Let's determine the vertices of the tetrahedron, given the midpoints of four of its edges!
A tetrahedron has six edges from which we can choose four in two different ways: either three of them lie in the same plane, or no three of them lie in the same plane. In the first case, the midpoints are not in the same plane, while in the second case, they not only lie in the same plane but also form a parallelogram. Accordingly, we distinguish between two cases.

![](https://cdn.mathpix.com/cropped/2024_05_02_c95e59371653796ece64g-1.jpg?height=1271&width=979&top_left_y=346&top_left_x=562)

a) The given points are not in the same plane. Choose any three of them and draw lines through the vertices parallel to the opposite sides in the triangle they define. This gives us the three edges that lie in the same plane. By connecting any vertex of the resulting face to the fourth given point and measuring the segment from the vertex to the point again on the resulting line, we obtain the fourth vertex of the tetrahedron. The correctness of our procedure and the fact that we obtain all possible tetrahedra is evident. We get a total of 12 tetrahedra, because we can choose from 4 cases first and 3 cases second.

b) The given points lie in the same plane. In this case, there is a solution only if the points form a parallelogram. By adding a segment that is not in the same plane as the diagonals and whose midpoint is the same as the midpoint of the diagonals, we obtain the two missing midpoints of the edges. Now we have three segments, \(PQ\), \(RS\), \(TU\), which are not in the same plane, but their midpoints are the same. Draw a line through \(P\) parallel to \(RU\), and measure the segment \(RU\) from \(P\) in both directions. Draw a line through \(Q\) parallel to \(RT\), and measure the segment \(RT\) from \(Q\) in both directions. Let the resulting endpoints be \(A\), \(B\), \(C\), \(D\), the midpoint of \(RU\) be \(D_1\), and the midpoint of \(ST\) be \(C_1\). Since \(C_1D_1\) also bisects \(PQ\) like \(RS\) and \(TU\), and is parallel to \(CD\), and half its length, \(C_1D_1\) is the midline of the triangle \(PCD\). Therefore, \(C_1\) bisects \(PC\), and \(D_1\) bisects \(PD\). Consequently, \(RU\) is the midline of the triangle \(ABD\), and \(ST\) is the midline of the triangle \(ABC\), as they bisect \(PD\) and \(PC\), respectively, are parallel to \(AB\), and are half the length of \(AB\). Thus, in the tetrahedron \(ABCD\), \(P\), \(Q\), \(R\), \(S\), \(T\), \(U\) are indeed midpoints of the edges.

In this case, there are infinitely many solutions.