Let's consider one vertex of a cube and look at the rays leading from it to the other vertices. How many different magnitudes of angles do we get if we pair these rays in every possible way?
Let one face of the cube be $ABCD$, the third edges starting from its vertices be $AE, BF, CG, DH$, and the selected vertex be $A$. Among the 7 rays leading to the other vertices, 3 are edges of the cube: $AB, AD$, and $AE$, 3 are face diagonals: $AC, AF$, $AH$, and the seventh, $AG$, is a space diagonal. The mentioned 3 edges can be transformed into each other by $120^{\circ}$ rotations around $AG$, and the same applies to the mentioned face diagonals, so it is sufficient to determine the magnitudes of the angles whose one side is $AB$ or $AC$.

$AB$ forms a right angle with $AD$ and $AE$, and similarly, the diagonal $AH$ does as well. The diagonals $AC$ and $AF$ form a $45^{\circ}$ angle, because, for example, $ABF$ is an isosceles right triangle. Finally, in the right triangle $ABG$, if $AB$ is taken as a unit, then $BG = \sqrt{2}$, so the tangent of the angle $BAG$ is $\sqrt{2}$. (This angle is clearly not $45^{\circ}$.)

The angles formed by $AC$ with the edges are already known from the above. The angle $CAF$ is $60^{\circ}$, because the triangle $CAF$ is equilateral (face diagonals). The angle $CAG$ is such that its tangent is $1 / \sqrt{2}$, since the triangle $CAG$ is congruent to the previously mentioned triangle $FAG$. According to this, this angle differs from the previous ones, because $\tan 60^{\circ} = \sqrt{3}$.

According to these, 5 different angles appear among the pairs formed by the 7 rays drawn from $A$ to the other vertices: $90^{\circ}, 60^{\circ}, 45^{\circ}$, and those whose tangents are $\sqrt{2}$ and $1 / \sqrt{2}$. (Their values in degrees are $54^{\circ} 44'$ and $35^{\circ} 16'$.)

György Ponácz (Székesfehérvár, József A. Gymnasium II. o. t.)