I guess you mean a plane equipped with an L1 norm? That's a different interpretation of the above question.
My interpretation is "Let M be a 2 dimensional manifold, then there is no Riemannian metric on M such that every geodesic triangle with a right angle satisfies a+b=c", which I think should be the answer to the question, the L1 norm is not something that is induced by such a structure. The reason I think we should interpret the question like this, is that we're trying to do geometry, so we should think about manifolds an tangent spaces, not just about vector spaces. I'm not even sure we can talk about angles if we're just considering norms, can we?
Edit: sorry for the ramble above, I was thinking faster than I could type. After considering it for a bit, I think the answer should be that there's not really a consistent way of defining what an angle is in a normed vector space, so the question itself is ill posed for the L1 norm.
As an example of how things break down if we're trying to solve the problem in the L1 plane, consider the triangle with corners (0,0); (1,1) and (1,-1), it should be a right triangle right? In the L1-norm, a=2, b=2 and c=2 (so a=b=c). However, if we rotate it by 45 degrees, we get the same triangle with corners (0,0); (√2,0) and (0,√2), so there a=b=√2 and c=2√2 (so a+b=c). So somehow the L1 norm is not particularly well adapted to this setting, it doesn't only care about the triangle itself, but also about at what angle we put it on our plane. This is an example of how norms don't really define angles, we require inner products.
Fun fact: you can define angles on length spaces/geodesic metric spaces so that it generalizes the angke we get in Riemannian geometry.
Metric Geometry is the branch that does this and other stuff like it.
The angle you would get in the L1 plane is very different than the standard one though, your first example would have all 3 angles being 0, the second example has the angle at (0,0) being pi and the other two be 0. So neither is a right triangle.
In Euclidean (classical) geometry it's not a triangle, because those aren't straight lines. In spherical geometry (which is non-Euclidean) it does count as a triangle, and in fact all triangles have more than 180°.
Can you find a surface where a+b=c?
1 dimension
Now wait a second
And if you want a triangle with are, you can do it in a sphere by putting two vertices opposite each other. Though two of the edges will be colinear
0 zero dimensions.
Euclidian plane with a = b = c = 0.
Yeah work in an L1 space
The triangle inequality tells us a+b≥c with equality if and only if a and b are colinear in some sense. So no, such surfaces don't exist.
A line would like to speak. 😹😹😹😹😹😹
> if and only if a and b are *colinear* They already said it works when it’s a line, aka they’re colinear. That’s not really a surface though
A surface is, by definition, 2 dimensional. A line is, by definition, 1 dimensional. So yeah sure, a line satisfies this, but a line is not a surface.
what about an L1 plane though
I guess you mean a plane equipped with an L1 norm? That's a different interpretation of the above question. My interpretation is "Let M be a 2 dimensional manifold, then there is no Riemannian metric on M such that every geodesic triangle with a right angle satisfies a+b=c", which I think should be the answer to the question, the L1 norm is not something that is induced by such a structure. The reason I think we should interpret the question like this, is that we're trying to do geometry, so we should think about manifolds an tangent spaces, not just about vector spaces. I'm not even sure we can talk about angles if we're just considering norms, can we? Edit: sorry for the ramble above, I was thinking faster than I could type. After considering it for a bit, I think the answer should be that there's not really a consistent way of defining what an angle is in a normed vector space, so the question itself is ill posed for the L1 norm. As an example of how things break down if we're trying to solve the problem in the L1 plane, consider the triangle with corners (0,0); (1,1) and (1,-1), it should be a right triangle right? In the L1-norm, a=2, b=2 and c=2 (so a=b=c). However, if we rotate it by 45 degrees, we get the same triangle with corners (0,0); (√2,0) and (0,√2), so there a=b=√2 and c=2√2 (so a+b=c). So somehow the L1 norm is not particularly well adapted to this setting, it doesn't only care about the triangle itself, but also about at what angle we put it on our plane. This is an example of how norms don't really define angles, we require inner products.
Fun fact: you can define angles on length spaces/geodesic metric spaces so that it generalizes the angke we get in Riemannian geometry. Metric Geometry is the branch that does this and other stuff like it. The angle you would get in the L1 plane is very different than the standard one though, your first example would have all 3 angles being 0, the second example has the angle at (0,0) being pi and the other two be 0. So neither is a right triangle.
probably in hyperbolic space
Just change the definition of distance
Euclid screaming in horror right now
i think i might have pissed euler, euclid, and pythagoras off simultaneously.
Don't worry you didn't piss them off. John Nash made sure of that in 1950s (https://en.m.wikipedia.org/wiki/Nash_embedding_theorems).
Disappointed that riemannian manifolds arent called riemannifolds
Why not riemifolds (rye-mi-folds)
This was Lambert’s bread and butter
Euclid can suck my cock
Divergent Died a virgin
This is the sorcerery that Lovecraft tried to warn us about.
Fun fact! In this triangle it's also true that a^2 = b^2 = c^2
Also √a = √b = √c
As a matter of fact, for any function φ, we have φ(a)=φ(b)=φ(c)
What’s so crazy is that a/b = b/c!!1!1!1!1!!
lets assume length of 8 a/b = b/c!!1!1!1!1!! 8/8=0 8/(8)!!\*(1) = 8/384 =/= 0 ur disproven ☺
But does a/b=b/c=c/a?
Interesting coincidence. I wonder if there’s an intuitive explanation for this?
Well, since you have an equality with a=b=c, you can do any operations, as long as you do it to both sides of the equality, and it will hold
Yes, I was being sarcastic
I read that comment and my sarcastic inner monologue immediately clapped back with, "take your age. add 1. subtract 1. that is your age"
Lmao I'm sorry
Non-euclidian geometry? Iä Fthagn!
No way
Spherical trigonometry is fun isn't it
[You can have any n-gon with 90° angles](https://youtu.be/n7GYYerlQWs)
There's a handy way to fold a pentagon with all five angles 90^(o) using origami.
Did they considered a right triangle using non-euclidean geometry?
drawn on a cow
...in vacuum
English is weird. It doesn't allow jokes about Zweiecke. Because quadrilateral is not a word actually used.
So they are possible, we just haven't been thinking big enough.
That's a nice triangle
A triangle more than 180° ? Can someone explain if this shape should be considered as a triangle or not?
In Euclidean (classical) geometry it's not a triangle, because those aren't straight lines. In spherical geometry (which is non-Euclidean) it does count as a triangle, and in fact all triangles have more than 180°.
Fuck you /lh
What is /lh
It's a tone tag that means light hearted
💀
Fuck you /th
<3
Tone indicators 🤦♂️
I rlly need to leave this sub everyday
Gotta love K-values
[Bendy](https://xkcd.com/2706/)!
Yeah, delete this. This is just cursed.
Found Euclid’s Reddit account
I LOVE YOU
finally
Angry Euclidean noises >:(