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My intuition tells me that there's an extra node not pictured, let's call it Z. There's a triangle ADZ that's out of view and a planar surface DZCF that mirrors DABE.
I'm sure my intuition is wrong too, but without pulling out some kind of CAD I can't really see why.
It appears to me to be 1 triangular face and 4 quadrilateral ones. The one they list as ABC should be ABCX where X is a hidden fourth corner on the base. Which would render the shape possible.
I personally don’t see the image as “impossible”, in terms of seeing it as a projection of a 3D object. I didn’t interpret ABDE as being flat (A, B, D, and E coplanar), and I didn’t expect G, I, and H to be intersections of lines in 3D.
It feels like it's a problem similar to spending to much time doing "2 + 5 = _" problems and thinking the equality symbol is directional, in this case, spending too much time looking at figures that do meet at a point and thinking that is obligatory for all figures.
I'd wonder if there is some topology theorem that generalizes hamiltonian paths into higher dimensions, where the difference between a graph and a solid is whether it has a hamiltonian path. e.g. this shape is "impossible" as represented because of the lack of a hamiltonian path.
I didn't really get it at first, in fact, I might still not be getting it, but if you draw a line from A to C, then you can see that the triangle ABC will not be similar to DEF... when it should be? I don't know.
For it to be a real pyramid, you only need the side edges lines to meet at the apex, which always happen when the top and bottom triangles are similar.
Similar triangles are a sufficient, but not necessary condition.
It becomes possible again if you cut the front face into two triangles, that being the "hidden edge" the author mentions.
The pyramidal explanation sounds like it's using a theorem that I didn't know before, but without proving it. You could say the explanation deserves its own explanation. It's an interesting fact, that the lines have to intersect, nonetheless.
I understand the argument some people are making as "my intuition tells me it's impossible because $X" and that makes sense. My intuition tells me that there's an extra node not pictured, let's call it Z. There's a triangle ADZ that's out of view and a planar surface DZCF that mirrors DABE. I'm sure my intuition is wrong too, but without pulling out some kind of CAD I can't really see why.
I personally don’t see the image as “impossible”, in terms of seeing it as a projection of a 3D object. I didn’t interpret ABDE as being flat (A, B, D, and E coplanar), and I didn’t expect G, I, and H to be intersections of lines in 3D.
If my "geometric intuition" is working properly, the "problem" is that the figure in the picture wouldn't meet in a point. There would be a line at the top, and it wouldn't be a pyramid. But there's nothing "impossible" about that. The impossibility simply seems to be an assertion of impossibility. It feels like it's a problem similar to spending to much time doing "2 + 5 = _" problems and thinking the equality symbo…
The proof given in the article seems fine. Assuming the figure has three flat faces, the arrangement of those faces is impossible. A figure such as you describe, with a line on the top, would not be ruled out by the proof, but the depicted figure cannot match that description.
For a quick summary-style restatement of the proof:
1. Consider the three sides (as opposed to the top and bottom) of the shape to be flat. Each of them will come to a separate point. Those three points are labeled G, H, and I.
2. We can easily show that the point G lies in the same plane as each side of the shape. We can symmetrically show that this is also true of H and of I.
3. When G, H, and I are the same point, this doesn't restrict the sides in any meaningful way - no matter what the "angles" of three planes are, you can always translate them such that they'll all intersect at an arbitrary point.
4. But when G, H, and I are all different points, there is only a single plane that contains them all. ("Three points determine a plane".) This tells us that the three faces of such a shape would all be coplanar, which obviously can't happen.
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(5. You are positing that, for example, G and H might coincide while I is a different, second point. But the depicted figure doesn't satisfy that description.)