# The Prisoner's Dilemma

### D6. Rectangle Pinwheel

Contributed by Arsalan Wares, Valdosta State University

We received solutions to this Dilemma from Matthew Helmer (Pacific Lutheran University), William Jones (PMP), and Jesse Waite (PMP), and partial solutions from Christopher Havens (PMP Founder) and Sonny Kim (PMP). As Jesse pointed out, the rectangles look like dominoes (length twice the width), but since we want the exact answer, we’d better not rely on that. Instead, let the length of each tile be $L$ and the width be $W$. We know the perimeter of each rectangle consists of two lengths and two widths, so we conclude that $L+W=13$.
Next we need to use the fact that the perimeter of the entire pinwheel is 136. Again following Jesse’s notes, since the pinwheel has fourfold rotational symmetry, we can look at just one quadrant of it, which must contribute $136/4=34$ to the perimeter. But the border of that quadrant consists of three widths, one length, and three segments formed by subtracting a width from a length. In symbols, this means $34=3W+L+3\left(L-W\right)=4L,$ so we conclude $L=17/2$. Now from the rectangle perimeter condition we get $W=9/2$ (slightly more than half of $L$, justifying our caution!).
Finally, the total area of the pinwheel is (in units of square centimeters, of course): $16LW=16\frac{17}{2}\frac{9}{2}=612.$