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Bang's Theorem

Stage: 4 Challenge Level: Challenge Level:1


Thank you Hyeyoun Chung, age 14, St Paul's Girls School, London, for your solution. In a tetrahedron each face, a triangle, shares a different side with every other triangle. Therefore every one of its sides has to be equal in length to one side of each of the other triangles and we can label the side lengths $a, b, c, p, q$ and $r$ as in the diagram. All the faces have the same perimeter (say $k$) so
\begin{eqnarray} a + b + c &= k \quad (1) \\ a + r + q &= k \quad (2) \\ b + p + r &= k \quad (3) \\ c + p + q &= k \quad (4). \end{eqnarray}
Adding the last three of these equations gives $$a + b + c + 2(p + q + r) = 3k$$ so $$p + q + r = k \quad (5).$$ From equations (2) and (5) we get $a = p$, from equations (3) and (5) we get $b = q$ and from equations (4) and (5) we get $c = r$. As we can see from the diagram all the faces of the tetrahedron have sides of lengths $a, b$ and $c$ so all four faces are congruent.