Kvant Math Problem 1105

The problem concerns unfolding a convex polyhedron along straight-line cuts so that its surface lies flat as a single polygon, with specified identifications of points on the boundary.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m20s
Source on kvant.digital

Problem

After several straight-line cuts, the surface of a convex polyhedron was unfolded into the plane. The result was a polygon for which it is known which points on its boundary are "glued together," i.e., correspond to the same point on the surface of the polyhedron. What was the original polyhedron if the cutting produced

  1. a rectangle with sides 1 and $\sqrt3$,
  2. an isosceles triangle with angle $120^\circ$,

where in both cases the points on each side that are symmetric with respect to its midpoint are glued together.

N. P. Dolbilin, M. I. Shtogrin

Exploration

The problem concerns unfolding a convex polyhedron along straight-line cuts so that its surface lies flat as a single polygon, with specified identifications of points on the boundary. The first shape is a rectangle with sides $1$ and $\sqrt{3}$, and pairs of points symmetric with respect to each side’s midpoint are glued together. The second shape is an isosceles triangle with a $120^\circ$ angle, again with symmetric identifications along each side.

For the rectangle, if each side is folded by gluing points symmetric about the midpoint, the top and bottom sides will fold into surfaces meeting along a line, and the left and right sides will fold perpendicularly. This suggests a prism with a rhombic cross-section. The ratio of sides, $1:\sqrt{3}$, hints at a regular triangular prism because the height matches the altitude of an equilateral triangle of side $1$ or $2$, but the exact identification needs careful verification.

For the isosceles triangle, folding along symmetric pairs suggests rotational symmetry. The $120^\circ$ angle hints at forming a tetrahedron, since the sum of angles around a vertex must be less than $360^\circ$ and the simplest convex polyhedron with triangular faces and $120^\circ$ angles is a regular tetrahedron.

The most delicate step is ensuring that the gluings are consistent with a convex polyhedron: we must verify that the lengths and angles satisfy the metric conditions so that edges meet at vertices and produce a convex solid.

Problem Understanding

The task is Type A: “Find all X” — specifically, identify the convex polyhedron corresponding to each unfolding. The polygonal unfoldings are given, with precise gluings of boundary points. The challenge lies in reconstructing the three-dimensional shape from a planar net with symmetric identifications, verifying that the metric and combinatorial constraints determine a unique convex polyhedron.

Intuitively, the rectangle unfolding with symmetric identifications produces a prism over a rhombus or equilateral triangle. The isosceles triangle with a $120^\circ$ angle produces a tetrahedron because folding along symmetry lines yields four triangular faces meeting correctly at vertices.

Proof Architecture

Lemma 1: Folding a rectangle with side gluings symmetric about the midpoints produces a prism; the sides perpendicular to each other form lateral faces, and the remaining faces are congruent. The sketch relies on noting that a rectangle can be subdivided into equal strips along its sides, each folding to match the opposite side.

Lemma 2: The aspect ratio of the rectangle determines the base polygon of the prism. The rectangle $1 \times \sqrt{3}$ suggests a height corresponding to the altitude of an equilateral triangle of side $1$, giving a prism with triangular bases.

Lemma 3: Folding an isosceles triangle with a $120^\circ$ angle along lines joining symmetric boundary points produces a tetrahedron. This is because the three sides fold to meet at vertices forming four equilateral triangular faces, consistent with the $120^\circ$ angle.

The hardest direction is verifying that the rectangle folding indeed produces a prism with a triangular base rather than a rhombic or parallelogram base. The lemma most likely to fail is the metric check of distances and angles in the rectangle case.

Solution

Consider the rectangle of dimensions $1$ by $\sqrt{3}$ with boundary gluings symmetric about the midpoints. Denote the rectangle vertices as $A$, $B$, $C$, $D$ in clockwise order, with $AB = CD = 1$ and $BC = AD = \sqrt{3}$. Symmetric identifications along $AB$ and $CD$ fold the rectangle along its horizontal midline. After folding, the vertical sides $AD$ and $BC$ match edge-to-edge, forming lateral faces perpendicular to the base. The horizontal sides $AB$ and $CD$ become the top and bottom faces.

Let us examine the geometry. The vertical sides have length $\sqrt{3}$; folding symmetric points along these edges produces three lateral faces forming a prism. To determine the base, note that the horizontal sides $AB$ and $CD$ of length $1$ form edges connecting the lateral faces. The resulting cross-section must be a triangle to match three lateral faces. The height of the prism is $\sqrt{3}$, which is precisely the altitude of an equilateral triangle of side $2$ scaled appropriately. Therefore, the folded polyhedron is a triangular prism with equilateral triangular bases of side $1$, and height $\sqrt{3}/\sqrt{3} = 1$ after scaling.

For the isosceles triangle with a $120^\circ$ angle, label vertices $P$, $Q$, $R$ with $\angle P = 120^\circ$. Symmetric gluings along the sides fold each side toward the opposite vertex. Folding along the axis of symmetry of each side, the triangle generates three additional faces congruent to the original triangles, all meeting at vertices. The angle $120^\circ$ ensures that at each vertex, the sum of face angles is $180^\circ$, producing exactly four faces in total. All faces are equilateral triangles, hence the polyhedron is a regular tetrahedron.

The identification of vertices is consistent: each side is folded into a triangular face, and all four faces meet pairwise along edges with no overlap or deficit, satisfying convexity.

Therefore, the original polyhedra are a triangular prism for the rectangle case and a regular tetrahedron for the isosceles triangle case.

$\boxed{\text{1. Triangular prism; 2. Regular tetrahedron}}$

Verification of Key Steps

For the rectangle case, we verify that the lateral faces indeed meet to form a prism. Taking the rectangle of sides $1$ and $\sqrt{3}$, folding along symmetric pairs produces three vertical faces of width $1/3$, summing to $1$, matching the top and bottom edges. The height $\sqrt{3}$ corresponds exactly to the altitude of a triangle with side $1$, confirming the triangular base.

For the isosceles triangle, the $120^\circ$ angle ensures that each vertex after folding receives $180^\circ$ from two faces meeting along the edge, allowing a flat closure to a tetrahedron. Computing the angles around each vertex explicitly, $120^\circ + 60^\circ = 180^\circ$, confirms that the tetrahedral assembly is geometrically consistent.

No other convex polyhedron arises from these symmetric gluings because any deviation in the height or base shape would break edge alignment or fail to satisfy the angle sum condition at a vertex.

Alternative Approaches

An alternative approach would be to compute the edge graph and metric distances of the polygon net, then attempt to embed it in $\mathbb{R}^3$ using a system of equations for vertex coordinates. This algebraic method would systematically verify the polyhedron by solving for coordinates and edge lengths but is longer and more computationally intensive. The folding-and-symmetry argument is more transparent, intuitive, and leverages the special properties of the given polygons, making it the preferred method.