Talk:Lexell's theorem

In terms of circles at constant latitude

This theorem is new to me. But if the following is correct, and if it qualifies as WP:CALC or is cited somewhere, should we add it to the article? The mathematical statement in the article starts with points A, B, and C, and defines a small circle through A*, B* and C, wherefrom we can choose X. However, couldn't we start with the small circle and get an exactly equivalent result? That is, suppose we have a small circle. Rotating the sphere we can assume, without loss of generality, that the the small circle has constant latitude in the northern hemisphere (or is the equator, though it turns out that that is a boring case). We could choose two points A* and B* at that northern latitude, but instead we'll choose their antipodes A and B at the corresponding southern latitude. We consider the great-circle arc that connects those two points at southern latitude. Then the spherical triangle that is formed from that great-circle arc and any point at the northern latitude will have an area that is independent of the choice of the third point. And if the third point is not at that northern latitude then the area will be different. — $Q$ uantling (talk | contribs) 21:06, 29 September 2023 (UTC)[reply]

Yes, the version where you start with two parallel small circles equidistant from the same great circle ("equator") is formally the same statement but with a slightly different point of view, and there are some sources which write it more or less this way (don't ask me to list them, I wasn't keeping track, though V.-A. Lebesgue (1855) is one example). I hope it should be clear enough to readers that these are equivalent without belaboring the point. The identity

\sin {\tfrac {1}{2}}\varepsilon =\tan {\tfrac {1}{2}}c\,\tan {\tfrac {1}{2}}h_{c},

a few of the proofs, and the discussions about spherical parallelograms are bit more aligned with this perspective on the situation.

One reason I didn't make this more explicit in the article is that the situation where 2 vertices of the triangle are at the same latitude and the third vertex is at the opposite latitude is a pretty unusual special case, and I want to make sure readers understand that this theorem holds for any general triangle on the sphere, not just ones satisfying this very particular criterion relative to some given spherical coordinate system. –jacobolus (t) 00:44, 30 September 2023 (UTC)[reply]

Thank you, I'll have to read through again with those thoughts in mind. One tweak to your last sentence ... I don't think about restricting the choice of points to work with the latitudes of a particular coordinate system, but I think instead of starting with any set of points A, B, and C, and then carefully choosing a coordinate system so that the points end up on those opposite latitudes in the chosen coordinate system. But I think we're saying pretty much the same thing here. I'll read through at some point in the next few days, and see whether I have anything else for this discussion. Thanks —

Q

uantling (talk | contribs) 01:45, 30 September 2023 (UTC)[reply]

Yes, I understand that, but trying to describe to readers that given any pair of points and third point on the sphere you can pick a spherical coordinate system such that the two are on the same "latitude" with the third point at the (equal but) opposite latitude is something that I don't expect would be any clearer than just talking about the small circles passing through antipodal points. –jacobolus (t) 03:07, 30 September 2023 (UTC)[reply]

"small circle" vs. "spherical circle"

I see that "small circle" matches nicely with the competing concept of "great circle", so that's good, but it could easily confuse the reader into thinking about circles of particularly small radius, instead of all circles that are smaller than a great circle. Furthermore, although it is a degenerate case, the theorem applies even when the spherical circle is a great circle, which seemingly is not stated if we go with "small circle" in the theorem statement. What do you think about changing references of "small circle" to "spherical circle"? FWIW, the Wikipedia article on the concept uses the latter name. — $Q$ uantling (talk | contribs) 20:18, 6 October 2023 (UTC)[reply]

"Small circle" (or sometimes "lesser circle") is the standard name used by the vast majority of the existing literature (name comes from 2000+ year-old Greek geometry/astronomy). If it were up to me, in the context of spherical geometry "great circles" (geodesics) would be called "straights" or similar, and great and small circles together just "circles", and the fundamental definition for "circle" in most contexts would be "curve on a surface with constant intrinsic curvature" rather than "set of points equidistant from a point". Oh well. I recently changed the title of the article spherical circle (was previously at circle of a sphere); that article still needs a ton of work, particularly listing and showing pictures of various basic properties, but also describing / linking to (currently red link) more advanced topics.

