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MattcAnderson 2021-10-16 15:45

17-gon
 
Fermat Search dot org

Assuming F0 = 3, and that is the smallest Fermat number,

F1 = 5, and F2 = 17.

This implies that a regular 17-gon can be constructed with
pencil and compass and straight-edge.

See a YouTube video "The Amazing Heptadecagon (17-gon ) - Numberphile
Brady Heron is usually the star of that channel
This video was made in 2015.

[URL="https://www.youtube.com/watch?v=87uo2TPrsl8"]17-gon[/URL]

Good fun.
Matt

xilman 2021-10-16 18:15

[QUOTE=MattcAnderson;590762]Fermat Search dot org

Assuming F0 = 3, and that is the smallest Fermat number,

F1 = 5, and F2 = 17.

This implies that a regular 17-gon can be constructed with
pencil and compass and straight-edge.

See a YouTube video "The Amazing Heptadecagon (17-gon ) - Numberphile
Brady Heron is usually the star of that channel
This video was made in 2015.

[URL="https://www.youtube.com/watch?v=87uo2TPrsl8"]17-gon[/URL]

Good fun.
Matt[/QUOTE]A 257-gon can also be so constructed. Never seen anyone do so.

a1call 2021-10-16 18:53

[QUOTE]
In 1894, Hermes completed his decade-long effort to find and write down a procedure for the construction of the regular [B]65537-gon[/B] exclusively with a compass and a straightedge. His manuscript, with [B]over 200 pages[/B], is today located at the University of Göttingen
[/QUOTE]

[url]https://en.wikipedia.org/wiki/Johann_Gustav_Hermes[/url]

Viliam Furik 2021-10-16 18:57

[QUOTE=MattcAnderson;590762]Brady Heron is usually the star of that channel
[/QUOTE]

Brady Haran

Dr Sardonicus 2021-10-17 02:07

[QUOTE=xilman;590789]A 257-gon can also be so constructed. Never seen anyone do so.[/QUOTE]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.

In [color=red][b]WARNING![/b] 33MB PDF file![/color] [url=http://pyrkov-professor.ru/Portals/0/Mediateka/18%20vek/nahin_p_j_dr_euler_s_fabulous_formula_cures_many_mathematica.pdf]Dr. Euler's fabulous formula cures many mathematical ills[/url], Notes to Chapter 1, we find[quote]18. Michael Trott, "cos(2n/257) à la Gauss," Mathematics in Education and Research 4, no. 2,1995, pp. 31-36.

You can read the details of what is involved in constructing the 257-gon, at a level a bit deeper than here, in

Christian Gottlieb, "The Simple and Straightforward Construction of the Regular 257-gon," Mathematical Intelligencer, Winter 1999, pp. 31-36.

The author's title is a bit of tongue-in-cheek humor, as he ends his essay with the line "I wish the reader who wants to pursue the construction in all details good luck."[/quote]

MattcAnderson 2021-10-18 05:05

constructable polygons and Fermat numbers
 
Thanks for the useful comments everyone.

See the Wikipedia article on [URL="https://en.wikipedia.org/wiki/Constructible_polygon"]Constructable polygon[/URL].

Assuming that there are only 5 Fermat primes, then

every constructible polygon has a number of sides s with

s = 3^e1 * 5^e2 * 17^e3 * 257^e4 * 65,537^e5.

where e1, e2, e3, e4, and e5 are in the infinite set 0,1,2,...

So polygons with number of sides like 9 (3*3) , 15 (3*5) , and 51 (17*3) are constructible.

Enjoy.

Matt

a1call 2021-10-18 05:28

[QUOTE=MattcAnderson;590932]Thanks for the useful comments everyone.

See the Wikipedia article on [URL="https://en.wikipedia.org/wiki/Constructible_polygon"]Constructable polygon[/URL].

Assuming that there are only 5 Fermat primes, then

every constructible polygon has a number of sides s with

s = 3^e1 * 5^e2 * 17^e3 * 257^e4 * 65,537^e5.

where e1, e2, e3, e4, and e5 are in the infinite set 0,1,2,...

So polygons with number of sides like 9 (3*3) , 15 (3*5) , and 51 (17*3) are constructible.

Enjoy.

