2006/07 AHSMC Part 1 Problem 8

Let $A$ and $B$ be the respective midpoints of two opposite edges of length $9$. Let $C$ and $D$ be the respective centres of the two $7 \times 8$ faces. (See the diagram below.) Then $AB$ and $CD$ intersect at the centre $O$ of the box. Place a coin of diameter $9$ on the plane determined by $AB$ and $CD$ with centre $O$. Its circumference will pass through $C$ and $D$, but $A$ and $B$ are not covered up since $AB = \sqrt{7^2 + 8^2} > 9$. Rotate the coin about $AB$ so that it just comes off $C$ and $D$. We can then expand the coin slightly and still have it fit inside the box. The answer is (e). We do not know the actual maximum value. That is why the question is phrased in its current form.