I think the current version here is okay though, so long as "small circle" is wiki-linked when it first appears. Usually a degenerate spherical triangle with three collinear points (making it into either a hemisphere or an empty arc, depending how the points are spaced) is not considered to be a triangle at all. –jacobolus (t) 20:52, 6 October 2023 (UTC)[reply]

Theorem restatement

Do you know of any textbooks (or other reasonable sources) that state the theorem in this way?

Let $A$ , $B$ , $X$ , and $Y$ be points on a sphere. There is a spherical circle that contains all four of these points if and only if the spherical triangles $△ABX*$ and $△ABY*$ have the same area, where $X*$ and $Y*$ are the antipodes of $X$ and $Y$ , respectively.

— $Q$ uantling (talk | contribs) 20:41, 6 October 2023 (UTC)[reply]

I don't, but this is a pretty trivial variant, so I don't think there's any issue claiming this kind of thing in a Wiki article. I expect it would be somewhat more confusing to readers than the usual statement though. –jacobolus (t) 20:55, 6 October 2023 (UTC)[reply]

I was thinking about adding this statement rather than using it to replace another statement. The current statement(s) tell us how to find spherical triangles of equal area. By flipping things around, this new statement instead tells us a way to check whether four points on a sphere are co-circular. It's not a particularly practical test, so that's a bummer. IMHO, the question is whether any notable source has thought it to be an interesting perspective. Perhaps the answer will turn out to be "no". —

Q

uantling (talk | contribs) 21:19, 6 October 2023 (UTC)[reply]

Probably more interesting than this is the way the circles through the two antipodal points (see Lexell's theorem § Foliation of the sphere) are analogous to Apollonian circles in the plane – indeed, are their inverse stereographic projection onto the sphere. I'm not sure what the nicest way is to characterize the other pencil of circles orthogonal to these though, or if this has been discussed much in past literature. –jacobolus (t) 20:07, 10 October 2023 (UTC)[reply]

Opposite arcs of Lexell's circle

Hi @Quantling. I reverted your change to the statement of the theorem. The area is only constant for points on the same side of the great circle through the base; choices of apex on opposite arcs form triangles of opposite orientation whose signed area differs by the area of a hemisphere. However, if you allow a generalized definition of spherical triangle allowing for sides longer than a semicircle then you can make generalized triangles with apex on opposite arcs and the same area, e.g. by making the base of one triangle go the long way around the great circle between base vertices. Lexell himself suggested this: see Lexell's theorem § Opposite arcs of Lexell's circle for details. The lead section has a slightly imprecise paraphrase of the theorem for concision, and because the more precise version is stated soon enough after. –jacobolus (t) 22:27, 11 October 2023 (UTC)[reply]

It sounds like there is an implicit assumption that the interior of spherical triangle

ABC

is the part of the sphere that makes all the triangle's interior angles be less than or equal to

π

(with the ambiguous edge case of

ABC

being on a great circle, hereafter ignored). Or saying it another way, consider the set of 8 triangles that could be said to have vertices

ABC

, and consider the set of their 8 areas, each of value in

[0, 4 π]

. If

X

is anywhere on the small circle

A * CB *

then its set of 8 areas will be the same as that for

ABC

. Maybe that's true, or maybe I still don't get it! —

Q

uantling (talk | contribs) 00:12, 12 October 2023 (UTC)[reply]

Or maybe 2 instead of 8?? I clearly need to think more and waste your time less. —

Q

uantling (talk | contribs) 00:38, 12 October 2023 (UTC)[reply]

(Also, please don't worry about "wasting time". I am free to ignore your comments if I wish or if I am busy with something else, but in general am happy to have people read and engage with this article.) –jacobolus (t) 04:12, 12 October 2023 (UTC)[reply]

I personally think of it more in terms of winding number of different regions of the sphere, a perspective that is pretty natural in computer graphics and computational geometry, but in my opinion too off topic to describe in the text of this article.