Matt[/QUOTE]
Hi Matt,
I don’t think that is quite correct. From your link:
[QUOTE]

A regular n-gon can be constructed with compass and straightedge if and only if n is the product of a power of 2 and any number of distinct Fermat primes (including none).

[/QUOTE]
So 9-gon is not constructible what are is:
3*2^n
5*2^n
…..

MattcAnderson 2021-10-18 06:16

You are correct A1call.

A square is constructible. Also an octagon is constructible (8 sides). So any n-gon with n=2^m is constructible.

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.

Here I try to work out (unsuccessfully) Why 31 - from article

my combinatorics skills are not that good

3, 5, 17, 257, and 2557 (one Fermat prime each) [5 count]

3*5, 3*17, 3*257, 3*2557 then
5*17, 5*256, 5*2557 then
17*256, 17*2557 then
257*2557 (two Fermat primes each) [4+3+2+1=10 count]

3*5*17, 3*5*257, 3*5*2557 then
3*17*257, 3*17*2557 then
3*257*2557 (three Fermat primes each ) [3+2+1 = 6 count]

3*5*17*257, 3*5*17*2557, 3*5*257*2557, 3*17*257*2557, 5*17*257*2557 (four Fermat primes each) [5 count]

3*5*17*257*2557 (five Fermat primes for sides of this n-gon)[1 count]

So add 5+10+6+5+1 = 26 errrrr I seem to have missed five somewhere. It should be 31. Oh well, going to post anyway.

oops the fourth Fermat prime is actually 65,537 so there is an error above in my workings out.

to be clear, 2557 should be 65537.

Who can show that there are only 31 known n-gons that are constructible?

Regards,
Matt

a1call 2021-10-18 06:30

[QUOTE]
A regular n-gon can be constructed with compass and straightedge if and only if n is the product of a power of 2 and any number of distinct Fermat primes (including none).
[/QUOTE]

[QUOTE]
Since there are 31 combinations of anywhere from one to five Fermat primes, there are 31 known constructible polygons with an odd number of sides.
[/QUOTE]

The 2 statements from Wikipedia are contradictory IMHO. The 2nd statement does not seem to be correct.

I too would like to have an expert weigh in. But I can tell you that a 9-gon is not constructible AFAIK.

[QUOTE]
Although [B]a regular nonagon is not constructible with compass and straightedge[/B] (as 9 = 3^2, which is not a product of distinct Fermat primes), there are very old methods of construction that produce very close approximations. It can be also constructed using neusis, or by allowing the use of an angle trisector.
[/QUOTE]

[url]https://en.wikipedia.org/wiki/Nonagon#:~:text=Although%20a%20regular%20nonagon%20is,use%20of%20an%20angle%20trisector[/url].

According to the 1st (correct statement) there can only be 5 known odd-numbered constructible regular polygons not 31.
Unless we discover more Fermat primes.

MattcAnderson 2021-10-18 06:48

1 Attachment(s)
Yes, an expert would help here.
Attached is an image from Wikipedia Constructible polygons.
It enumerates all the odd numbers n such that that n-gon is constructible.
I counted the numbers in the file and there are 31 as there should be.
So we can agree that there are 31 odd numbers n such that those n-gons are geometrically constructible.

Good night.

Matt

a1call 2021-10-18 06:52

Thank you very much Matt I stand corrected. So as long as the Fermat primes have a power of less than 2 then the polygon is constructible.
I did not know that.

Ok, I think I can see how a 15-gon can be constructed. I assume similar processes can be used for other combinations.
1/3-1/5= 2/15
So by centering the 2 angles you would get 1/15th of a circle on each side (or you can just bisect it).

a1call 2021-10-18 08:01

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.

axn 2021-10-18 08:40

[QUOTE=a1call;590936]The 2 statements from Wikipedia are contradictory IMHO. The 2nd statement does not seem to be correct[/QUOTE]

I assume from the subsequent posts that you no longer consider the wikipedia statements as contradictory?

a1call 2021-10-18 08:46

[QUOTE=axn;590945]I assume from the subsequent posts that you no longer consider the wikipedia statements as contradictory?[/QUOTE]

Nah, I misunderstood the 1st statement. I interpreted distinct as distinct in general and not per polygon. :smile:

retina 2021-10-18 08:53

[QUOTE=MattcAnderson;590934]So add 5+10+6+5+1 = 26 errrrr I seem to have missed five somewhere. It should be 31.[/QUOTE]To make the set enumerating easier for you next time:

Since there are 5 known Fermat primes then you can take all the 5-bit binary numbers from 00000 to 11111 (32 in total) and replace each 0 with nothing, and each 1 with a Fermat prime. And excluding the all zeros, then that leaves 31 possible combinations.

