You have an unlimited supply of regular polygons, each with side length 1: triangles, squares, pentagons, etc. Some combinations of these polygons may be arranged in a “cycle” so that they all share a common vertex and any consecutive pair of polygons share exactly a common edge. (See the example below.) What is the largest *n*-gon in such a cycle?

Example using 3 triangles and 2 squares.

Hint (sort of): There is a much better title for this post, but unfortunately it would give away the solution to the problem.

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Assuming the cycle must contain at least 3 polygons (otherwise any two arbitrarily large n-gons would do), I’m going to say the largest is a dodecagon (12), where the cycle is two 12-gons and a triangle.

You are right that the cycle must contain at least 3 polygons. It is also not too difficult to show that there can be *at most* 6 polygons in a cycle, making an exhaustive enumeration of possible configurations feasible but interesting. As an additional hint, it turns out that the maximum number of sides is realized with just 3 polygons, but each with a different number of sides.

A square, a pentagon and a 20 sided polygon (forgive me for not knowing what to call it). It results in a 23 sided polygon if I am not mistaken.

Opps. Doing this in my haed. I got a 29 sided polygon. A triangle, an octagon and a 24 sided polygon.

Wow, I certainly can’t do this in my head! This is very close– among all possible cycles, this one involves the second-largest polygon. But there is one larger, see the following week’s post here for the complete solution.