Geometry of the Tetrahedron
Steven Dutch, Natural and Applied Sciences, University
of Wisconsin - Green Bay
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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
- 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 to
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Created 29 August 2002, Last Update
25 January 2012
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