In standard Euclidean geometry, are all equiangular polygons with an odd number of sides also equilateral?

It is easy to prove that all equiangular triangles are also equilateral using basic trogonometric rules.

On the other hand, it is easy to conceive of an equiangular quadrilateral that is not equilateral, i.e. a rectangle.

Extending this further, I can easily conceive of an equiangular hexagon that is not equilateral, but I haven't been able to visualize an equiangular pentagon that is also equilateral:

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Is this true that all equiangular polygons with an odd number of sides also equilateral? If so, is there a straightforward way to prove it? If not, is there a counterexample, an equiangular polygon with an odd number of sides that is not equilateral?

  • $\begingroup$ As a note, all cyclic equiangular odd polygons are equilateral. $\endgroup$ – user706071 Jun 21 '14 at 11:20
  • $\begingroup$ And, all cyclic convex equilateral polygons are equiangular. $\endgroup$ – Oscar Lanzi Jun 23 at 19:50

No they're not. Just take your regular pentagon, and pull its base down while preserving the angles. It will get shorter as its adjacent edges get longer.

  • $\begingroup$ Curious thought when I read this: what about pulling two sides of the pentagon out, preserving all angles. Will the lengths of the four growing sides remain the same relative length? $\endgroup$ – Hidde Jan 2 '12 at 0:12
  • $\begingroup$ @Hidde: No, they won't. If you only care about relative lengths, your operation (pulling the two uppermost edges) is exactly the same as mine. $\endgroup$ – TonyK Jan 2 '12 at 0:31

No. If you take your regular pentagon, for example, you can slide one of the sides away from the center without changing its direction, and extend its two neighbor sides appropriately. This makes the chosen side shorter, two other sides longer, but leaves the angles intact.


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