 mersenneforum.org 17-gon
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R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

62516 Posts Quote:
 Originally Posted by Dr Sardonicus 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 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.   2021-10-20, 05:11 #24 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 11·103 Posts 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   2021-10-20, 17:54 #25 alpertron   Aug 2002 Buenos Aires, Argentina 22×3×112 Posts 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.   Thread Tools Show Printable Version Email this Page

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