Given 4 points in space, which are not in the same plane. How many different parallelepipeds are there, for which these four points are vertices?
From the given 4 points of the sought parallelepiped,

a) a corner consisting of one vertex and the adjacent vertices,

b) a snake line consisting of three connected edges,

c) a kite consisting of 3 vertices of one face and a vertex adjacent to the fourth, or
d) a tetrahedron formed by 4 non-adjacent vertices

can be constructed. In the corner, the corner element itself plays a distinguished role, and any of the 4 points can be chosen, thus yielding 4 cases.

![](https://cdn.mathpix.com/cropped/2024_05_02_8646c68fa02b0a6a49c4g-1.jpg?height=519&width=1717&top_left_y=483&top_left_x=193)

The ends of the snake line can be chosen in $\binom{4}{2}=6$ ways, and the other two can be connected in 2 ways, giving 12 cases. For the kite, the isolated point can be chosen in 4 ways, and the corner from the other three in 3 ways, again yielding 12 cases. Finally, the tetrahedron gives only 1 case, and in total we have 29 cases. It is easily seen that the general position of the points ensures the existence and uniqueness of each of these cases.

Sparing László (Szombathely, Nagy Lajos Gymnasium)

Remark. If we add to the cases a), b), c), d) in the solution the case of 4 vertices lying on one face or on a diagonal plane, and also distinguish between right and left-handed snakes, we get 7 cases, which was one of the tasks in the first round of the 1976 student competition. In our case, the orientation of the snakes is determined by the position of the 4 given points, and we were only interested in the topological relationship of the points. We could have calculated as follows: from the 8 vertices of the parallelepiped, 4 can be chosen in $\binom{8}{4}=70$ ways, of which 6 are on one face and 6 on a diagonal plane, leaving 58. Each selection and its complement play the same role, so there are 29 cases. This reasoning may be correct, but as we said, it is not complete, it hangs in the air, and proving the correctness of this simple calculus would probably be more work than the above enumeration.