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-   -   17-gon (https://www.mersenneforum.org/showthread.php?t=27230)

R. Gerbicz 2021-10-18 20:17

[QUOTE=Dr Sardonicus;590832]You might like [url=http://facstaff.susqu.edu/brakke/constructions/big-gon.htm]Constructing 17, 257, and 65537 sided polygons[/url]

See also the references in Wolfram Mathworld's [url=https://mathworld.wolfram.com/257-gon.html]257-gon[/url]. The one by author Richelot, F. J. looks like it's right up your alley.[/QUOTE]

Still in high school's math camp we constructed a regular 17-gon.

[QUOTE=a1call;590942]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.[/QUOTE]

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.

MattcAnderson 2021-10-20 05:11

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

alpertron 2021-10-20 17:54

If you open my [URL="https://alpertron.com.ar/POLFACT.HTM"]Polynomial factorization and roots calculator[/URL], 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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