7. (4 points) Four out of six midpoints of the edges of a certain tetrahedron form a regular tetrahedron with an edge length of 1. Find the edges of the original tetrahedron.
Answer: Five edges of length 2, one edge of length $2 \sqrt{2}$.

Solution:

Let the vertices of the tetrahedron be $A, B, C, D$. Denote the midpoint of edge $A B$ as $M_{A B}$, and similarly for the other edges.

Suppose the midpoints of the skew edges of the original tetrahedron do not fall into the regular tetrahedron, without loss of generality, we can assume these are $M_{A B}$ and $M_{C D}$. Note that the segment $M_{B C} M_{B D}$ is the midline of triangle $B C D$, so it is half the length of edge $C D$ and parallel to this edge. The same can be said about the edge $M_{A C} M_{A D}$, meaning these segments are parallel and equal, thus forming a parallelogram. Therefore, the four vertices that should form a regular tetrahedron lie in the same plane.

Thus, the midpoints of adjacent edges of the tetrahedron did not fall into the regular tetrahedron, without loss of generality, we can assume these are $M_{B D}$ and $M_{C D}$. Triangle $M_{A B} M_{A C} M_{B C}$ is the medial triangle of triangle $A B C$, and thus similar to it with a coefficient of $\frac{1}{2}$, from which $A B=A C=B C=2$. Additionally, $B D=2 M_{A B} M_{A D}=2$ and $C D=2 M_{A C} M_{A D}=2$ since $M_{A B} M_{A D}$ and $M_{A C} M_{A D}$ are the midlines of triangles $A B D$ and $A C D$ respectively.

It remains to find the length of edge $A D$. Let $X$ be the point of intersection of the medians of triangle $A B C$ (and, obviously, of triangle $M_{A B} M_{A C} M_{B C}$ as well). Then $M_{A D} X$ is the height of the regular tetrahedron with edge 1, that is, equal to $\sqrt{\frac{2}{3}}$; at the same time, $A X$ is $\frac{2}{3}$ of the median of triangle $A B C$, that is, $\frac{2}{\sqrt{3}}$. Moreover, they are perpendicular, since $M_{A D} X$ is the height of the tetrahedron and perpendicular to the entire plane $A B C$, therefore, by the Pythagorean theorem $M_{A D} A^{2}=M_{A D} X^{2}+A X^{2}=\frac{6}{3}=2$. And $A D=2 M_{A D} A=2 \sqrt{2}$.