4
$\begingroup$

If such a pyramid exists, could someone provide the coordinates of its vertices?

$\endgroup$
5

3 Answers 3

21
$\begingroup$

This picture of a quadrirectangular tetrahedron is from István Lénárt, `The Right Triangle as the Simplex in 2D Euclidean Space, Generalized to $n$ Dimensions' enter image description here

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.