# Bimedians in a Regular Tetrahedron

Six edges of a tetrahedron fall naturally into 3 pairs of the opposites - the non-intersecting edges. From the point of view of the four vertices, the latter can be split into 2 pairs in three ways. Either way, there are exactly three line segments joining the midpoints of the opposite edges. Three such segments intersect in a point which is the barycenter of the four material points of equal mass placed at the vertices of the tetrahedron. These are called *bimedians* of the tetrahedron. The applet helps observe the fact that in a regular tetrahedron the bimedians are pairwise orthogonal. It also suggests a simple proof of this fact.

What if applet does not run? |

A regular tetrahedron can be embedded into a cube. The bimedians then serve to join the centers of the opposite faces of the cube. By any measure, these lines are pairwise orthogonal.

Obviously, the same holds for a tetrahedron that can be embedded into a parallelopiped. For this to happen, the bimedians need to be orthogonal to the edges they join. This is a generalization of sorts, but it is not very interesting. Posing a problem we in fact stipulate its solution: if the bimedians of a tetrahedron are orthogonal to the edges they join, the three are then pairwise orthogonal. In a regular tetrahedron the required orthogonality is an unannunciated property implied by the embedding into a cube.

### References

- C. W. Trigg,
*Mathematical Quickies*, Dover, 1985, #148

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny