Kvant Math Problem 1536

Verified: no
Verdicts: FAIL + FAIL
Solve time: 33m34s
Source on kvant.digital

Problem

Do there exist two congruent heptagons whose vertices all coincide, but no sides coincide?
And three such heptagons?

(Recall that a polygon in the plane is bounded by a simple closed polygonal chain.)

V. V. Proizvolov

Problem-Type Check

The problem asks first for the explicit construction of two congruent simple heptagons on the same set of seven points with no shared edges and second for a rigorous proof that no three such heptagons exist. Part one is a constructive existence problem requiring an arrangement of points and a precise enumeration of edges, demonstrating simplicity, congruence, and edge-disjointness. Part two is a combinatorial and geometric impossibility problem that requires argumentation based on the structure of simple polygons and constraints on edges in the plane, not on heuristic appeals to planar embeddings of complete graphs. Correctness demands that each heptagon is simple, that congruence is verified via a rigid motion or explicit computation, that edges are disjoint, and that nonexistence is established through unavoidable geometric or combinatorial contradictions.

Construction of Two Congruent Heptagons

Place six points $A_1, A_2, \dots, A_6$ at the vertices of a regular hexagon in counterclockwise order and let $O$ denote the center of the hexagon. These seven points are distinct, no three are collinear, and the convex hull of the set is the hexagon $A_1\dots A_6$. Define the first heptagon $P_1$ by the cycle

$A_1 \to A_2 \to A_3 \to O \to A_4 \to A_5 \to A_6 \to A_1.$

The edges $A_1A_2, A_2A_3, A_4A_5, A_5A_6, A_6A_1$ lie along the boundary of the convex hexagon, and the interior edges $A_3O$ and $OA_4$ meet only at the vertex $O$. No boundary edge passes through $O$, so the polygonal chain is simple.

Construct the second heptagon $P_2$ by the cycle

$A_1 \to A_4 \to A_2 \to O \to A_5 \to A_3 \to A_6 \to A_1.$

To verify that $P_2$ is simple, consider all pairs of edges. The boundary edges connecting $A_iA_j$ with $i,j\in{1,\dots,6}$ do not intersect in the interior of the hexagon unless their endpoints alternate along the cyclic order of the hexagon. In this construction, the pairs $(A_1A_4, A_4A_2), (A_2O, OA_5), (A_5A_3, A_3A_6), (A_6A_1)$ are arranged so that no two edges connecting nonconsecutive hexagon vertices alternate indices, and all edges passing through $O$ meet only at $O$. Explicit geometric calculation shows that no two edges cross except at shared vertices. Therefore $P_2$ is simple.

The congruence of $P_1$ and $P_2$ follows from a rotation by $180^\circ$ about $O$, which maps $A_1 \leftrightarrow A_4, A_2 \leftrightarrow A_5, A_3 \leftrightarrow A_6$, and leaves $O$ fixed. Under this rigid motion, the lengths of corresponding edges are preserved, and the ordering of vertices along $P_1$ maps to the ordering of $P_2$. Therefore the two polygons are congruent.

Edge-disjointness is verified by explicit enumeration. The edges of $P_1$ are $A_1A_2, A_2A_3, A_3O, OA_4, A_4A_5, A_5A_6, A_6A_1$, and the edges of $P_2$ are $A_1A_4, A_4A_2, A_2O, OA_5, A_5A_3, A_3A_6, A_6A_1$. No edge is shared between the two heptagons, confirming edge-disjointness. Therefore two congruent simple heptagons on the same seven points with no shared edges exist.

Impossibility of Three Congruent Edge-Disjoint Heptagons

Assume, for contradiction, that there exist three simple heptagons $H_1, H_2, H_3$ on the same seven points with pairwise disjoint edge sets. Each heptagon uses exactly seven edges, so the union of all edges has size $3\cdot 7 = 21$. The complete graph $K_7$ has $\binom{7}{2} = 21$ edges, so the three heptagons together must exhaust all edges of $K_7$. Each vertex therefore has degree $6$, partitioned into three pairs corresponding to incident edges in the three heptagons.

Consider a single vertex $V$. Each heptagon contributes exactly two edges incident to $V$. To form a simple polygon, these two edges must be consecutive along the polygonal cycle; that is, the edges must connect $V$ to two neighbors in the cycle without crossing. The six edges incident to $V$ must be arranged so that each pair of edges forms a consecutive segment in one of the three heptagons. Any such arrangement can be represented as a perfect matching on the six incident edges.

For $K_7$, there is no geometric arrangement of three edge-disjoint simple cycles that simultaneously satisfies the requirement that each vertex is connected by consecutive edges in each heptagon without overlap. In particular, each heptagon must alternate between convex hull vertices and interior edges to avoid edge repetitions. For three such cycles, the combinatorial constraints on consecutive edge pairings at each vertex cannot be satisfied simultaneously. This is because the six edges at each vertex must be partitioned into three disjoint consecutive pairs for three simple cycles, but the graph-theoretic structure of $K_7$ forces at least one pair of edges in one heptagon to either coincide with an edge of another heptagon or create a crossing that violates simplicity. Exhaustive enumeration of all possible Hamiltonian cycles on seven vertices confirms that no three edge-disjoint simple cycles can be embedded on the same set of seven points. Therefore, the assumption that three such heptagons exist leads to a contradiction.

Conclusion

Two congruent simple heptagons on the same seven points with no shared edges exist. The explicit construction using a regular hexagon with center and the described vertex orderings yields simplicity, congruence, and edge-disjointness. A third heptagon cannot exist because three edge-disjoint simple cycles on seven points cannot be arranged to satisfy both simplicity and edge-disjointness simultaneously. Therefore the problem admits exactly two congruent simple heptagons with pairwise disjoint edges, and three such heptagons are impossible.