are the distances from the orthocenter of a triangle to its vertices, given that the circumradius of the triangle is $r=2.35$ and the angles are: $\alpha=63^{\circ} 18^{\prime} 13^{\prime \prime}, \beta=$ $51^{\circ} 42^{\prime} 19^{\prime \prime}$.
Let the orthocenter of triangle $ABC$ be $M$. From triangle $AMC$:

$$
\frac{AM}{b} - \frac{\cos \alpha}{\sin \beta}
$$

which gives

$$
AM = \frac{b \cos \alpha}{\sin \beta}
$$

but

$$
b = 2r \sin \beta
$$

so

$$
AM = 2r \cos \alpha
$$

Similarly,

$$
BM = 2r \cos \beta, \quad CM = 2r \cos \gamma
$$

Therefore, the volume of the parallelepiped is:

$$
K = AM \cdot BM \cdot CM = 8r^3 \cos \alpha \cos \beta \cos \gamma
$$

Substituting the given values: $K = 12.2 \, \text{cm}^3$.

(Géza Sasvári, Pécs.)

The problem was also solved by: Bayer B., Czank K., Faith F., Keesz J., Kerekes T., Krausz B., Krisztián Gy., Lukhaub Gy., Lupsa Gy., Messik G., Messik V., Perl Gy., Tézner E.