Talk:Hyperbolic triangle

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Picture?

picture?

Done. Gandalf61 08:05, 14 February 2007 (UTC)

At a request, here is another:

Hyperbolic triangle (yellow) and hyperbolic sector (red) corresponding to hyperbolic angle u, constructed in 2d Cartesian coordinates from a rectangular hyperbola (equation y = 1/x) and a line (equation y = x). The lengths of the triangle are the hyperbolic cosine and sine functions.

I'm not 100% sure which is correct by the description (in this article and on my talk page...), or if the vertices are all wrong - could someone please clarify? Thanks and apologies this is wrong first time, the correct version will be drawn (i.e. not two diagrams, just the correct one)... Maschen (talk) 01:01, 4 December 2012 (UTC)

The hyperbolic triangle for the section Euclidean geometry is a right triangle with hypothenuse being the radius from the origin to a point on the hyperbola y = 1/x . Thus it shares two sides with the hyperbolic sector currently displayed. In fact, the angle at the origin of the hyperbolic sector is the same "hyperbolic angle" φ used to determine cosh φ and sinh φ. Dropping the perpendicular to y = x from the point on the hyperbola makes the two legs of the right triangle. Note that the diagram at hyperbolic functions has φ/2 for angle magnitude. That is a "normalized diagram" since the legs of the right triangle are just the hyperbolic functions cosh φ and sinh φ. The diagram here uses a hyperbola with minimum distance to the origin √2. The angle is not divided by two, but the legs are scaled up by √2. Remember that for a circle, a radius of √2 is necessary for sector area to equal arc length. So the confusing √2 and /2 really are consistent with circular practice.Rgdboer (talk) 20:09, 4 December 2012 (UTC)

Thank you very much for clarifying my confusion - I think I have the right idea now but there are still errors in the newest version, I'll fix them now. Maschen (talk) 21:14, 4 December 2012 (UTC)

Done - better? Maschen (talk) 21:25, 4 December 2012 (UTC)

Yes, that provides the requisite triangle. Very good caption too. Including the hyperbolic sector provides useful context. Note that the value of u is well away from zero so the triangle begins to approach the isosceles shape since sinh and cosh get closer (tanh goes to one).Rgdboer (talk) 23:12, 4 December 2012 (UTC)

Thanks, although the caption I wrote is a mouthful, but it had to explain everything... Thanks for inserting the √2 also! Maschen (talk) 07:23, 5 December 2012 (UTC)

should be split into two articles

This article seems to be about 2 different things. It should be split into two articles unless there is some connection I'm not seeing. Bhny (talk) 21:22, 1 November 2013 (UTC)

If you're referring to the hyperbolic geometry part and the Euclidean geometry part, I agree.

The Euclidean geometry section talks about constructing hyperbolic triangles in Euclidean space, although not sure where best to put this content. Creating a separate article just on hyperbolic triangles in Euclidean space would not be much use (topic is too narrow?). Hyperbolic functions? Coordinate geometry? Hyperbolic sector? M∧Ŝc²ħεИ_τlk 12:54, 2 November 2013 (UTC)

The two main articles for this topic are hyperbolic geometry and hyperbolic function. This hyperbolic triangle article is of the nature of disambiguation, letting the reader know which way to turn when seeing the phrase. Splitting the article would defeat the disambiguation and not add anything substantive to our Project. The Euclidean geometry version of the hyperbolic triangle is a fundamental concept that introduces hyperbolic angle, a topic that is not generally considered to belong to hyperbolic geometry.Rgdboer (talk) 20:12, 3 November 2013 (UTC)

A disambiguation page should be marked as such, and have links to separate articles. At the moment this is neither an article nor a disambiguation page. Bhny (talk) 20:50, 3 November 2013 (UTC)

As seen at WP:MOSDAB the style of a disambiguation page is quite rigid. In this case two ideas of triangle are compared where each is modified by the adjective "hyperbolic". Some detail is required to divide he ideas: in the hyperbolic plane the sides are geodesics, a subtle concept from differential geometry. The other triangle is of the Euclidean variety but in a particular position. This article is a type of disambiguation that requires more detail than found in an ordinary disambiguation page. IMO the present article has important content that should not be split.Rgdboer (talk) 21:12, 4 November 2013 (UTC)

From my reading of the article there is no comparing of triangles. These are different topics that share the same name. I don't know what you mean by "type" of disambiguation page, as you said the dab style is rigid. Bhny (talk) 21:55, 4 November 2013 (UTC)

Considering the methods of disambiguation used here, I decided to move the Euclidean geometry section to hyperbolic sector per suggestion of Maschen.Rgdboer (talk) 00:18, 5 November 2013 (UTC)

Formula for the area of a hyperbolic triangle

Currently, there is a "citation needed" template on one of the formulas for the area of a hyperbolic right triangle, namely:

${\displaystyle {\textrm {Area}}=2\arctan(\tanh({\frac {x}{2}})\tanh({\frac {y}{2}}))\,.}$

There is also a link to a blog which appears to provide a proof, but the proof is extremely complicated and has some notations which I did not understand, so I could not follow it. However, I found a much simpler proof myself.

