If such a pyramid exists, could someone provide the coordinates of its vertices?
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asked Sep 26, 2023 at 15:52
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$\begingroup$ Here’s a related Concept : en.m.wikipedia.org/wiki/Trirectangular_tetrahedron $\endgroup$– Sidharth GhoshalCommented Sep 26, 2023 at 16:00
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$\begingroup$ See also math.stackexchange.com/questions/3214139/… $\endgroup$– Marco RipàCommented Sep 26, 2023 at 16:06
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4$\begingroup$ "All four faces" seems to imply you mean a tetrahedron. If you had said "tetrahedron" rather than "pyramid", this would be clearer. $\endgroup$– Michael HardyCommented Sep 26, 2023 at 23:33
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$\begingroup$ If three right angles are at one vertex, then you just have the corner of one octant of the usual $(x,y,z)$-Cartesian coordinate system. Are there points $(x>0,0,0),\,\, (0,y>0,0),\,\, (0,0,z>0)$ that are the vertices of a right triangle? No. Now the alternatives are: two right angles and one acute angle at one vertex, or at most one right angle at each vertex. $\endgroup$– Michael HardyCommented Sep 26, 2023 at 23:38
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2$\begingroup$ en.wikipedia.org/wiki/Schl%C3%A4fli_orthoscheme $\endgroup$– Robin HoustonCommented Sep 27, 2023 at 0:34
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3 Answers
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This picture of a quadrirectangular tetrahedron is from István Lénárt, `The Right Triangle as the Simplex in 2D Euclidean Space, Generalized to $n$ Dimensions' 
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If by pyramid, you mean the shape with a square base and four triangular sides, and you want the right angles all where the triangles meet, there is a solution, but it is degenerate.
Sample Vertices: Set A: (0,0,0), (1,0,0), (0,1,0), (-1,0,0), (0,-1,0) Set B: (0,0,0), (1,1,0), (1,-1,0), (-1,-1,0), (-1,1,0)
By degenerate, I mean that all of the vertices are in the same plane (z=0, in my example), and the shape has zero volume. It's flat.
It is obvious that this must be the case. You have four right angles meeting at a point. That's 360 degrees, exactly what you need for them to be in the same plane.
If the first point in my example A was (0,0,1), then all of the edges would be the same length (the square root of 2), so the triangles would all be equilateral--and all of their angles would be 60 degrees. The more you raise that point, the smaller the angles there get.
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Giving this StackExchange one last try. In addition to the answer posted with triangular base, here is an example with a square base:
Use two 3-4-5 right triangles and two 3-5-sqrt(34) triangles.
