The outermost polygon shown in figure 2 as the region shaded in pink consists of 16 congruent rectangles placed so that their edges align (and vertices of four of the rectangles coincide at the center of the polygon). We call this polygon a *28-gon* because it has 28 *edges*: the line segments that make up its boundary. The perimeter of each rectangle is 26 cm and the perimeter of the 28-gon is 136 cm. What is the area of the 28-gon?

### D6. Rectangle Pinwheel

*Contributed by Arsalan Wares, Valdosta State University*

*This problem originally appeared in the* Prisoner’s Dilemma *in the 2022 Summer issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.*

**Solution**.

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 $$34=3W+L+3(L-W)=4L,$$ 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): $$16LW=16\frac{17}{2}\frac{9}{2}=612.$$