Is there a similar triangle on the sphere?

Question

It has been thought that there are no similar triangles on the sphere, but in fact they are not. There are also similar triangles on the sphere, the similar conditions are: the corresponding sides are parallel and proportional, and the corresponding angles are equal. The similarity on a sphere is not exactly the same as that on a plane. However, the similarity on a plane is a special case of the similarity on a sphere. am I correct?

Triangles whose sides are formed by great arcs are triangles. In my definition, triangles whose sides are formed by arcs are also triangles. My definition is more general.

In addition, there are similar squares on the sphere. It should be noted that the sum of the inner angles of the squares on the sphere is greater than 2 π.

@Arthur For example, the latitude is parallel to the equator. — z.qmpx, Nov 29 '19 at 13:48
@z.qmpx The latitudes usually aren't considered lines, though. Apart from the equator itself, of course. — Arthur, Nov 29 '19 at 13:49
@z.qmpx I can, but my question was which triangles you are drawing. If their sides are "latitudes" (parallels?), then like Arthur says those aren't really "straight" in any reasonable sense. — Moisés, Nov 29 '19 at 14:02
@Arthur At present, people don't think the latitude line is a straight line, but in the picture and description I give, two equilateral triangles are similar. — z.qmpx, Nov 29 '19 at 14:03
Like Matthew said, the area is determined by the angles. For example, on Earth an equilateral triangle whose vertices are on the equator has 180 degree angles, while a very small equilateral triangle has almost 60 degree angles. — Moisés, Nov 29 '19 at 14:05
@Moisés Don't you think that equilateral triangle in the picture is similar? — z.qmpx, Nov 29 '19 at 14:05
A spherical triangle has as sides, by definition, three arcs of great circles. If you have in mind some other kind of triangle, then please define it and then ask your question. — Intelligenti pauca, Nov 29 '19 at 14:09
@z.qmpx I don't think they are both triangles. I think that if, for instance, the larger one is a triangle, the smaller one is something like a Reuleaux triangle. — Arthur, Nov 29 '19 at 14:10
@z.qmpx I think they look similar right now, and if you make the small one smaller and the big one bigger at one point they will either stop looking similar or stop looking like triangles (their sides will stop looking straight). — Moisés, Nov 29 '19 at 14:12
The triangles you have drawn are not even similar by your definition. Either the sides are not "parallel" (according to your definition) or the angles are not equal. — , Nov 29 '19 at 14:27
@Aretino Triangles whose sides are formed by great arcs are triangles. In my definition, triangles whose sides are formed by arcs are also triangles. My definition is more general. — z.qmpx, Nov 29 '19 at 16:31
"Your definition" should then be clearly stated at the beginning of your question. — Intelligenti pauca, Nov 29 '19 at 17:23
@Arthur In my definition, the sides of a spherical triangle are made of arcs, so large and small equilateral triangles are triangles. — z.qmpx, Nov 29 '19 at 22:51
@Moisés No matter how the two equilateral triangles become larger or smaller, as long as they are equilateral triangles, they are similar. Although it looks like their edges are more curved. — z.qmpx, Nov 29 '19 at 22:59
@Rahul I use professional software to draw the pictures. I guarantee that they are parallel and the angles are equal. You don't need to doubt this. — z.qmpx, Nov 29 '19 at 23:04
In that case, can you give the formulas for the arcs of the two triangles and the numerical values of their angles, and explain how you calculated them to verify that they are equal? Or do we just have to accept your guarantee without evidence? — , Nov 30 '19 at 05:39
Up to my understanding, it seems you call two great/small arcs on sphere are parallel when the planes holding the arcs are parallel. For this definition of parallelness, the three conditions "corresponding sides are parallel", "corresponding sides are proportional" and "corresponding angles are equal" are in general, not compatible with each other. You need to pick one (and probably only one) of these condition in your definition of "similarity". — achille hui, Nov 30 '19 at 07:25
@Rahul You think, because they are all equilateral triangles, proportioning the corresponding sides is not a problem. And because the corresponding edges are parallel, the corresponding angles are equal, don't you object to this? There are of course formulas for spherical curves. It can be provided to you if needed. — z.qmpx, Nov 30 '19 at 13:35
@achillehui See my answer to Rahui for an explanation of the incompatibility. — z.qmpx, Nov 30 '19 at 13:41
"And because the corresponding edges are parallel, the corresponding angles are equal, don't you object to this?" Yes, I do object to this. As someone mentioned elsewhere on this page, "You have to be careful with our habits in plane geometry." That is why I asked, have you actually computed the angles and verified whether they are really equal? — , Nov 30 '19 at 13:59
@Rahul You have to understand a property: on a sphere, if two arcs are parallel and intersect the third arc, then the angle of apposition is equal. According to this property, the corresponding angles are equal. Do you understand what I mean? — z.qmpx, Nov 30 '19 at 15:25
This is again an assertion you are making without a proof. Why have you not computed the angles of your original triangles and compared them? Are you afraid that they will turn out to be different, showing your claim to be false? — , Nov 30 '19 at 15:33
I don't see where you have calculated the angles between the edges of the triangles. — , Nov 30 '19 at 15:55

