From: firstname.lastname@example.org (James "Kibo" Parry) Date: Wed, 12 Aug 1998 07:27:19 GMT Newsgroups: sci.math,alt.religion.kibology Subject: Re: HEPTADECAGON, regular: Help please!
In sci.math, Bill Taylor (email@example.com) wrote: > > There is a particularly slick construction of a regular 17-gon that > I would like to see again; it used to be in my office somewhere, but > seems to have gone. > > Dunno if it was Gauss' original, I tend to doubt it. > > [...] > > P.S. I think it was REALLY mean of the town council of Gottingen, or > the executors of his will, or whoever, to NOT put a regular 17-gon > on his gravestone, as he requested. Especially as that was his most > proud theorem; rather like Archimedes and his cylinder-sphere grave icon. > > And the truly *pathetic* reason given, that it was "too hard" for the > stone-cutter to do, was *really* feeble. What are these bloody masons > paid to do anyway, goddammit? Geez, what a mutant... Okay, here's my secret plan to take over the world. Don't tell anybody. When I die, I want a scalene triangle of unspecified scalenity on my tombstone. The Master Masons carving my monument will say "Hey! Kibo made this easy for us 'cause he liked us! So we like him! Let's give him an extra-special monument!" and then they'll carve it out of solid diamond using a golden chisel, the most difficult expensive combination other than Archimedes Plutonium's proposed solid plutonium pyramid tomb. At sunrise on solstice morning, New Agers from around the world would come to stare at the sun through my lens-shaped diamond tombstone. Also, all the lettering on the tombstone would be not just incised so you could feel it, but flavored so you could taste it. All this from a humble scalene triangle, the cheapest kind to make, unless you're using Legos. And speaking of scalene triangles, it bothers me that you can have a perfectly equilateral triangle but mathematicians have not yet agreed on the dimensions of a perfectly scalene triangle. I was going to propose one where the sides are 1, 2, and 3 units long, as I think that would be a pretty cool triangle, what with side B being twice side A and side C being three times A and 50% improved over B. Also the angle opposite C would have over ten trijillion times as many degrees (plus some radians, too) as any of the other two angles, which would be so sharp you wouldn't be able to see them before you accidentally tried to pick the triangle up. But then I realized that what makes a 1:2:3 triangle so excitingly scalene is that two sides are much longer than one of the other sides. Wouldn't it be even better if side C was to B as B is to A? Therefore, let's try 1:2:4. Success! The triangle is so scalene that the mind of Man cannot even conceive of it! And carrying this scheme to its omega we reach the pinnacle of scalenitude: 1 : 100000000 : 10000000000000000000000000000000000 The technical term for this triangle is "the equisilly triangle" because no one angle in it is any sillier than the whole. I have applied for a patent on the equisilly and am awaiting the patent office's approval and the return of my working model. > The time you enjoy wasting is not wasted time. - Bertrand Russell Yay! It's okay for me to waste other people's time as long as I'm happy! -- K. OH I AM EVER SO HAPPY!
From: firstname.lastname@example.org (Kevin Brown) Date: Thu, 13 Aug 1998 07:49:00 GMT Newsgroups: sci.math Subject: Re: HEPTADECAGON, regular: Help please!
On 12 Aug 1998 email@example.com (Bill Taylor) wrote: > There is a particularly slick construction of a regular 17-gon that > I would like to see again... the construction of the regular > heptadecagon I am looking for... may have even been so simple > as the above [constrtuction of pentagon], but with quarter-sections > replacing the bisections... You may be thinking of Richmond's construction (1893), as reproduced in Stewart's "Galois Theory". The proof begins with two perpindicular radii OA and OB in a circle centered at O. Then locate point I on OB such that OI is 1/4 of OB. Then locate the point E on OA such that angle OIE is 1/4 the angle OIA. Then find the point F on OA (extended) such that EIF is half of a right angle. Let K denote the point where the circle on AF cuts OB. Now draw a circle centered at E through the point K, and let N3 and N5 denote the two points where this circle strikes OA. Then, if perpindiculars to OA are drawn at N3 and N5 they strike the main circle (the one centered at O through A and B) at points P3 and P5. The points A, P3, and P5 are the zeroth, third, and fifth verticies of a regular heptadecagon, from which the remaining verticies are easily found (i.e., bisect angle P3 O P5 to locate P4, etc.). On 12 Aug 1998 firstname.lastname@example.org (Bill Taylor) wrote: > Dunno if it was Gauss' original, I tend to doubt it. Gauss's Disquisitiones gives only the algebraic expression for the cosine of 2pi/17 in terms of nested square roots, i.e., cos(2pi/17) = -1/16 + 1/16 sqrt(17) + 1/16 sqrt[34 - 2sqrt(17)] + 1/8 sqrt[17 + 3sqrt(17) - sqrt(34-2sqrt(17)) - 2sqrt(34+2sqrt(17)] which is just the solution of three nested quadratic equations. Interestingly, although Gauss states in the strongest terms (all caps) that his criteria for constructibility (based on Fermat primes) is necessary as well as sufficient, he never published a proof of the necessity, nor has any evidence of one ever been found in his papers (according to Buhler's biography). On 12 Aug 1998 email@example.com (Bill Taylor) wrote: > I think it was REALLY mean of the town council of Gottingen, or > the executors of his will, or whoever, to NOT put a regular 17-gon > on his gravestone, as he requested. Especially as that was his > most proud theorem; rather like Archimedes and his cylinder-sphere > grave icon. I've heard that this story is apochryphal (about Gauss, not about Archimedes), but it's apparently true that Gauss's discovery of the 17-gon's constructibility (which had been an open problem from antiquity) at or before the age of 19 led to his decision to follow a career in mathematics rather than philology.