An AP rectangle is one whose area is numerically equal to its perimeter. Michael of Madras College, St Andrews sent the following solution and one nearly the same came in from Prav and Sheli from the North London Collegiate School Maths Puzzle Club.

If the sides of the rectangle have lengths $A$ and $B$ then

\[ AB = 2A + 2B \]

so

\[ A = \frac{2B}{B-2} \]

If you are given the length of one side, say $B$, then for any value of $B$ greater than 2, you can always find an AP rectangle with one side the given length. As the sides of the rectangle cannot have negative length $B$ cannot be less than 2.