score 2 · Answer 1 · answered Nov 29 '19 at 13:46

2

The area of a spherical triangle with radius $R$ and spherical angles $A,B,C$ is $R^2(A+B+C-\pi)$. Given that, two triangles with the same spherical angles must have the same area. I wouldn't necessarily say that they cannot be similar, but rather that they must be congruent.

answered Nov 29 '19 at 13:46

You have to be careful with our habits in plane geometry. – z.qmpx Nov 29 '19 at 14:10
1

@z.qmpx Which habits are those? – Nov 29 '19 at 14:12
@z.qmpx If you mean habits like the fact that triangles are defined by axiomatically-defined lines and not arbitrary curves that can have any angle we like, then I agree. – Nov 29 '19 at 14:15
What is the formula for calculating the area of small equilateral triangles in my graphs? – z.qmpx Nov 29 '19 at 22:31

z.qmpx · Answer 2 · 2019-12-02T15:41:21.983

-2

Some researchers need to know how my spherical similar equilateral triangle graph comes from. Now I will publish the code about running in MATLAB, hoping to help you with your research.

I have received the help of netizens in solving the relevant equations and modifying the code. I would like to express my deep thanks here.

$% Similarity of equilateral triangles on sphere
%Small equilateral triangle

[x y z]=sphere(50);
mesh(x,y,z);
hold on
axis equal

K=1; 
% K is the slope of the spherical circle, tangent of the angle between the plane of the spherical circle and the plane of the equator
R=1;
C=0.3;  
% In the direction perpendicular to the equator, the offset of the center of the circle. If C is zero, the circle is big, otherwise it is small.

alpha =pi/6: pi/50: pi*5/6;
A=(1+K^2*(sin(alpha)).^2).^0.5; 
B=C*(1+K^2);
beta = 2*pi+acos(-B./A)-asin(1./A); 
[X,Y,Z] = sph2cart(alpha,beta,R);
plot3(X,Y,Z, 'r','linewidth',2);
grid on;
axis square;
hold on;

%------------
alpha =-pi/2: pi/50: pi/6;
A=(1+K^2*(sin(alpha-pi/3)).^2).^0.5; 
B=C*(1+K^2);
beta = 2*pi+acos(-B./A)-asin(1./A); 
[X,Y,Z] = sph2cart(alpha,beta,R);
plot3(X,Y,Z, 'g','linewidth',2);
grid on;
axis square;
hold on;

alpha =-pi*7/6: pi/50: -pi/2;
A=(1+K^2*(sin(alpha-2*pi/3)).^2).^0.5; 
B=C*(1+K^2);
beta = 2*pi+acos(-B./A)-asin(1./A); 
[X,Y,Z] = sph2cart(alpha,beta,R);
plot3(X,Y,Z, 'b','linewidth',2);
grid on;
axis square;
hold on;


% Similarity of equilateral triangles on sphere
%Large equilateral triangle

[x y z]=sphere(50);
mesh(x,y,z);
hold on
axis equal

K=1; 
% K is the slope of the spherical circle, tangent of the angle between the plane of the spherical circle and the plane of the equator
R=1;
C=0.1;  
% In the direction perpendicular to the equator, the offset of the center of the circle. If C is zero, the circle is big, otherwise it is small.

alpha =pi/6: pi/50: pi*5/6;
A=(1+K^2*(sin(alpha)).^2).^0.5; 
B=C*(1+K^2);
beta = 2*pi+acos(-B./A)-asin(1./A); 
[X,Y,Z] = sph2cart(alpha,beta,R);
plot3(X,Y,Z, 'r','linewidth',2);
grid on;
axis square;
hold on;

%------------
alpha =-pi/2: pi/50: pi/6;
A=(1+K^2*(sin(alpha-pi/3)).^2).^0.5; 
B=C*(1+K^2);
beta = 2*pi+acos(-B./A)-asin(1./A); 
[X,Y,Z] = sph2cart(alpha,beta,R);
plot3(X,Y,Z, 'g','linewidth',2);
grid on;
axis square;
hold on;

alpha =-pi*7/6: pi/50: -pi/2;
A=(1+K^2*(sin(alpha-2*pi/3)).^2).^0.5; 
B=C*(1+K^2);
beta = 2*pi+acos(-B./A)-asin(1./A); 
[X,Y,Z] = sph2cart(alpha,beta,R);
plot3(X,Y,Z, 'b','linewidth',2);
grid on;
axis square;
hold on;$

It can be seen from this code that since $K$ is constant, the corresponding edges are parallel, and the corresponding angles are equal. Since $K$ is an equilateral triangle, the corresponding edges are also proportional.