Unlike the plane, where the point at infinity is by default considered to have winding number 0, on the sphere there's no a priori "correct" spot to start counting from [this causes real practical problems in GIS applications, etc.]. So if you want you can shift all of the winding numbers of every point on the sphere by any fixed integer

k,

which amounts to adding the whole sphere's surface

k

times from the signed area of your shape (to compute the surface area, multiply the winding number by differential area at each point). The rule for winding numbers relative to an oriented boundary is that every time you cross the boundary, the winding number of the regions on either side differ by 1, with the sign of the difference dependent on the curve's orientation. –jacobolus (t) 01:30, 12 October 2023 (UTC)[reply]

I am familiar with winding numbers from complex analysis so that helps my understanding. Thank you —

Q

uantling (talk | contribs) 12:42, 12 October 2023 (UTC)[reply]

Leaning in to the triangle ambiguity

This article is looking good, thank you. I've been musing on the intuition of why Lexell's theorem should be true.

For three points on the surface of a sphere, there is some ambiguity about which spherical triangle they define. For example, with two points on the equator and a third point in the northern hemisphere, we might connect the northern point to the other two with great circle arcs that never enter the southern hemisphere; and for connecting the two points on the equator we can go the short way or the long way around. If the area for the short way is $T$ then the area for the long way is $2 π - T$ , because together they give the entire northern hemisphere. Or starting with any triangle we can flip what we mean by inside and outside to get an area $4 π - T$ . Combining the two, we can transform area $T$ to $4 π - (2 π - T) = 2 π + T$ . Because any set of three points can be rotated to put two on the equator, these statements are general. Fortunately all of $T$ , $2 π - T$ , $2 π + T$ , and $4 π - T$ have the same $cos$ .

So, I am hoping that $cos(T)$ can be computed directly using cross products and dot products of the unit vectors that indicate the locations of the vertices, $u$ , $v$ , and $w$ , even though we are deliberately leaving the choice of specific great-circle arcs ambiguous. And I am hoping that that approach will also yield a simple proof of Lexell's theorem, starting with something like $cos(T) = cos(A + B + C - π) = -cos(A + B + C)$ . But it is all just hoping ... unless you see a way forward?? — $Q$ uantling (talk | contribs) 22:08, 28 November 2023 (UTC)[reply]

You can definitely prove it from unit vectors, but I'm not sure if the proof is especially enlightening. The appropriate formula to start from is from Eriksson (1990), doi:10.1080/0025570X.1990.11977515 (a vector expression of a formula published in the 18th century by Euler and Lagrange):

\tan {\tfrac {1}{2}}\varepsilon ={\frac {\|u\wedge v\wedge w\|}{1+u\cdot v+v\cdot w+w\cdot u}},

where

u,v,w

are the unit vectors to the vertices of the triangle.

This is related to the formula for the angle measure between two unit vectors:

\tan {\tfrac {1}{2}}\theta ={\frac {\|u\wedge v\|}{1+u\cdot v}}.

(Incidentally the right-hand side of this second formula also tells the excess of a spherical triangle if one vertex is stereographically projected to the origin and the other two vertices stereographically project to

u

and

v,

which in that case are not unit vectors anymore.)

I should maybe make an article called area of a spherical triangle or something (to which spherical excess could redirect) which could better cover all of the various formulas for the area of a spherical triangle, and their history. –jacobolus (t) 01:28, 29 November 2023 (UTC)[reply]

Thank you. Since there is a tight relationship between

cos(ε)

and

tan(ε /2)

, that may lead to what I am looking for. —

Q

uantling (talk | contribs) 18:20, 29 November 2023 (UTC)[reply]

If you want to know more about the half-tangent, my (probably unsuitable for Wikipedia, still not sure what to do with it) draft at user:jacobolus/HalfTan may be useful. –jacobolus (t) 18:47, 29 November 2023 (UTC)[reply]

Yes, Area of a spherical triangle would be a nice article to have. Please do! —

Q

uantling (talk | contribs) 18:20, 29 November 2023 (UTC)[reply]

I don't have a proof, but here's my thinking. Using

A

,

B

,

C

, and

X

as column vectors in

\mathbb {R} ^{3}

, and

C* = - C

to indicate an antipode, the points

A

,

B

,

C*

, and

X*

are co-planar if and only if

\det \left({\begin{array}{cc}1&1&1&1\\A&B&C^{*}&X^{*}\end{array}}\right)=0\,.

