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, 2011

Combinatorics Difficulty 5.0 AIME, harder Prove it South Africa

Can you enumerate the edges of a tetrahedron with numbers from $1$ to $6$ so that all four triangular faces have the same sum, where the sum of a face is calculated by adding the numbers on its edges?

Solution

Label the vertices as shown.

Let $f(XY)$ denote the number on edge $XY$ . Notice that
$\begin{align*} f(BA) + f(AD) + f(DB) &= f(AD) + f(DC) + f(AC) \\ &= f(BC) + f(CD) + f(BD) \\ &= f(AB) + f(BC) + f(AC). \end{align*}$
according to the condition. This means that $f(BC) + f(CD) = f(BA) + f(AD)$ and $f(AD) + f(DC) = f(AB) + f(BC)$ . Subtracting these two equalities from one another we find that $f(BC) - f(AD) = f(AD) - f(BC)$ . Hence $f(AD) = f(BC)$ . This means that the numbers cannot be distinct and so it cannot be done.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.