# Tetrahedra Tester

An irregular tetrahedron is composed of four different triangles.

Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?

How many different tetrahedra can be constructed?

Explain how you know you have found all the possible tetrahedra.

What are the limitations on the lengths of the sides of any triangle?

Look at different cases.

If the base is a 4 by 5 by 6 triangle, what size can the other faces be?

How many different tetrahedra can have as its base a 4 by 5 by 6 triangle?

Can you find 8 or more possible tetrahedra?

How will you know when you have them all?

Using the combination nCr,

6C3 = all the possible triangles = 20.

If you take each possible triangle as a "base", then use nPr to find the number of tetrahedra that can be constructed from one base.

3P3 = 6

We must realize that using these figures, one tetrahedron will be listed four times, each time with a different triangle on the base, but still the same tetrahedron. Therefore, we must divide the figure by 4.

The lengths 4, 5 and 9 cannot form a triangle, so tetrahedra with this value must be removed. There are 6 tetrahedra with base 4, 5 and 9, so there 4 times (24) this value listed. Therefore, we must subtract this value from the figure.

6 x 20 = 120 = all hypothetical tetrahedra

120 - 24 = 96 = all possible tetrahedra (inc. tetrahedra counted more than once)

96 / 4 = 24 = total number of tetrahedra that can be made

ans = 24

A challenge that requires a systematic approach and the ability to visualise in 2 and 3 dimensions.