Tetrahedron with Orthogonal Perpendiculars to Opposite Edges

Tetrahedron in which common perpendiculars of the opposite edges are pairwise orthogonal is isosceles; and vice versa.

Proof

Let $AB$ and $CD$ be a pair of opposite edges of a tetrahedron $ABCD.$ Let $\alpha$ be the plane through $AB$ parallel to $CD,$ and $\gamma$ the plane through $CD$ parallel to $AB.$ The common perpendicular, say $h,$ to $AB$ and $CD$ is perpendicular to both of these planes.

These two planes define the opposite faces of the parallelepiped associated with the tetrahedron. Similar arguments apply to the other two pairs of opposite faces.

As a general rule, the normals to two planes are perpendicular only if the planes are perpendicular. Thus the common perpendiculars of the opposite edges are pairwise orthogonal iff this is true of the pairs of opposite faces of the parallelepiped associated with the tetrahedron.

It follows that the parallelepiped is a cuboid, which is equivalent to its bing isosceles.

This problem has been given in 1984 at a Romanian selection test for the Balkan Mathematical Olympiad.

References

  1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.8 (p. 25)

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