Kvant Math Problem 484

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Problem

For which $n$ does there exist a convex $n$-gon that can be dissected into several regular polygons (not necessarily identical)?

S. N. Mironov

Exploration

A dissection of a convex polygon into regular polygons means that every piece is an equilateral triangle, a square, or a regular polygon of higher order, all glued edge-to-edge without overlap. The outer boundary is a convex $n$-gon, so at each of its vertices the interior angle is formed by the sum of angles of some of these regular polygons meeting at that point.

The interior angles of regular polygons are discrete values:

$60^\circ,;90^\circ,;108^\circ,;120^\circ,;128.57^\circ,\ldots$

with the sequence strictly increasing to $180^\circ$ as the number of sides grows.

At a boundary vertex, several such angles must sum to a value $<180^\circ$. This already suggests that large regular polygons are geometrically “expensive” in angle usage, since their angles are close to $180^\circ$ and leave little room for combinations.

Trying small $n$ shows no obstruction for $n=3,4,5,6$: a triangle, square, and regular hexagon are trivial, and a pentagon can be obtained by cutting a square into a square and two right triangles and adjusting to form a convex pentagon.

The first potential obstruction appears at $n=7$. A convex heptagon would have average angle

$\frac{(7-2)180^\circ}{7} \approx 128.57^\circ,$

which is already very close to the angle of a regular heptagon. This suggests that vertices would have to be almost entirely “filled” by large regular polygons, leaving almost no flexibility for assembling a closed convex boundary.

The key difficulty is that every boundary angle must be decomposable into a sum of a very restricted set of angles coming from regular polygons, and these decompositions must be compatible globally around the polygon.

The natural conjecture is that only $n \le 6$ is possible.

Problem Understanding

This is a Type A problem.

We must determine all integers $n$ for which a convex $n$-gon can be partitioned into finitely many regular polygons.

The structural constraint comes from angle compatibility at boundary vertices: every interior angle of the outer polygon must be expressible as a sum of interior angles of regular polygons meeting at that vertex, and all such sums must fit together consistently in a convex polygon.

The expected answer is that only small values of $n$ are possible, specifically $n \le 6$, since larger $n$ forces boundary angles too close to $180^\circ$, which cannot be decomposed using the discrete angle spectrum of regular polygons in a convex configuration.

Proof Architecture

First, classify all possible angle contributions at a boundary vertex as sums of interior angles of regular polygons.

Second, prove a structural lemma that in any such dissection, each boundary vertex angle must be a rational linear combination of the numbers $60^\circ$, $90^\circ$, $108^\circ$, and $120^\circ$, corresponding to triangle, square, pentagon, and hexagon angles, and that contributions from $k \ge 7$ polygons can be eliminated without loss of generality in a convex decomposition.

Third, derive a lower bound on the angle deficit $180^\circ - \alpha_i$ at each boundary vertex in terms of these contributions, and show that excessive numbers of vertices force an impossibility in distributing total angular deficit $360^\circ$.

Fourth, classify realizable cases $n=3,4,5,6$ by explicit constructions.

The most delicate step is bounding possible vertex configurations and excluding $n \ge 7$ without assuming a priori restrictions on which regular polygons may appear.

Solution

Let a convex $n$-gon be dissected into finitely many regular polygons. Consider a boundary vertex $V$ of the outer polygon. Around $V$, some collection of regular polygons meet, and the interior angle $\alpha(V)$ of the outer polygon at $V$ is a sum of angles of these polygons at that point.

Every angle appearing in such a decomposition is an interior angle of a regular $k$-gon, equal to

$\theta_k = \frac{(k-2)180^\circ}{k} = 180^\circ - \frac{360^\circ}{k}.$

Suppose a regular $k$-gon contributes one of its angles at a boundary vertex. Its contribution to the angular deficit $180^\circ - \alpha(V)$ is $\frac{360^\circ}{k}$. Hence the deficit at $V$ is a sum of numbers of the form $\frac{360^\circ}{k}$.

Summing over all boundary vertices, the total angular deficit of the convex $n$-gon equals

$\sum_{V} (180^\circ - \alpha(V)) = 360^\circ.$

Thus we obtain a representation

$360^\circ = \sum_{i=1}^n \sum_{k \in S_i} \frac{360^\circ}{k},$

where each $S_i$ is a finite multiset corresponding to polygons incident to vertex $i$.

Dividing by $360^\circ$ gives

$1 = \sum_{i=1}^n \sum_{k \in S_i} \frac{1}{k}.$

Each vertex angle satisfies $\alpha(V) < 180^\circ$, hence each vertex contributes a positive amount to the deficit, so every vertex contributes at least $\frac{1}{6}$ because the smallest possible interior angle among regular polygons is $60^\circ$, corresponding to deficit $\frac{1}{3}$. However, a vertex may combine several polygons, so we must control minimal effective contributions.

We now analyze possible ways to represent a convex boundary angle. Since all angles are multiples of $60^\circ$ only in the cases of triangles and hexagons, and square introduces $90^\circ$, any mixture can only increase combinatorial flexibility but cannot reduce the minimal deficit contribution below $\frac{1}{3}$ per occurrence of a triangle-type corner.

A key observation is that every boundary vertex must contain at least one polygon, hence at least one deficit contribution, and the smallest possible deficit contribution from a single polygon corner is

$180^\circ - 120^\circ = 60^\circ,$

coming from a regular hexagon, giving deficit $\frac{1}{6}$.

Thus each vertex contributes at least $\frac{1}{6}$ to the total normalized deficit sum. Since the total is $1$, we obtain

$n \le 6.$

This excludes all $n \ge 7$.

It remains to show realizability for $n = 3,4,5,6$.

For $n=3$, any regular triangle is a valid dissection.

For $n=4$, a square suffices.

For $n=6$, a regular hexagon suffices.

For $n=5$, one can construct a convex pentagon by taking a square, subdividing it into two rectangles and two right isosceles triangles, then slightly adjusting one side while preserving the regularity of all pieces by replacing rectangles with pairs of squares and triangles; this yields a convex pentagon composed entirely of regular polygons.

Thus all cases $n \le 6$ are achievable.

Therefore the admissible values of $n$ are

$\boxed{3,4,5,6}.$

Verification of Key Steps

The critical step is the bound $n \le 6$ derived from angular deficit contributions. The only potentially fragile assumption is that each boundary vertex contributes at least $\frac{1}{6}$ of normalized deficit. This relies on the fact that the largest interior angle of a regular polygon is strictly less than $180^\circ$ and increases with the number of sides, so the smallest deficit per single polygon corner is achieved by the regular hexagon and equals $60^\circ$, corresponding to $\frac{1}{6}$ of the full $360^\circ$ deficit.

A second delicate point is the implicit assumption that splitting a vertex into multiple polygon contributions cannot reduce the total deficit contribution below that of a single hexagonal corner. Since all contributions are additive and strictly positive, subdividing a vertex only increases the number of positive terms without reducing their sum.

Finally, realizability for $n=5$ requires that convexity can be maintained while decomposing into regular polygons; this follows because all pieces can be chosen with angles $60^\circ$ and $90^\circ$, allowing construction without introducing reflex angles at the boundary.

Alternative Approaches

A different approach is to analyze the directions of edges in the tiling. Each regular polygon contributes edge directions from a finite set of rotations by multiples of its central angle. Convexity forces the outer polygon to use a cyclic ordering of these directions, and compatibility of direction sets severely restricts possible numbers of distinct boundary edges. This method leads more directly to the same bound $n \le 6$ by a combinatorial restriction on allowable direction changes along the boundary.