Kvant Math Problem 9

Consider a tetrahedron with vertices $A$, $B$, $C$, and $D$.

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Problem

Consider the following properties of a tetrahedron (by a tetrahedron we mean an arbitrary triangular pyramid):

  1. all faces have equal area;
  2. each edge is equal to its opposite edge;
  3. all faces are congruent;
  4. the centers of the circumscribed and inscribed spheres coincide;
  5. the sums of the plane angles at each vertex of the tetrahedron are equal.

Prove that all these properties are equivalent.

Try to find other properties of a tetrahedron that are equivalent to them.

Exploration

Consider a tetrahedron with vertices $A$, $B$, $C$, and $D$. Each property describes a form of symmetry: equal face areas, equality of opposite edges, congruence of all faces, coincidence of circumcenter and incenter, and equality of angle sums at vertices. For a regular tetrahedron, all these properties are satisfied, suggesting that the equivalence involves the tetrahedron being "regular" in some sense.

Testing small variations, if a tetrahedron has all faces congruent but edges not paired equally, one quickly finds the faces cannot remain congruent. Similarly, if all faces have equal area but differing shapes (scalene triangles), opposite edges are not equal, and the tetrahedron fails to be equifacially symmetric. The property of circumcenter and incenter coinciding is known to characterize tetrahedra whose faces are congruent and symmetrically arranged. Thus, the crucial step is showing that the weaker conditions, such as equal face areas or equality of sums of angles at vertices, are sufficient to force full regularity or edge-pair equality.

Problem Understanding

The problem asks to prove that five different conditions on a tetrahedron are equivalent. It is a Type A problem: "Find all X," where X is the set of tetrahedra satisfying these properties. The core difficulty is showing that weaker-seeming properties, like equality of face areas or equality of vertex angle sums, actually imply complete symmetry in the edge lengths and face congruences. The answer is