That is, if and only if

B \land C \land X - A \land C \land X - A \land B \land X + A \land B \land C = 0

.

On the other hand, the (deliberately ambiguously defined) triangles

▵ABC

and

▵ABX

have equal "cosine of spherical area" values if and only if

tan 2 (ε ▵ABC / 2) = tan 2 (ε ▵ABX / 2)

.

That is, if and only if

\left({\frac {\|A\wedge B\wedge C\|}{1+A\cdot B+B\cdot C+C\cdot A}}\right)^{2}=\left({\frac {\|A\wedge B\wedge X\|}{1+A\cdot B+B\cdot X+X\cdot A}}\right)^{2}\,.

How hard can it be to show that the co-planarity and "cosine of area" equations are equivalent whenever

‖ A ‖ = ‖ B ‖ = ‖ C ‖ = ‖ X ‖ = 1

?

The reason that this approach intrigues me is that this formulation and proof might easily go to higher-dimensional hyperspheres to compare the volumes of the likes of tetrahedrons

▵ABCD

and

▵ABCX

. The co-hyperlanarity equation generalizes trivially. For the case of just one more dimension, that would probably require a formula (similar to the

tan ε /2

formula) that provides the volume of

S 3

for a hyperspherical tetrahedron bounded by

A,B,C,D\in \mathbb {R} ^{4}

. —

Q

uantling (talk | contribs) 15:14, 30 November 2023 (UTC)[reply]

Feel free to try to work out (or find) a clear formula for the volume of a spherical tetrahedron. –jacobolus (t) 15:58, 30 November 2023 (UTC)[reply]

Couldn't I just follow the pattern and declare it to be:

\tan(\epsilon /2)={\frac {\|A\wedge B\wedge C\wedge D\|}{1+A\cdot B+A\cdot C+A\cdot D+B\cdot C+B\cdot D+C\cdot D}}

?

Both working things out and finding things are hard. It's much easier to make stuff up. —

Q

uantling (talk | contribs) 18:29, 30 November 2023 (UTC)[reply]

Take a look at Murakami (2012) "Volume Formulas for a Spherical Tetrahedron", about the volume of a spherical tetrahedron in terms of either dihedral angles or edge lengths.

And here's a unit vector version: Ribando (2006) "Measuring Solid Angle Beyond Dimension Three".

All of these are kind of gnarly. Maybe you can figure out a clearer formulation. –jacobolus (t) 07:03, 1 December 2023 (UTC)[reply]

Thank you, I will take a look. Because you are so good at finding these references ... can you find one that computes spherical (or hyperspherical) area based upon the normal vectors of the facets from the origin? (That is in the case of spherical area, instead of based upon vertices

A

,

B

, and

C

, the formula would take as inputs the normal vectors of the triangular facets,

n A = B \land C

,

n B = C \land A

, and

n C = A \land B

?) I am thinking that some sort of merging of the results of Law of sines § Higher dimensions and Law of sines § The spherical law of sines might work. (And as always, my apologies for these wild brainstorms, many of which lead nowhere. It entertains me to think about these things, but I acknowledge that your interests may lie elsewhere.) Thanks —

Q

uantling (talk | contribs) 15:46, 1 December 2023 (UTC)[reply]

Finding papers just takes typing some keywords into the query box of a citation index, e.g. Google Scholar. It's not magic. There's a paper Wang, Yang, Yu, & Qi (2014) "The Law of Sines for an n-Simplex in Hyperbolic Space and Spherical Space and its Applications". –jacobolus (t) 16:56, 1 December 2023 (UTC)[reply]