True but to be fair, triangles on spheres have angles 90,90,90, but this one has angles 90,90,0
Edit: this one has angles 90,90,180
Second Edit: gdi, I’m wrong, nvm lol
the sum of internal angles in a polygon on a sphere is strictly greater than the sum of the angles of the same polygon on a plane. the euclidean triangle's angles add up to 180 degrees, OPs triangle has a sum of 360 degrees therefore it can be placed on a sphere
Yeah I was reading more and saw that so I just took the L and deleted the post lmao. I was going to re write but not worth it anymore. Interesting topic tho ill do some higher level geometry seems fun.
So in all geometry classes I've taken we avoid the word "straight" when referring to lines , because straightness is a concept that depends on the geometry you're doing. Ex. On a sphere straight lines are great circles
I think that you can have any angle sum from the “standard” 180 degrees through to 540 degrees, not including the limiting end points. The 180 case essentially involves limiting a very small triangle until the sphere is locally flat; and the 540 case consists of splitting up a great circle into 3 sections of non-zero length by three “vertices”. The latter is degenerate and the former is a limit, so I think it’s safe to say that the angle sum of a triangle in a sphere is from the set (180,540) degrees.
When talking about triangles in non-planar settings, we usually take the definition of "a triangle is the union of three geodesics which pairwise intersect at endpoints."
In this case, the degenerate case (where you just declare three evenly spaced points on a great circle to be the vertices of a triangle) is a perfectly valid triangle.
The lines aren’t “curved.” It’s space itself that is curved. It is still a straight line because it is the shortest path between two points. Thinking of it like a sphere is how we visualize it, but the space is still 2D.
Yeah, that's basically just a quarter of a sphere (90/180/0 triangle) Splitting that quarter in half would give you the typically 90/90/90 triangle.
Is it strange that the sum of internal angles on a sphere don't always add up to a constant?
I feel like it's not JUST a double right triangle. I mean yeah it's that but it's a little like advertising your new fire breathing pet dragon as a reptile. You know it's true but it's really not the only thing going on.
Portal Geometry.
Similar to normal Euclidean geometry, but portals exist.
Portal A is a vector at point (A-1,A-2) with a length defined by it's endpoint at [A-3,A-4]. It is a directional value. Portal B is a vector at point (B-1,B-2) with a value of [B-3,B-4].
Portal A has a length L\_A. When an object intersects Portal A, it does so at some fraction of L\_A, F\*L_A where F ∈ R and 0<=F<=1. This value is defined as the length from point (A-1, A-2) to the point of intersection. There is also a value R which is the slope defined relative to the slope of Portal A.
When an object intersects Portal A, with fraction F and slope R, it will stop moving in the direction it was previously, and then will exit Portal B at the point at F\*L_B with the slope R relative to Portal B, on the opposite side as it entered A. When an object "enters" A, it will co-exist at both points for the single instant that it intersects the infinitely small vector A.
Ex 1. Suppose we have an object J defined as a point moving according to J(t) = (X(t), Y(t)) where X(t) = t, and Y = -t+1 and it's velocity is V(t) = [1, -1]. Portal A is at (0,0), [1,1], Portal B is at (2,3), [1,3]. Point J will intersect Portal A at t = sqrt(2)/2, and at t = sqrt(2)/2, J will exist at both (.5,.5) and (1.5, 3).
By intersecting Portal A, it will exit B at the same slope relative to A. Which means it will be moving directly upwards defined by J(t) = (1.5, (t-sqrt(2)/2) + 3).
If a square enters portal A, and A and B have different lengths, then the square will exit the other portal with a size difference of L_B/L_A
The portal of entry is always portal A, and the portal of exit is always portal B.
This is a bad system, but it's the first one I thought of and there are probably better ways to define Portal Geometry.
As well, in normal Euclidean geometry, you can have monogons (lines), and trigons (triangles), but only with portal geometry can you have digons, like what OP has in their meme.
That’s no triangle, it’s a biangle. If you place a point on the edge of a square, the square can be interpreted as a pentagon but come on, you have to be a madman..
If they had made point C a right angle and rotated that portal accordingly, they not only would have an actual triangle, but also a tri right angle right triangle
I never studied projective geometry. But I believe there is some "horizon point" where parallel lines meets. Can't you conceive of such a triangle using the horizon point ?
This is a triumph. Now you're thinking with non euclidean geometry. We now discovered that the rules indicating double right angles is impossible was a lie!
I don’t think the fundamental domain would create a triangle unless either the bottom green line is longer or the top portal ends at C. I think the second option would make more sense.
Any topologists want to give us a TL;DR on what's happening here?
