Geometry of the Tetrahedron

Steven Dutch, Professor Emeritus, Natural and Applied Sciences, Universityof Wisconsin - Green Bay

Since the tetrahedron occurs so often in crystallography, it's useful to summarize its geometry.

In the diagram above left,

Edges are defined as equal to unity: AB=BC=CA=AO=BO=CO=1
AD=DB=BE=EC=1/2
ABC is an equilateral triangle, so AE=CD = sqrt(3)/2, etc.
The medians of a triangle all intersect at a point that divides the medians in ratio 1:2. Thus FE = AE/3 and AF = 2AE/3, etc.
Since AE = sqrt(3)/2, AE = sqrt(3)/3 and FE = sqrt(3)/6, etc.
Finally, since AO=1 and AF = sqrt(3)/3, OF = sqrt(2/3) = sqrt(6)/3 = 0.8165.

The altitude of a tetrahedron = sqrt(2/3) = sqrt(6)/3 = 0.8165

The diagram above right illustrates how to find FX.

OF is perpendicular to AF and AG is perpendicular to OG
Since they are both right triangles and have angle O in common, triangles OFB and OXG are similar.
Therefore XG/fB=OG/oF
By symmetry, XF = XG, so XF/fB = OG/oF
XF = FB*OG/oF
FB = sqrt(3)/6, OG = sqrt(3)/3 and OF = sqrt(6)/3
Therefore XF = sqrt(6)/12 = 0.2041
Since OF = sqrt(6)/3 and XF = sqrt(6)/12, XF=OF/4

The centroid of a tetrahedron is one-fourth of the way from base toopposing vertex.

Created 29 August 2002, Last Update 31 May 2020