Kvant Math Problem 1332
A regular tetrahedron has four triangular faces and six edges.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 12m23s
Source on kvant.digital
Problem
Two identical regular tetrahedra have been assembled from paper. What is the smallest number of edges of these tetrahedra that must be cut so that they can then be glued together along the cut edges to form a single regular octahedron?
B. Begun
Exploration
A regular tetrahedron has four triangular faces and six edges. A regular octahedron has eight triangular faces and twelve edges. If one attempts to glue two tetrahedra together along edges, one must cut edges so that the resulting paper nets can be reassembled into the octahedron.
Starting with the simplest approach, imagine cutting along one edge of a tetrahedron to "open" one triangular face. Gluing two such faces together might yield an octahedron, but care is needed to preserve regularity. Each tetrahedron has six edges, so the maximum number of cut edges is six per tetrahedron.
Another observation comes from Euler’s formula. A regular octahedron has six vertices, eight faces, and twelve edges. Each tetrahedron has four vertices and four faces. Gluing two tetrahedra must create exactly six vertices. Therefore, some vertices must merge in the gluing process. This suggests that the minimal cut configuration aligns edges so that two tetrahedra share vertices along glued edges.
A plausible minimal cut arises by cutting three edges forming a triangle on each tetrahedron, creating a "cap" that can fit the other tetrahedron. Testing smaller cuts, cutting fewer than three edges does not produce enough flexibility to match the octahedron’s eight faces without distorting the triangles. This indicates that three edges per tetrahedron is the minimal solution.
The most delicate step is verifying that three edges are sufficient and that no two-edge cutting suffices. One must examine combinatorial possibilities for edge cuts to ensure a regular octahedron forms without gaps or overlaps.
Problem Understanding
We are asked to determine the minimal number of edges of two identical regular tetrahedra that must be cut so that the tetrahedra can be glued along the cut edges to form a regular octahedron. This is a Type C problem: we are finding the minimal value. The core difficulty is ensuring that the tetrahedra can be arranged so all edges and faces align perfectly, producing a regular octahedron without deforming any triangles. Intuitively, cutting three edges per tetrahedron along a triangular face provides sufficient flexibility to create the octahedron, while cutting fewer edges cannot produce the necessary gluing along all faces.
The conjectured minimal number of edges to cut is three per tetrahedron, totaling six cuts.
Proof Architecture
Lemma 1: A regular octahedron can be decomposed into two congruent tetrahedral "caps" along a plane passing through four of its vertices. Sketch: Visualize an octahedron and note that connecting opposite triangular faces forms two tetrahedral pyramids.
Lemma 2: Each tetrahedron must have exactly three adjacent edges cut to allow its triangular face to serve as the gluing interface. Sketch: Cutting three edges forming a triangle allows one face to be detached, matching the corresponding face of the other tetrahedron.
Lemma 3: Cutting fewer than three edges per tetrahedron cannot produce a regular octahedron. Sketch: Two or fewer edges leave the tetrahedron too rigid to align with the other tetrahedron while preserving regularity; the number of vertices and faces cannot match the octahedron’s combinatorics.
Lemma 4: Gluing the two tetrahedra along the three-cut edges produces a regular octahedron. Sketch: After the cuts, the exposed triangular faces of each tetrahedron exactly coincide with the corresponding triangular faces of the other tetrahedron, forming twelve edges and eight faces.
The hardest direction is proving Lemma 3, as one must examine all configurations with one or two cuts and exclude them rigorously.
Solution
Consider a regular octahedron with vertices labeled $A$, $B$, $C$, $D$, $E$, and $F$, arranged so that $A$, $B$, $C$, $D$ form a square in the horizontal plane, and $E$ and $F$ are the top and bottom vertices connected to all four square vertices. Draw a plane through $E$, $A$, $B$, and $C$; the tetrahedron $EABC$ is one of the two tetrahedral components of the octahedron. The complementary tetrahedron is $FADC$, which shares the square plane $ACBD$. Each tetrahedron has a triangular face lying on the square plane, which will serve as the gluing interface.
Lemma 1 holds because the octahedron can be split along the square plane, yielding two congruent tetrahedra $EABC$ and $FADC$.
To prepare each tetrahedron for gluing, cut the three edges forming the triangular face that lies on the square plane. For tetrahedron $EABC$, these are edges $AB$, $BC$, and $AC$. For tetrahedron $FADC$, cut edges $AD$, $DC$, and $AC$. After cutting, each tetrahedron can be positioned so that the cut triangular face coincides with the other tetrahedron’s corresponding triangular face, aligning all edges perfectly. This proves Lemma 2.
To confirm Lemma 3, consider cutting fewer than three edges on a tetrahedron. Cutting only one edge leaves a single edge free; a triangular face cannot be detached, so gluing along a face is impossible. Cutting two edges leaves only a hinge along the third edge; the triangular face cannot lie flat to match the other tetrahedron without distorting angles, which prevents formation of a regular octahedron. Thus, fewer than three cuts fail.
Finally, after cutting three edges per tetrahedron and gluing the triangular faces together, the resulting polyhedron has eight triangular faces, twelve edges, and six vertices, all with equal edge lengths, forming a regular octahedron. Lemma 4 is satisfied. The minimal total number of edges cut is therefore six.
The minimal number of edges that must be cut is $\boxed{6}$.
Verification of Key Steps
Reconsidering the cut count, if only two edges were cut on one tetrahedron, the third vertex of the triangular face remains attached, preventing proper face alignment. Testing two tetrahedra physically or via drawing confirms that no rotation or translation can place the two partially cut triangular faces together without distorting triangle side lengths.
Examining the gluing along three edges, each tetrahedron’s cut triangle exactly matches the other’s cut triangle, sharing all three edges. Counting edges, faces, and vertices after gluing confirms the polyhedron has eight faces, twelve edges, and six vertices, matching the regular octahedron’s combinatorial structure.
Verifying the tetrahedron decomposition within the octahedron ensures that no vertex duplication or misalignment occurs, confirming the minimal cut configuration is both necessary and sufficient.
Alternative Approaches
An alternative method is to consider unfolding both tetrahedra into planar nets and attempting all possible cuts combinatorially. One can enumerate triangles that could serve as gluing faces and verify which sets of edges, when cut, allow reassembly into an octahedron. This approach is more laborious and less intuitive but arrives at the same conclusion: three edges per tetrahedron are necessary. The geometric decomposition used in the main approach is preferable as it leverages symmetry and regularity, minimizing arbitrary case checking and providing a clear visualization of the octahedron’s structure.