When exploring penrose triangle, it's essential to consider various aspects and implications. geometry - Why is the Penrose triangle "impossible"? As another additional bonus point, although it's not possible to embed the Penrose triangle into normal, flat, Euclidean 3D space, it is possible to embed it into curved three dimensional space. The video below, by @ZenoRogue on Twitter, shows Penrose triangles embedded into something called "nil geometry". Another key aspect involves, can we compute the surface area and volume of Penrose Triangle?. The Penrose triangle is not logically self-contradictory. Nonetheless, with a suitable definition involving three equal-length sides of square cross section meeting perpendicularly, a Penrose triangle is incompatible with three-dimensional Euclidean geometry.
Cohomology, Penrose Triangle, and Shepards Tone. This is the fact that the Penrose Triangle represents a nontrivial cohomology class in the "group of distances from the observer" of the circle. Another key aspect involves, penrose has an article on this here: On The Cohomology of Impossible Figures.
Penrose Triangle as a 4D object - Mathematics Stack Exchange. It's important to note that, the first step towards answering this would be to clarify the precise sense in which the 2D Penrose Triangle image is not the projection of a 3D object (capturing the idea that it depicts an impossible object). This seems tricky, since in a simpler sense every 2D figure is a projection of a 3D figure, and we can even make 3D models that look like the Penrose Triangle from certain directions. Penrose Triangle and Umbilic Torus - Mathematics Stack Exchange.
The figure on the right shows the Penrose triangle with a square cross-section and also a single twist. Each possesses a single surface. Finally, the figure at the bottom shows a 3-D printing of the Penrose triangle. You can find more images and some animations at A New Twist on Möbius.
Penrose's remark on impossible figures - Mathematics Stack Exchange. The way Penrose used $\tau$ is to identify a 1-cocycle (the "distance cocycle") which is not a coboundary, the key fact being $\tau : C^1 (X) \to \Bbb R$ being 1 on coboundaries. I suppose it's kind of like integration to me, in the same way that integral over exact forms are $0$ can be used to come up with closed forms which are not exact. Moreover, how can I understand is the picture $2D$ or $3D$.
The Penrose triangle is not globally consistent in this sense; physical sculptures may consist of three straight bars not forming a spatial triangle, or may have curved sides that appear straight from some direction, but a closed triangle made of straight sides having square cross section and meeting spatially "as they appear to" is impossible. soft question - What is the true relationship between impossible .... The drawing domain of most impossible figures is a circle or an annulus. The reason they're impossible is that there is no single "height function" that covers all of it, even though there is a "steepness function". In other words, there exists a function that looks like it should be a derivative / gradient, but isn't.
That's exactly what non-trivial cohomology is all about. From another angle, this is impossible ... Equally important, law of cosines with impossible triangles - Mathematics Stack Exchange. This is not a directly a matter of hyperbolic geometry but of complex Euclidean geometry.
📝 Summary
Important points to remember from this discussion on penrose triangle demonstrate the relevance of understanding this topic. By applying these insights, one can make informed decisions.