Problem 11-3. From $n$ regular hexagons with side 1, a polygon on the plane was made by gluing hexagons along their sides. Any two hexagons either have exactly one common side or have no common points at all. There are no holes inside the polygon. Moreover, each hexagon has at least one side lying on the boundary of the polygon. What is the smallest perimeter that the polygon can have under these conditions?
Answer: $2 n+6$ for $n \geq 2$, $6$ for $n=1$.

Solution. It is easy to understand that if a point is an endpoint for at least one side of a hexagon, it is an endpoint for two or three sides: more than three is impossible because all angles are $120^{\circ}$.

Let's draw a bit. Imagine that in each point that is an endpoint for three sides of a hexagon, we place a red dot. And each path along the sides of the hexagons from a red dot to a red dot we outline in blue. Thus, blue lines run along the sides of the hexagons and through vertices from which only two sides emerge (in other words, which are vertices of exactly one hexagon). Note that blue lines do not intersect except at their endpoints, which are red dots. It is easy to understand that if a point is an endpoint for at least one side of a hexagon, it is an endpoint for two or three sides: more than three is impossible because all angles are $120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed678a66967d87c9ed7cg-04.jpg?height=465&width=877&top_left_y=898&top_left_x=561)

Fig. 1: Example of a polygon made from hexagons.

Thus, we have obtained an image of a multigraph on the plane. For it, Euler's formula $F-E+V=2$ holds, where $F, E, V$ are the numbers of faces, edges, and vertices, respectively. Since all vertices have degree 3, $3V = 2E$. Additionally, $F = n+1$, since these are all the hexagons and the external face. From these three equations, we derive $E = 3n-3$. Let's walk along the outer cycle. In doing so, we walked along all $n$ hexagons, meaning we changed the hexagon we were walking on at least $n$ times (note: this statement is false if $n=1$: we just walked around one hexagon, never changing it). Thus, in the outer cycle, there are at least $n$ edges, meaning in the rest of the graph, there are no more than $2n-3$ edges. Each of these consists of exactly one segment that was a side for two hexagons, since inside the polygon, there cannot be points that are endpoints for exactly two side segments. Thus, inside, there are no more than $4n-6$ sides of hexagons, glued in pairs, meaning on the boundary, there are at least $2n+6$ sides.

| Rating | Score | Content Criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Minor errors in the proof |
| $+/-$ | 10 | Significant gaps in the proof (e.g., existence of a hexagon touching others by no more than 2 sides is not proven) |
| $+/2$ | 6 | There is a claim that the minimum perimeter for $n+1$ differs by 2, but the proof is missing. Example is provided. |
| $-/+$ | 4 | There is an example with $2n+6$ or ideas as in the $+/2$ criterion, but no clear example. |
| - | 0 | Solution is completely incorrect/only the answer |