Yes, I knew that power-of-2-greater-than-4-gons
could not have
rational vertices. I was just observing that
the equilateral oddgon
theorem gets us most of the way there. That
with the regular
octogon theorem gets us home.
----- Original Message -----
Sent: Wednesday, June 11, 2003 9:45 AM
Subject: Re: [math-fun] What's your proof?
On Wed, 11 Jun 2003, David Wilson wrote:
> With the
additional step, Fred's proof coincides with mine.
>
> So, there is
no such thing as an equilateral oddgon with rational vertices.
>
>
Here's a quick application:
>
> If a regular polygon has rational
vertices, it's a power-of-2-gon.
>
Yes,
since in fact it's a 4-gon, and 4 is indeed a power of 2.
John
Conway
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