All faces of a tetrahedron are right-angled triangles. It is known that three of its edges have the same length $s$. Find the volume of the tetrahedron.
The three equal edges clearly cannot bound a face by themselves, for then this triangle would be equilateral and not right-angled. Nor can they be incident to the same vertex, for then the opposite face would again be equilateral.
Hence we may name the tetrahedron \(ABCD\) in such a way that \(AB = BC = CD = s\). The angles \(\angle ABC\) and \(\angle BCD\) must then be right, and \(AC = BD = s\sqrt{2}\). Suppose that \(\angle ADC\) is right. Then by the Pythagorean theorem applied to \(ACD\), we find \(AD = s\). The reverse of the Pythagorean theorem applied to \(ABD\), we see that \(\angle DAB\) is right too. The quadrilateral \(ABCD\) then has four right angles, and so must be a square.
From this contradiction, we conclude that \(\angle ADC\) is not right. Since we already know that \(AC > CD\), \(\angle CAD\) cannot be right either, and the right angle of \(ACD\) must be \(\angle ACD\). The Pythagorean theorem gives \(AD = s\sqrt{3}\). From the reverse of the Pythagorean theorem, we may now conclude that \(\angle ABD\) is right. Consequently, \(AB\) is perpendicular to \(BCD\), and the volume of the tetrahedron may be simply calculated as

\[
\frac{AB \cdot BC \cdot CD}{6} = \frac{s^3}{6}
\]