Three sides of an equilateral triangle:

$K=1$

$C=0.2$

$\alpha =\cfrac{\pi}{6}: \cfrac{\pi}{50}: \cfrac{5\pi}{6};$

$\beta = 2\pi+acos(-\cfrac{C(1+K^2 )}{\sqrt{1+K^2 (\sin\alpha)^2 } })-asin(\cfrac{1}{\sqrt{1+K^2 (\sin\alpha)^2 }} ); $

$\alpha =-\cfrac{\pi}{2}: \cfrac{\pi}{50}: \cfrac{\pi}{6};$

$\beta = 2\pi+acos(-\cfrac{C(1+K^2 )}{\sqrt{1+K^2 (sin(\alpha-\frac{\pi}{3}))^2 }} )-asin(\cfrac{1}{\sqrt{1+K^2 (sin(\alpha-\frac{\pi}{3}))^2 }} ); $

$\alpha =-\cfrac{7\pi}{6}: \cfrac{\pi}{50}: -\cfrac{\pi}{2};$

$\beta = 2\pi+acos(-\cfrac{C(1+K^2 )}{ \sqrt{1+K^2 (sin(\alpha-\frac{2\pi}{3}))^2 }} )-asin(\cfrac{1}{\sqrt{1+K^2 (sin(\alpha-\frac{2\pi}{3}))^2 } });$

If K does not change, C can be changed to ensure the parallelism of the corresponding sides of the triangle.

edited Dec 02 '19 at 15:41

answered Nov 30 '19 at 15:45

z.qmpx

121

@Rahul It's very simple. I can't help it if you don't understand the code. – z.qmpx Nov 30 '19 at 16:00
@Rahul This is a basic geometric concept. How can you not understand it? – z.qmpx Nov 30 '19 at 16:03
1

Nobody asked you for the code. I am only asking a simple question. What is the angle of the small triangle, in degrees? What is the angle of the large triangle? And how did you calculate them? – Nov 30 '19 at 17:57
@Rahul Calculation is certainly possible. But in fact you don't need to calculate to know that the corresponding angles must be equal. why? Because the corresponding sides are parallel and they all intersect with the third side, the isoposition angles are equal, so the corresponding angles are equal. – z.qmpx Nov 30 '19 at 22:37
2

I don't believe you. Find the actual angles and tell me how you calculated them. – Dec 01 '19 at 04:44
@Rahul This angle is easy to calculate. First, the equation of circular arc plane is solved, and then the angle between planes is calculated by plane normal.If you study mathematics, the calculation is very simple. With the plane equation, there is a formula to calculate the angle between planes. – z.qmpx Dec 02 '19 at 14:28
2

I see, so you are using a wrong definition of angle as well. Here's a question for you: if two circles on the sphere meet tangentially, should their angle of intersection be zero or nonzero? – Dec 02 '19 at 15:59
@Rahul Notice that K in my formula, K, is the angle between the arc plane and the equatorial plane (of course, K is the tangent of this angle). Whether the arc has only one contact point with the equator or not. That is, even if there is only one contact point between an arc and an arc, their angles are not equal to zero. – z.qmpx Dec 03 '19 at 02:29
1

Then it is not really measuring the angle between the two curves on the surface of the sphere, is it? – Dec 03 '19 at 12:12
@Rahul You should know that spherical and plane angles are different. A spherical angle is a dihedral angle. It is a plane-to-plane relationship. – z.qmpx Dec 04 '19 at 22:54
1

