Timeline for Who discovered integer triangles with one angle trisecting another?
Current License: CC BY-SA 3.0
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| Oct 24, 2017 at 17:36 | comment | added | Bassam Karzeddin | It seems clear then, that any integer degree angle $n$, where $n$ is not divisible by $3$ is non-existing angle, which implies two-thirds of our known angles are actually fictional and non-existing angles, however, the set of this integer degrees angles are more interesting than many other lists or sets of many fictional angles, since they are impossible to fit in any existing and imaginable triangle, for sure, where the trick is really interesting and lies beyond our naive believe of the simple existence of any named angle from our mind, it is really more thrilling since it reveals more | |
| Jan 18, 2017 at 19:05 | comment | added | user4900 | The OP tries to make himself interesting, by disseminating some bits and never the full story. For example he did this post in Jan 3 2016 here, but another post from Sep 3 2015 at this link math.stackexchange.com/q/1419425/4414 shows already the polynom. But I don't think this leads to a ruler and compass method for trisection, which shouldn't exist anyway. | |
| Jul 20, 2016 at 2:23 | history | edited | Conifold | CC BY-SA 3.0 |
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| Feb 8, 2016 at 23:25 | comment | added | Conifold | @René Yep, that's how it goes, it does give all the triangles by the "converse" law of sines. I realized later that the detour through the implicit equation is unnecessary, and trig formulas give polynomial parametrization directly as Chebychev polynomials of the second kind with 2cos α as parameter. | |
| Feb 6, 2016 at 19:08 | comment | added | R.P. | Given any n, one can just start from any α with cosα∈Q. For such α, a triangle exists with angles α, nα, 2π−(n+1)α and with rational sides. Put β:=nα and γ:=2π−α−β. Let (a,b,c) be the sides of any triangle with angles α,β,γ. Then both a/b = sin α/sin β and b/c = sin β/sin γ are rational: indeed, they are rational functions in sin(α) and cos(α) by standard addition formulae for trig functions, and since they are also even functions of α, they can be written in terms of cos α alone. I think this is the easiest construction, although it's not clear to me that it gives all such triangles. | |
| Feb 4, 2016 at 23:19 | history | edited | Conifold | CC BY-SA 3.0 |
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| Jan 15, 2016 at 0:55 | comment | added | Conifold | @bassam karzeddin I take it back, one can get self-n-secting families of triangles for any n by essentially the same method. Which means that the corresponding curves are highly degenerate and non-generic (that's why I thought otherwise). @ Franz Lemmermeyer Schappacher does suggest that Poincare gave geometric interpretation to what Diophantus did, albeit vaguely. | |
| Jan 12, 2016 at 8:14 | comment | added | user2255 | Schappacher does not suggest that the parametrization of the circle is due to Poincare. | |
| Jan 12, 2016 at 7:53 | comment | added | Bassam Karzeddin | Yes there are with more rotations for any angle, I shall try to remember the derivations | |
| Jan 12, 2016 at 2:18 | comment | added | Conifold | @bassam karzeddin You mean finding integer sided triangles with one angle n-secting another? I suspect that there are at most finitely many for generic n. | |
| Jan 11, 2016 at 22:59 | history | edited | Conifold | CC BY-SA 3.0 |
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| Jan 11, 2016 at 12:45 | comment | added | Bassam Karzeddin | Very rich answer and historical content indeed, even my simple understanding to the problem & deriving it was completely unaware of all those valuable knowledge that Greek Euclidean had, however there are much more to add to this basic simple triangle, with multi section of the angle in general | |
| Jan 11, 2016 at 4:13 | history | edited | Conifold | CC BY-SA 3.0 |
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| Jan 10, 2016 at 22:12 | history | answered | Conifold | CC BY-SA 3.0 |