😄 Really? What is your background?

Thanks so much. I will definitely check out that paper.

“Plane and simple” 😂😂😂 humanity has fallen

Please see comment below.

Perhaps a very strong magnet.

You don't need a coordinate system to do geometry or topology. That is what Euclidean geometry is mainly based on--we want to discover intrinsic properties of the shapes themselves. Coordinate systems are introduced in algebra to analyze functions and then later on used for calculus. You can change the length of the sides as long as they still obey the triangle inequality: x+y>z.

@@learning.isnt.linear OK, that is just the problem I face ;-) If I have a line of length 5, another one of length 4, one endpoint of each line coincides, and there is a third line of length 3 and the free endpoints of the first two lines coincide with the third line, what makes this a triangle? Or worded differently: If I have 3 lines of length 3, 4, 5, what do I need to make a triangle?

@@user-fl5nv7oh3z [Line segments, not lines] Ok, so what you're getting at is the definition of a polygon. You are concerned about creating a piece-wise linear function looking-type shape. You can look this up on Wiki, but in short one can say we have this shape with straight sides (edges), and each node (or vertex) of an edge must have valence 2. And the angle between any two consecutive edges cannot be 180 degrees. There are other classifications, but this is one.

@@user-fl5nv7oh3z Check out my book which has a chapter on geometry where I talk extensively about congruence maps.

@@learning.isnt.linear "is not linear" ;-) sometimes it even goes forward and backward. I desperately try to be understood, so I ask the same question in different ways, but feel misunderstood. So I may ask: if there is a Pythagorean triple, does this imply there is Euclidean Norm? If there is a right angled triangle (geometry) I can construct the squares over the sides and then see, that the smaller areas are equal to the larger one. Now, seen from the other direction: there are three numbers A + B = C, is this equivalent to have a right angled triangle with the side length equal to the square roots of A, B, C respectively?

You have three similar right triangles: the big one and the left and right smaller ones. The corresponding sides of similar polygons (in this case triangles) are proportional. Drawing extra angles might help. Make alpha the angle between side 'a' and side 'u' and beta the other angle in the left smaller triangle. Then beta is also the angle between side 'b' and side 'v'. Take out those two triangles: (1) w/ sides a,b,c and the other other (2) w/ sides u,a. Align the corresponding sides and angles and we see that a/u = c/a ---> a²=uc.

​@@learning.isnt.linear thanks, now it's clear

Oh yes, of course. It is just a way to approximate a square in the curve and a class could practice constructing simple closed curves in Desmos and finding a square in their partner's curve. But I was thinking about this point earlier and I wondered if it could be the case that what _seems_ like good approximation for the square turns out not to work at all🙄, and the "true" square is in a completely different position on the curve. Though there can be more than one square for some curves Also, your question reminded of the solving of the equations part. I just wanted to say in general nonlinear equations are undecidable to solve, but perhaps I should have mentioned the fact that on a compact domain (which we have in this case [0,2\pi]) there are some nuances, but I don't think it helps. We do notice we will always have the trivial solution (0,0,0,0) or more generally (p,p,p,p).

Also, for smooth curves (C^2, even C^1 [piece-wise smooth] ) it has been proven we can always find at least one square. But, yes for certain C^0 curves this would not work because it is hard to tell where the points really are. The problem is open for C^0 curves mainly in the form of some monstrosus non-symmetric fractal which is differentiable nowhere.

Thank you for the honest feedback.

Do you use Desmos a lot?

Yeah, research is hard.

Nice copy paste comment bot

@@dgafbrapman688 nah dude, I'm just giving an alternative somewhere between eating plastic particles in your food and going out and buying a new pepper grinder, while explaining why.

This is a good insight.

Haha.

Glad I could help.

Thank you. It took a lot of work.

Glad you enjoyed it.

That is false. The largest circle lies within the hexagon.

@@learning.isnt.linear Yup, I'm wrong about it. The largest one is the hexagon inside the cube.

@@jakeaustria5445 Yes, thank you. That was very humble of you. Then you need to develop these rotation matrices in order to embed the circle inside this regular hexagonal plane. Also, check out my book on amazon.com.

@@learning.isnt.linear Sorry, I haven't watched the whole video yet when I wrote that.

@@jakeaustria5445 Where are you from?

Thank you. Totally agree!