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Professor Erasmus has constructed a special tetrahedron that he modestly calls the "Professor-Erasmus-tetrahedron". The professor claims that all four faces of his tedrahedron are right-angled and congruent to each other.

Has the professor once again made one of his mathematical blunders, or do such tetrahedrons indeed exist?

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No, it's not possible.

For any vertex, the three edges coming out of it must have distinct length, since otherwise the triangle must be isosceles as well as right, which forces an equilateral triangle for the opposite face. Therefore, the three angles at each vertex must correspond to the three angles of the right triangle; call them $\alpha$, $\beta$, and $90^\circ$. But since $\alpha + \beta = 90^\circ$, the faces must join flat at the vertex, forming the degenerate flat tetrahedron shown below.

enter image description here

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    $\begingroup$ +1. I had an answer, but this answer made me check my math. Sure enough, the angle up from the base was 0 (i.e. my tetrahedron was flat). $\endgroup$ – Joel Rondeau Feb 2 '15 at 14:51

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