Baltic Way competition problems, 2013

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
$$\frac{AB \cdot BC \cdot CD}{6} = \frac{s^3}{6}.$$