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March 4th, 2018, 11:58 AM  #1 
Newbie Joined: Mar 2018 From: Europe Posts: 5 Thanks: 0  When were constructible polygons actually constructed?
Gauss proved in 1796 that 17gon could be constructed  but did not actually show how to do so. The first construction of heptadecagon was by Erchinger, several years later  and it does not appear to be the best, because several constructions are offered with later dates. In 1801, Gauss showed that 257gon and 65 537gons also could be constructed. But also without giving actual way to construct them. First construction of 257gon was discovered 21 years later, in 1822, by Paucker, and first construction of 65537gon only 93 years later, in 1894, by Hermes. So much about the primes. But how about the multiples? The products like 65537*257 or 65537*17 are also constructible. Are these constructions trivial given the constructions of the primes? Or are they also difficult to actually make? 
March 4th, 2018, 11:07 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
Choose one of the products and attempt the construction. If it’s “trivial”, you’ll succeed quite easily.

March 5th, 2018, 09:19 AM  #3 
Newbie Joined: Mar 2018 From: Europe Posts: 5 Thanks: 0 
That was not a useful reply. But I did find the key to the approach, and it indeed was trivial. Apparently the constructions of constructible polygons are, or can easily be converted to, constructing the polygon with vertices on a given circle. Now, the factors contain each prime just once. Therefore they do not have any common factors. If you want to construct, say, a 85gon: Construct a 17gon and a circle containing the vertices of the 17gon. Then construct a 5gon with vertices on the same circle and 1 of the 5 vertices identical with 1 of the 17 vertices of 17gon. Since 5 and 17 have no common factor, but both are factors of 85, the remaining 4 vertices of 5gon will be 4 of the 68 as yet missing vertices of 85gon. From that point, there are 2 possible approaches: 1) From the shared vertex, first vertex of pentagon is 17/85 of circle away, between vertices of 17gon at 15/85 and 20/85. But the second vertex of pentagon is 34/85 of circle away, next to vertex of 17gon at 35/85. So you get an angle of 1/85 of circle, and can carry it over around the circle to mark the remaining 64 vertices. 2) Or you can repeat construction of 5gons sharing each of the vertex of 17gon, and so get all vertices of 85gon. Since every odd constructible number is a product of two odd numbers with no common factors, then if you can construct the prime factors, constructing the composites is indeed trivial. Last edited by skipjack; March 5th, 2018 at 06:21 PM. 
March 5th, 2018, 06:41 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
A regular 85gon is constructible if an angle of 2$\pi$/85 is constructible, and it is, as 2$\pi$/85 = 7(2$\pi$/17)  2(2$\pi$/5). 
March 5th, 2018, 08:31 PM  #5  
Senior Member Joined: Oct 2009 Posts: 406 Thanks: 141  Quote:
 

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