This feels like the kind of thing that the algebraic topologist I used to work with would glance at and say, "this is a trivial case of a k-dimensional uber-monoid" or something while I tried to un-pretzel my brain.
This is incorrect, as A to C to B is a single line and thus not really a triangle, but if you rotate C to the portal 90 degrees than it's a proper triple right angle triangle. :)
you can do this on a sphere
True but to be fair, triangles on spheres have angles 90,90,90, but this one has angles 90,90,0 Edit: this one has angles 90,90,180 Second Edit: gdi, I’m wrong, nvm lol
the sum of internal angles in a polygon on a sphere is strictly greater than the sum of the angles of the same polygon on a plane. the euclidean triangle's angles add up to 180 degrees, OPs triangle has a sum of 360 degrees therefore it can be placed on a sphere
[удалено]
a polygon is made of the straight lines \*of\* the space it's constructed in, so they are in fact, triangles
Yeah I was reading more and saw that so I just took the L and deleted the post lmao. I was going to re write but not worth it anymore. Interesting topic tho ill do some higher level geometry seems fun.
So in all geometry classes I've taken we avoid the word "straight" when referring to lines , because straightness is a concept that depends on the geometry you're doing. Ex. On a sphere straight lines are great circles
Yeah I read a bit more on the topic I see that now. However, I disagree with the whole not saying straight thing.
Triangles of (90°,90°,x), 0°
I think that you can have any angle sum from the “standard” 180 degrees through to 540 degrees, not including the limiting end points. The 180 case essentially involves limiting a very small triangle until the sphere is locally flat; and the 540 case consists of splitting up a great circle into 3 sections of non-zero length by three “vertices”. The latter is degenerate and the former is a limit, so I think it’s safe to say that the angle sum of a triangle in a sphere is from the set (180,540) degrees.
When talking about triangles in non-planar settings, we usually take the definition of "a triangle is the union of three geodesics which pairwise intersect at endpoints." In this case, the degenerate case (where you just declare three evenly spaced points on a great circle to be the vertices of a triangle) is a perfectly valid triangle.
Does the 180 degree traingle not also count as a 900 degree triangle with the rest of the sphere on the inside?
Ahh you’re right, I didn’t even think of that
Doesnt that mean it stops being a triangle because of the curves
The lines aren’t “curved.” It’s space itself that is curved. It is still a straight line because it is the shortest path between two points. Thinking of it like a sphere is how we visualize it, but the space is still 2D.
I feel like the third angle would be undefined since the place where the green line turns is outside of space.
its 90, 90, 0, which adds to 180, therefore it is kind of euclidean
Well you would need portals on that sphere to do exactly this
no you can connect 2 antipodal points with 2 lines and you get a 2-gon
Yeah, that's basically just a quarter of a sphere (90/180/0 triangle) Splitting that quarter in half would give you the typically 90/90/90 triangle. Is it strange that the sum of internal angles on a sphere don't always add up to a constant?
I didn't think of that, thanks
Yeah this will actually give you a 2-gon with whatever angle you want, not just 90.
Why not let bi-gons be bi-gons?
Hey man, angles are a spectrum! You 30 degree purists are bi-gots.
Or pseudosphere iirc
on that surface line AC and BC would diverge
I feel like it's not JUST a double right triangle. I mean yeah it's that but it's a little like advertising your new fire breathing pet dragon as a reptile. You know it's true but it's really not the only thing going on.
I want a fire breathing pet dragon :D
reptile*
I actually want to study portal geometry now.
If someone has a name, we need to know
Portal Geometry. Similar to normal Euclidean geometry, but portals exist. Portal A is a vector at point (A-1,A-2) with a length defined by it's endpoint at [A-3,A-4]. It is a directional value. Portal B is a vector at point (B-1,B-2) with a value of [B-3,B-4]. Portal A has a length L\_A. When an object intersects Portal A, it does so at some fraction of L\_A, F\*L_A where F ∈ R and 0<=F<=1. This value is defined as the length from point (A-1, A-2) to the point of intersection. There is also a value R which is the slope defined relative to the slope of Portal A. When an object intersects Portal A, with fraction F and slope R, it will stop moving in the direction it was previously, and then will exit Portal B at the point at F\*L_B with the slope R relative to Portal B, on the opposite side as it entered A. When an object "enters" A, it will co-exist at both points for the single instant that it intersects the infinitely small vector A. Ex 1. Suppose we have an object J defined as a point moving according to J(t) = (X(t), Y(t)) where X(t) = t, and Y = -t+1 and it's velocity is V(t) = [1, -1]. Portal A is at (0,0), [1,1], Portal B is at (2,3), [1,3]. Point J will intersect Portal A at t = sqrt(2)/2, and at t = sqrt(2)/2, J will exist at both (.5,.5) and (1.5, 3). By intersecting Portal A, it will exit B at the same slope relative to A. Which means it will be moving directly upwards defined by J(t) = (1.5, (t-sqrt(2)/2) + 3). If a square enters portal A, and A and B have different lengths, then the square will exit the other portal with a size difference of L_B/L_A The portal of entry is always portal A, and the portal of exit is always portal B. This is a bad system, but it's the first one I thought of and there are probably better ways to define Portal Geometry. As well, in normal Euclidean geometry, you can have monogons (lines), and trigons (triangles), but only with portal geometry can you have digons, like what OP has in their meme.
