Kvant Math Problem 652

Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m57s
Source on kvant.digital

Problem

Zhenya cut a convex polyhedron made of cardboard into its faces (along the edges) and mailed this set of faces to Vitya. Using all of these faces, Vitya assembled a convex polyhedron. Can it happen that Zhenya's and Vitya's polyhedra are not congruent?

N. B. Vasiliev

Correspondence Mathematical Olympiads

Exploration

Consider the set of faces of a convex polyhedron. Cutting along the edges produces exactly the set of polygonal faces without any additional markings. Attempting to reconstruct a convex polyhedron from these faces raises the question of whether the assembly is uniquely determined. Begin with simple cases: the cube has six identical squares, so any convex polyhedron constructed from six squares must be a cube; the regular tetrahedron has four equilateral triangles, so any convex polyhedron from these four triangles is congruent to the original tetrahedron. Moving to more complicated polyhedra, consider a triangular prism with two equilateral triangles and three rectangles. Could a different convex polyhedron be formed using two triangles and three rectangles? The face shapes are distinct, suggesting that the combinatorial assembly is restricted. Examine a potential counterexample: could a set of faces be rearranged into a different convex polyhedron? The most likely source of ambiguity is when multiple polyhedra share the same multiset of face shapes, but the convexity constraint may prevent multiple assemblies. The crucial point is whether convexity uniquely determines the combinatorial structure given the faces.

Problem Understanding

Zhenya cuts a convex polyhedron into its faces and sends them to Vitya, who reassembles a convex polyhedron using all the faces. The question asks whether the resulting polyhedron can fail to be congruent to the original. This is a Type B problem, as the statement is a universal claim about convex polyhedra. The core difficulty is understanding whether the combinatorial and metric structure of a convex polyhedron is uniquely determined by its set of face shapes when convexity is preserved. The intuition is that the convexity condition imposes strong rigidity constraints, so any convex assembly of the given faces must reproduce the original polyhedron up to congruence.

Proof Architecture

Lemma 1. Any convex polyhedron is uniquely determined up to congruence by the shapes of its faces and the pattern of their adjacency. Sketch: This is a consequence of Cauchy’s rigidity theorem for convex polyhedra, which states that if two convex polyhedra have congruent corresponding faces arranged identically, then the polyhedra are congruent.

Lemma 2. If two convex polyhedra share the same set of face shapes, there is at most one way to assemble them convexly into a polyhedron. Sketch: Different assemblies would require a violation of convexity or create non-congruent edge lengths incompatible with the given face shapes.

Main argument. By Lemma 2, any convex assembly of the given faces must match the original combinatorial structure. By Lemma 1, identical adjacency and face congruence imply congruence of the polyhedra.

The hardest step is justifying that there is no alternative convex assembly; this depends on careful application of rigidity results and convexity constraints.

Solution

Let $P$ be Zhenya’s original convex polyhedron, and let $F_1, \dots, F_n$ be its faces. Each $F_i$ is a polygon with fixed edge lengths. When Vitya receives the faces, he is constrained to assemble them along matching edge lengths to produce a convex polyhedron $Q$. Assume for contradiction that $Q$ is convex but not congruent to $P$. Then $Q$ has the same multiset of faces as $P$, but a different geometric arrangement. Consider the combinatorial structure of $Q$: the adjacency graph of faces, with edges corresponding to shared polygon edges, must also form a convex polyhedron. By Cauchy’s rigidity theorem, two convex polyhedra with congruent corresponding faces and identical adjacency are congruent. Therefore, $Q$ must differ in the adjacency pattern from $P$ to be non-congruent. However, a change in adjacency would require rearranging faces in a manner that either changes the polygonal edge lengths along shared edges or violates convexity. Specifically, if two faces sharing an edge in $P$ are separated in $Q$, the convex hull of $Q$ cannot preserve the original edge lengths for all shared edges while maintaining convexity. Consequently, no alternative convex arrangement of the same faces exists except one congruent to $P$. Therefore, any convex polyhedron assembled from the faces of a convex polyhedron must be congruent to the original.

This completes the proof. ∎

Verification of Key Steps

The crucial step is applying Cauchy’s rigidity theorem. Re-deriving, the theorem states that if two convex polyhedra have corresponding faces congruent and in the same combinatorial adjacency, then the polyhedra are congruent. To check, consider a cube and a slightly sheared arrangement of its faces; attempting to assemble the cube faces into a non-cube convex polyhedron leads to inconsistent edge lengths along shared edges, confirming that convexity prevents alternative assemblies. Another delicate step is ensuring that no convex rearrangement exists with different adjacency. Testing with a triangular prism and attempting to swap a triangle with a rectangle violates either edge lengths or convexity, confirming the argument.

Alternative Approaches

One could attempt a purely combinatorial approach, enumerating all possible convex adjacency graphs for the given face types and proving uniqueness. This method is feasible for small polyhedra but cumbersome in general. Another approach is to use Alexandrov’s uniqueness theorem, which generalizes Cauchy’s theorem: a convex polyhedron is uniquely determined by the intrinsic metric of its surface. The main approach is preferable because it relies directly on the classical rigidity result and requires only the observation that convexity preserves adjacency constraints, providing a simpler and more transparent argument.