First, let the right triangle have legs x and y and hypotenuse r. If we hold x fixed and allow y to increase from zero, I can show that the area is

${\displaystyle {\textrm {Area}}=\int _{0}^{y}{\frac {\sinh x\,\mathrm {d} y}{\cosh x\cosh y+1}}\,.}$

In a circular sector, the ratio of the area of the sector to the length of the circular arc bounding it (along with two radii) is

${\displaystyle {\frac {(\cosh r-1)}{\sinh r}}\,.}$

The length of the infinitesimal circular arc associated to dy is the projection of dy onto a line perpendicular to the hypotenuse where it meets the leg y. The angle between dy and its projection is the complementary angle of the angle between the hypotenuse and the leg y. So the cosine of the first angle is the sine of the second and that sine is

${\displaystyle {\frac {\sinh x}{\sinh r}}\,.}$

Then notice that

${\displaystyle {\frac {(\cosh r-1)}{\sinh r}}\,{\frac {\sinh x}{\sinh r}}\,\mathrm {d} y={\frac {(\cosh r-1)\,\sinh x\,\mathrm {d} y}{\sinh ^{2}r}}={\frac {(\cosh r-1)\,\sinh x\,\mathrm {d} y}{(\cosh r-1)\,(\cosh r+1)}}={\frac {\sinh x\,\mathrm {d} y}{\cosh x\cosh y+1}}\,.}$

To show that a formula for the area is correct, it is sufficient to show that y=0 implies that the area is zero and that

${\displaystyle {\frac {\partial {\textrm {Area}}}{\partial y}}={\frac {\sinh x}{\cosh x\cosh y+1}}\,.}$

I was able to show this for the well known formula for the area

${\displaystyle {\textrm {Area}}={\frac {\pi }{2}}-\arctan {\frac {\tanh y}{\sinh x}}-\arctan {\frac {\tanh x}{\sinh y}}\,,}$

but the proof was rather lengthy and off topic, so I will not give it here.

Now, let us apply the test to the formula at issue. First notice that if y=0, then

${\displaystyle 2\arctan(\tanh({\frac {x}{2}})\tanh({\frac {y}{2}}))=2\arctan(\tanh({\frac {x}{2}})0)=2\arctan(0)=0}$

as desired. Next, differentiate with respect to y to get

${\displaystyle 2\,{\frac {1}{1+\left(\tanh({\frac {x}{2}})\tanh({\frac {y}{2}})\right)^{2}}}\,\tanh({\frac {x}{2}})\,\operatorname {sech} ^{2}({\frac {y}{2}})\,{\frac {1}{2}}\,.}$

From the hyperbolic half-angle formulas, we know that

${\displaystyle \cosh \left({\frac {y}{2}}\right)={\sqrt {\frac {\cosh(y)+1}{2}}}}$

and thus

${\displaystyle \operatorname {sech} ^{2}({\frac {y}{2}})={\frac {2}{\cosh(y)+1}}\,.}$

And also

${\displaystyle \tanh \left({\frac {x}{2}}\right)={\frac {\sinh(x)}{\cosh(x)+1}}}$

${\displaystyle \tanh ^{2}\left({\frac {y}{2}}\right)={\frac {\cosh(y)-1}{\cosh(y)+1}}\,.}$

Cancelling the two and the one-half and substituting the half angle formulas into the derivative gives

${\displaystyle {\frac {1}{1+\left({\frac {\cosh(x)-1}{\cosh(x)+1}}{\frac {\cosh(y)-1}{\cosh(y)+1}}\right)}}\,{\frac {\sinh(x)}{\cosh(x)+1}}\,{\frac {2}{\cosh(y)+1}}\,.}$

This simplifies to

${\displaystyle {\frac {2\sinh x}{(\cosh(x)+1)(\cosh(y)+1)+(\cosh(x)-1)(\cosh(y)-1)}}\,.}$

The cross terms in the denominator cancel and we can then cancel a factor of two between the numerator and the denominator to get the desired result, proving the formula for the area. JRSpriggs (talk) 07:01, 8 July 2016 (UTC)

This is the formula for right triangle. However, it is put under "The area for any other triangle is:"

Should it be moved up 2 lines? Bruce Chen 0010334 (talk) 18:07, 28 May 2023 (UTC)

Hi! I noticed this problem as well. If no one minds, I will go ahead and move it up two lines, for the sake of clarity. Mediator64 (talk) 20:27, 7 September 2024 (UTC)

Namings

The names a, b, c, A, B, C are used but not defined anywhere. They might be standard names, but at least a graphic which shows them would be a nice thing. 193.175.95.251 (talk) 13:53, 14 February 2018 (UTC)

discussion on a new section to the article

i would like to reach a consencus on adding an "occurrences in nature" section in which some examples of hyperbolic triangles in the natural world are listed.

any opinions? — Preceding unsigned comment added by RJJ4y7 (talk • contribs) 16:38, 2 December 2020 (UTC)

The natural world does not have the geometry of the hyperbolic plane or hyperbolic space. Therefore, hyperbolic triangles cannot appear in it. You must be thinking of some other kind of triangle that resembles in some way a hyperbolic triangle but is not actually hyperbolic. —David Eppstein (talk) 18:12, 2 December 2020 (UTC)