4

Suppose I have a regular polygon whose sides each measure $n$. I want to cut it up into smaller equilateral (but not necessarily regular) polygons whose sides each measure $1$.

Is this possible? If yes, what's a simple (easy to implement) algorithm that can generate the subdivision?

Blue
  • 75,673

1 Answers1

2

You can use this strategy for regular polygons with an internal angle greater than $120$ (that is, with $7$ or more sides): use $n$ equilateral triangles of side length $1$ to cover each side of the polygon, so that the uncovered region also forms an equilateral polygon of side length $1$.

Here's an example for a regular heptagon:

enter image description here

The importance of the internal angle greater than $120$ is, of course, so that the equilateral triangles don't overlap at the corner.

If your regular polygon have $3$, $4$ or $6$ sides, the situation is easy to handle. If it have $5$ sides, a similar strategy will work, so the answer to your question is: yes, it's always possible.

Alma Arjuna
  • 3,759
  • Why exclude the regular hexagon? – Blue Jul 19 '20 at 00:43
  • 1
    Your main strategy works fine for the regular hexagon. There's no reason to insist that the angle is strictly greater than $120^\circ$, only that it be at least $120^\circ$. – Blue Jul 19 '20 at 00:48