@z.qmpx: Certainly, you are allowed (even encouraged) to define spherical angles as dihedral angles (and lines as small circles) and see where that intellectual exercise takes you, but it is inappropriate to state such definitions as if they were the accepted convention in spherical geometry, since they are not. BTW: One can easily construct small-arc "squares" with "dihedral" angle sums of $2\pi$ —intersect the sphere with a centered square cylinder— contradicting your question's "should be noted" note that such angle sums are greater. – Blue Dec 05 '19 at 04:57
@Blue When non-Euclidean geometry was first published, people did not accept non-Euclidean geometry. Now I say that the small circle is also a straight line geometry, and people don't understand it. Everything is being explored, and it takes time for people to understand. – z.qmpx Dec 05 '19 at 10:14
@Blue: Actually that's a very interesting point. This way one can construct a spherical square with all sides being equal and all "angles" being $\pi/2$ -- thus, it would actually be "similar" to a planar square according to z.qmpx's definition :) I wonder if we can make all sorts of spherical squares that are "similar" but look totally different. – Dec 05 '19 at 11:02
@z.qmpx: It's not so much that people don't understand your bold geometrical ideas, but that you present those ideas in a way that tends to suggest that you don't understand the standard ones. Stop saying stuff like "spherical angles are dihedral angles" when what you mean is "in my geometry, spherical angles are dihedral angles"; the former causes readers to want to correct you, leading to lots of back-and-forth and frustration all around; the latter invites them to consider your perspective. As a teacher of your ideas, you bear responsibility for your students' (lack of) understanding. – Blue Dec 05 '19 at 11:05
@z.qmpx: As for comparisons to non-Euclidean geometry ... Note that we call non-Euclidean geometry "non-Euclidean geometry", so as not to confuse folks when terminology comes into conflict with standard/Euclidean usage. Saying "the maximum area of a triangle is $\pi$" is nonsense without proper context. (The proper context is hyperbolic geometry, specifically on a surface of curvature $-1$.) That's why I suggested long ago that you talk about "E.wei curvature", and more recently "quasi-similarity". Until your ideas become standard, you can't expect people to know how you've re-defined terms. – Blue Dec 05 '19 at 11:17
@Blue I don't know what your square looks like. You should give a picture in your answer. – z.qmpx Dec 05 '19 at 11:18
1

@z.qmpx: "It's very simple. I can't help it if you don't understand the [concept of a square cylinder]." ... But seriously: Take a(n infinite) cylinder with a square cross-section; it consists of (regions of) four planes that make right dihedral angles along its four edge-lines. Let a sphere have its center on the cylinder's axis, and let its diameter be no shorter than the cross-section's diagonal. The cylinder's edge-lines determine vertices of a quadrilateral (actually, two of them) with congruent small-arc sides and four right dihedral angles. You should be able to draw this yourself. – Blue Dec 05 '19 at 11:37
@Blue This is also possible. I say everything is under exploration. – z.qmpx Dec 05 '19 at 11:54
@Rahul: More simply, consider a cylinder with any particular plane triangle as a cross-section. This will meet the sphere in a (small-arc) "triangle" whose (dihedral) "angles" naturally match those of the cross-section. Is it possible to position the cylinder so that the (small-arc) "sides" of the "triangle" have lengths proportional to the sides of the cross-section? It's certainly so when the cross section is equilateral, but in general? Hmmm ... (@ z.qmpx: See how "qualifying" our terms allows us to have a perfectly satisfying conversation in your geometry without fear of confusion?) – Blue Dec 05 '19 at 11:56
@Blue Everything is being explored and I keep an open mind. – z.qmpx Dec 05 '19 at 12:19
@z.qmpx: "This is also possible." ... I'm glad you agree. ... But don't forget my point: Defining "spherical angles" as dihedral angles for these squares contradicts the "greater than $2\pi$" angle-sum declaration in your question. Is it any wonder that "people don't understand" your ideas? Making absolute statements, and then chastising people for being "wrong" when they disagree or get confused by your own erroneous explanations, doesn't make you look like an explorer with an open mind. ("Extended discussion" warning. I don't chat, so I won't reply further.) Good luck! – Blue Dec 05 '19 at 12:28
@Blue I said everything is being explored, so I make mistakes. But I ca n’t stop exploring because I make mistakes. – z.qmpx Dec 05 '19 at 13:17
@z.qmpx: (One last reply.) I'm not suggesting that you stop exploring. (Mathematics research is exceedingly satisfying, and I want to encourage everyone to do it!) Rather, I'm suggesting that you show more humility and consideration when discussing your ideas with others. That includes everything I've mentioned to you, from taking care to clarify your re-definitions of standard terminology instead of expecting everyone to just know what you mean, to taking your fair share of the blame when someone misunderstands you instead of treating them like they're somehow inferior, etc. Good luck. – Blue Dec 05 '19 at 13:57
@Blue You are making mistakes, you always say last, but you always break your promise. – z.qmpx Dec 05 '19 at 23:22
1

@Blue Your thoughts and suggestions are of reference significance. – z.qmpx Dec 06 '19 at 00:04

Is there a similar triangle on the sphere?

2 Answers2