Nick 2021-10-18 09:55

The mathematical background is that you can construct the regular polygon with n sides using ruler and compass
if and only if ϕ(n) is a power of 2 (where ϕ is Euler's function).
And that is true if and only if the prime factorization of n consists of 0 or more 2's and 0 or more distinct Fermat primes.

Dr Sardonicus 2021-10-18 12:10

Let f(x) be a monic irreducible polynomial in Z[x]. Let G be the Galois group of f(x). The order of G is always divisible by the degree d of f(x). The roots of f(x) = 0 are constructible with compass and straightedge if, and only if, the Galois group G of f(x) is a 2-group; that is, if the order of G is a power of 2.

Let n be a positive integer. The minimum polynomial for the primitive n[sup]th[/sup] roots of unity (cyclotomic polynomial) has degree [tex]\varphi(n)[/tex], the number of positive integers not exceeding n which are relatively prime to n. In Pari-GP this is eulerphi(n).

If p is a prime factor of n, then p-1 divides eulerphi(n). If p^2 divides n, then p divides eulerphi(n).

Thus, the degree eulerphi(n) is a power of 2 if, and only if, for each odd prime factor p of n, p - 1 is a power of 2, and p^2 does [i]not[/i] divide n.

Luckily, the Galois group of the cyclotomic polynomial has order equal to the degree; in fact [tex]G\;=\;\(\mathbb{Z}/n\mathbb{Z}\)^{\times}[/tex] so if every odd prime factor p of n is a Fermat prime, and p^2 does not divide n, the Galois group is in fact a 2-group if the degree is a power of 2, and the n[sup]th[/sup] roots of unity are constructible with compass and straightedge.

a1call 2021-10-18 13:32

And to think that Gauss had to reach the ripe old age of 19 before he could even prove that 17-gon was constructible. :no:

[QUOTE]
His breakthrough occurred in 1796 when he showed that a regular polygon can be constructed by compass and straightedge if the number of its sides is the product of distinct Fermat primes and a power of 2.[a]
[/QUOTE]

[url]https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss[/url]

:no:

Xyzzy 2021-10-18 13:45

[url]https://upload.wikimedia.org/wikipedia/commons/d/d1/Regular_Heptadecagon_Inscribed_in_a_Circle.gif[/url]

:mike:

sweety439 2021-10-18 14:32

[QUOTE=xilman;590789]A 257-gon can also be so constructed. Never seen anyone do so.[/QUOTE]

You can try to construct F33 (=2^(2^33)+1)-sided polygon, in fact, if you calculated cos(2*pi/F33) and....

* If this number only contain square roots and no roots beyond square roots, then F33 is prime.
* If this number contain roots other than square roots, then F33 is composite.

Dr Sardonicus 2021-10-18 16:43

[QUOTE=sweety439;591001]You can try to construct F33 (=2^(2^33)+1)-sided polygon, in fact, if you calculated cos(2*pi/F33) and....
<snip>[/QUOTE]:missingteeth: :missingteeth: :missingteeth: :missingteeth:

Proposing to attempt the construction without knowing whether F[sub]33[/sub] is prime, is epically inane.

The cosine can be calculated fairly precisely, however, using cos(x) = 1 - x[sup]2[/sup]/2 approximately, if x is small and in radians. If I did the arithmetic right,

cos(2*pi/F[sub]33[/sub]) = 1 - 2.128362442349688837*10[sup]-5171655945[/sup], approximately.

MattcAnderson 2021-10-18 17:14

[QUOTE=retina;590949]To make the set enumerating easier for you next time:

Since there are 5 known Fermat primes then you can take all the 5-bit binary numbers from 00000 to 11111 (32 in total) and replace each 0 with nothing, and each 1 with a Fermat prime. And excluding the all zeros, then that leaves 31 possible combinations.[/QUOTE]

Hi all,

You are correct Retina. The Binary counting from 00000 to 11111 and subtract the 0 case is an easy way to see that 31 is the right answer. 1+2+4+8+16 = 31. Thanks for that.

Regards,
Matt

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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