Reach for Polydron

A tetrahedron has two identical equilateral triangles faces, of side length 1 unit. The other two faces are right angled isosceles triangles. Find the exact volume of the tetrahedron.

Tetra Inequalities

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.

The Dodecahedron Explained

What is the shortest distance through the middle of a dodecahedron between the centres of two opposite faces?

Tetra Perp

Age 16 to 18Challenge Level

 We can show that the edges $AD$ and $BC$ of a tetrahedron $ABCD$ are mutually perpendicular if and only if $AB^2 + CD^2 = AC^2 + BD^2$. A tetrahedron has three pairs of opposite edges. We have to prove that one pair are mutually perpendicular if and only if the sums of the squares of the lengths of the other two pairs are equal. Edward of Graveney School, Tooting, London sent in an excellent solution to this question using vectors.
It is easiest to follow if the essential symmetry is made obvious by the notation. In order to do this we use the position vectors ${\bf a, b, c}$ and ${\bf d}$ for the vertices of the tetrahedron. The scalar product ${\bf a.a} = a^2$ gives the square of the length of the vector ${\bf a}$.

We know that

\eqalign { AB^2 + CD^2 &= ({\bf b - a})^2 + ({\bf d - c})^2 \cr &= a^2 + b^2 + c^2 + d^2 - 2 {\bf b.a} - 2{\bf d.c}}

and

\eqalign { AC^2 + BD^2 &= ({\bf c - a})^2 + ({\bf d - b})^2 \cr &= a^2 + b^2 + c^2 + d^2 - 2 {\bf c.a} - 2{\bf d.b}.}

Therefore

\eqalign { (AB^2 + CD^2)- (AC^2 + BD^2) &= 2({\bf c.a + d.b - b.a - d.c}) \cr &= 2({\bf c - b}).({\bf a - d}) }

The right hand side of this expression is just twice the scalar product of the vectors $\mathbf{BC}$ and $\mathbf{DA}$ and we know that the edges $BC$ and $DA$ are mutually perpendicular if and only if this scalar product is zero. The result follows.