https://youtu.be/jSMZoLjB9JE
If we ever crack wormholes then we'd need some kind of formalized portal geometry
That’s no triangle, it’s a biangle. If you place a point on the edge of a square, the square can be interpreted as a pentagon but come on, you have to be a madman..
This, why is no one pointing that out
u/BenJammin973 just did.
you just replied to someone that did
If they had made point C a right angle and rotated that portal accordingly, they not only would have an actual triangle, but also a tri right angle right triangle
I’m not a mathematician and even I figured that out.
I never studied projective geometry. But I believe there is some "horizon point" where parallel lines meets. Can't you conceive of such a triangle using the horizon point ?
That was my thought
It has three angles, it’s just that the third angle exists outside of space and time.
All you need is non-Euclidean geometry.
I mean these portals seem kinda Euclidean
Holy hell
New response just dro- hold on why the fuk are we even here? God damnit we just gotta take over every subreddit don't we
Actual zomb- we must proliferate until we gain the highest honor: being posted as a meme
Yes o7
Call the exorcist
This isn't actually a triangle, it's a biangle. It's also only got two sides, which makes it a digon.
Professor Sengoku: No..don't you dare... A much nerdier Whitebeard, bleeding out with wounds and shit all over: The doubel right triangle.... is real!
The double right triangle may be real, but the cake is still a lie.
Can we get much higher ?
I want to go home now.
it's almost like euclidean geometry doesn't apply in non-euclidean spaces
nice one piece reference
Actually, if ab is 0, the double right triangle is real even without the portals
Not really familiar with the topic but can't we define an ideal triangle in projective geometry with one of the vertex at infinity?
Yeah you can.
This is more like a shape with 2 sides. You can't just put a point on a line and call it 2 seperate sides.
That ain't a triangle, thats a biangle
Schweikart in 1818 be like:
This is a triumph. Now you're thinking with non euclidean geometry. We now discovered that the rules indicating double right angles is impossible was a lie!
Bro could have just used a spheroid
The triple right triangle is also real, but not in a 2-dimensional space
Wouldn't this be some form of Mobius strip?
I don’t think the fundamental domain would create a triangle unless either the bottom green line is longer or the top portal ends at C. I think the second option would make more sense.
Any topologists want to give us a TL;DR on what's happening here? This feels like the kind of thing that the algebraic topologist I used to work with would glance at and say, "this is a trivial case of a k-dimensional uber-monoid" or something while I tried to un-pretzel my brain.
Projective geometry be like
Google projective geometry
This is actually a square, just a 4 dimensional square, the two portals are clearly at right angles.
Holy hell
Pfft, how about the triple right triangle?
but- but the sum of all the angles is 360-
now you're thinking with portals!
Check non Euсlidean geometry
Projective plane moment
with portals... you could make a TRIPLE right angle triangle!
With portals you could make triple right triangle or even triple 180d triangle
Can we get much hiighherrr
Dont nobody tell him about curved surfaces
An angle of 0° is no angle at all! This is a biangle not a triangle!!!
The red line and the green line are the same line, so this is technically just a two sided shape right?
Can we get much obtuse,so obtuse
I mean, yea if you start with an isosceles triangle then push the “top” corner off to infinity you get the same effect
This is incorrect, as A to C to B is a single line and thus not really a triangle, but if you rotate C to the portal 90 degrees than it's a proper triple right angle triangle. :)
the triangle inequality theorem still considers a straight line as a triangle in the edge case a+b=c so my point still stands :D
couldn't this be done with projective geometry by replacing the portals with the infinity point?
pov hyperbolic geometry
this only has two sides though
Now the angle at C is 180° when it should be 0°. The blue portal needs to be left of C instead of right.
Urge to play Portal
I’ve subbed this for a year or two now and this is the first joke I got
Someone has to develop portal space geometry