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Old 2021-10-18, 20:17   #23
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
Hungary

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Quote:
Originally Posted by Dr Sardonicus View Post
You might like Constructing 17, 257, and 65537 sided polygons

See also the references in Wolfram Mathworld's 257-gon. The one by author Richelot, F. J. looks like it's right up your alley.
Still in high school's math camp we constructed a regular 17-gon.

Quote:
Originally Posted by a1call View Post
Similarly for 3*17=51 gon
1/3-3/17= 8/51 of a circle. Then you can bisect 3 times to get 1/51 of a circle.
There is an elementary way to show that if you can make a regular m and n-gon [using straightedge and compass] and gcd(m,n)=1 then you can make a regular m*n-gon. Because making a regular k-gon is equivalent with constructing a 2*Pi/k angle.

So we can make a 2*Pi/n and 2*Pi/m angle.

We assumed that gcd(m,n)=1 so with extended Euclidean algorithm there exists x and y integers:

n*x+m*y=1 divide this equation by m*n

x/m+y/n=1/(m*n) multiplie by 2*Pi

x*(2*Pi/m)+y*(2*Pi/n)=2*Pi/(m*n), what we needed.
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Old 2021-10-20, 05:11   #24
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
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Hi all,

From before,

According to Wikipedia (constructible polygon article), there are infinitely many constructible polygons, but only 31 with an odd number of sides are known.

5 Fermat primes are known.

I worked out why 31 different regular polygons with an odd number of sides are constructible.
We have nCk, read n choose k, defined as
nCk = n!/(k!*(n-k)!)

So we want combinations of 5 things taken 1,2,3,4, and 5 at a time without repetition. Hence

5C1 = 5
5C2 = 10
5C3 = 10
5C4 = 5 and
5C5 = 1

So 5+10+10+5+1 = 31.
So we see that there are 31 ways of, among 5 things, taking 1,2,3,4 or all 5 of them without repetition.

And all is right with the world.

Regards,
Matt
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Old 2021-10-20, 17:54   #25
alpertron
 
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Aug 2002
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If you open my Polynomial factorization and roots calculator, enter x^255-1 and press Factor, you will see after a few seconds the 255 roots of that polynomial, and as explained before, only square roots are needed, because 255 = 3 * 5 * 17, which is the product of three different Fermat primes.
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