There is a number N such that if you inscribe an N-gon in a circle, with all its angles an equal number of integral degrees, that N-gon will necessarily be a regular N-gon, with equal side lengths.

Also, if M is the number of possible divisors of N, including 1 and N itself, then the number M also meets the same criteria: If you inscribe an M-gon in a circle, with all its angles an equal number of integral degrees, that M-gon will necessarily be a regular M-gon, with equal side lengths.

What are N and M?

well obviously an equiangular triangle is a regular 3-gon so 3 is a canidate for m or n.  However if n=3 then the divisors of n are 1 and 3 thus m would be 2 and there is no such thing as an 2-gon, thus 3 would only work as m.  So if m=3 then we have that n=p^2 where p is any prime.

now for the angles to be integral we need
180(p^2-2)/p^2 to be integral
180(p^2-2)/p^2 = 180 - (360/p^2)
thus p^2 must be a divisor of 360=2^3*3^2*5
thus p=2 or p=3 which gives n=4 or n=9
now since it is possible to inscribe a rectangle (non regular equiangular 4-gon) inside a circle then n=4 does not work, that leaves us with n=9.
So now that leaves us with proving n=9 fits the criteria, which is beyond my geometric abilities, but for my part I'm guessing that n=9 and m=3 is the solution.

Posted by Daniel on 2010-05-28 12:26:47