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April 11

Parametrization of the set of triangles upto similarity

Is there any simple way to map a triangle to a pair of real parameters such that:

The parameters are invariant under (a) relabelling of the vertices, (b) translation, (c) rotation, (d) reflection and (e) scaling of the triangle
The mapping is continuous and infinitely differentiable, not piecewise - no "take the longest side"
The mapping is either invertible or otherwise provably one-to-one?

If there is no easy answer to that, then what topological surface is defined by the set; a sphere, a torus? Thanks, catslash (talk) 00:54, 11 April 2019 (UTC)[reply]

@Catslash: If you allowed 'take the two smallest angles' (or two largest angles), than such a pair is a mapping you want. Taking arbitrary two angles would work, but in one direction only: (30°, 60°) and (90°, 30°) would map to the same set of triangles.

BTW, what do you mean by continuous? That requires some distance between arguments and between values of a function. How would you define a distance between arbitrary two triangles? --CiaPan (talk) 09:33, 11 April 2019 (UTC)[reply]

Here, continuous means that the parameters are continuous functions of the real coordinates of the vertices (at least as long as the area of the triangle remains non-zero).

I had, perhaps erroneously, discounted taking two angles as being piecewise and therefore non-smooth. This is however clearly continuous at the joins where a smallest angle jumps from one vertex to another and the triangle passes through being isosceles into the mirror image of its former self. catslash (talk) 11:33, 11 April 2019 (UTC)[reply]

By the way, what do you mean by "being piecewise"? "Piecewise" has to modify another adjective rather than a noun. Anyway, the nature of the problem somewhat depends on what you mean by "triangle". That is, can its points coincide? Can they be collinear? The answer to those questions determine what kind of configuration space your domain is. –Deacon Vorbis (carbon • videos) 12:05, 11 April 2019 (UTC)[reply]

By piecewise, I meant having different maps for different pieces of the domain. It raises the issue of continuity and smoothness on passing from one piece to another. If implemented in a computer program, it necessitates conditional branching. catslash (talk) 19:26, 11 April 2019 (UTC)[reply]

Yes, but that's not what "piecewise" means. A function on the reals is piecewise blah (where blah might be continuous, or differentiable, etc.) if you can divide its domain up into a number of intervals such that the function is blah on each interval. But for more variables, it's a bit less clear what even, e.g., piecewise continuity would entail. Not only that, but as I mentioned above, the "space of triangles" isn't just

\mathbb {R} ^{6}

– you may or may not want to remove parts that correspond to degenerate triangles, and you're implicitly taking the quotient by the group action of the symmetric group which permutes the vertices of the triangle. Also, mathematically, there's really no such thing as being "piecewise" without being "piecewise blah". For example, the standard way to describe the absolute value function:

|x|={\begin{cases}-x&{\text{if }}x\leq 0\\x&{\text{if }}x\geq 0\end{cases}}

may be superficially given in a piecewise fashion, but it's still perfectly continuous. It could just as easily be written as

|x|={\sqrt {x^{2}}}.

So it really doesn't make sense to call a function "piecewise" without saying "piecewise blah" (still not forgetting that for functions of multiple variables, some extra caution is needed to use that sort of term). This is more than semantic quibbling; it really gets at what it means to specify a function. –Deacon Vorbis (carbon • videos) 21:44, 11 April 2019 (UTC)[reply]

Then I should have said either not merely piecewise, or else everywhere continuous and differentiable. I was unaware that piecewise had a technical meaning - thank you for explaining it. Irrespective of how |x| for real x is written, it is not smooth -has a discontinuous derivative- at x = 0.

Triangles of zero area are of zero interest - they can be included or excluded from the domain, whichever makes the problem tractable. Please let me know whether the question requires further clarification in order for you to offer an answer. catslash (talk) 23:10, 11 April 2019 (UTC)[reply]

What's wrong with your parameters

(a,b)

being mapped to a triangle with two angles

(\alpha ,\beta )

where

$\alpha =\arctan(a)+{\frac {\pi }{2}}$
$\beta =(\pi -\alpha ){\frac {\arctan(b)+{\frac {\pi }{2}}}{\pi }}$ ?

I think this maps the reals to the set of triangles up to similarity.--Leon (talk) 13:33, 11 April 2019 (UTC)[reply]

@Catslash: I suppose the pair of ratios of some global 'linear size' of a triangle to a similar measure of the circumscribed and the inscribed circle could work. For example:

{\text{triangle}}\mapsto \left({\frac {\text{triangle circumference}}{\text{circumradius}}},{\frac {\text{triangle circumference}}{\text{inradius}}}\right)

Areas (which are proportional to a square of linear sizes) should work as well:

{\text{triangle}}\mapsto \left({\frac {\text{triangle area}}{\text{circumcircle area}}},{\frac {\text{triangle area}}{\text{incircle area}}}\right)

This seems smooth and it's independent on translation, permutation, rotation and scaling. It doesn't cover degenerated triangles (which would have infinite or zero circumcircles and degenerated incircles). However, I didn't make sure yet it is one-to-one. --CiaPan (talk) 06:39, 12 April 2019 (UTC)[reply]

@CiaPan: I’m curious–how can you prove that the two ratios triangle circumference / circumradius and triangle circumference / inradius together specify a unique triangle (up to similarity)? Loraof (talk) 18:03, 14 April 2019 (UTC)[reply]

@CiaPan: Those are exactly the sort of parameters I was hoping for: normalized to the triangle area, two quantities out of; circumradius, inradius, second moment, perimeter, sum of squares of sides etc. But how to prove that the two parameters together specify a unique triangle (up to similarity)? catslash (talk) 22:33, 14 April 2019 (UTC)[reply]