Square $ABCD$ in Figure 1 has been dissected into eight equal-area rectangles. The width of the rectangle shaded red in Figure 1 is 35 units. What is the area of square $ABCD$ $?$ For a bonus challenge, can you devise and solve a similar problem that dissects a square into eleven pieces?

### D11. Octosection

*Contributed by Arsalan Wares, Valdosta State University*

*This problem originally appeared in the* Prisoner’s Dilemma *in the 2023 Spring issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.*

**Solution**.

This problem received submissions from PMP participants William Keehn, Blazej Kot, and Robert Noll. All solutions proceeded roughly as so: Let $s=BC$ be the side of the square. Since the rightmost white rectangle has area ${s}^{2}\text{/}8$, its width is $s\text{/}8$, and so the length of the blue rectangle is $7s\text{/}8$. Hence the width of the blue rectangle is $s\text{/}7$. That makes the height of the white rectangle below it $6s\text{/}7$, corresponding to a width of $7s\text{/}48$ (since again its area is ${s}^{2}\text{/}8)$. Subtracting the widths of the two white rectangles on the right shows the length of the grey rectangle is $35s\text{/}48$, corresponding to a width of $6s\text{/}35$. Subtracting the widths of the blue and gray rectangles shows the height of the red rectangle is $24s\text{/}35$, yielding a width of $35s\text{/}192$. But we are given that the width of the red rectangle is 35, so we know $s=192$ and hence the area of the outer square is ${192}^{2}=36,864$.

The bonus challenge catalyzed a burst of creativity: Every submitter proposed at least one 11-piece variant, some devising multiple possibilities. And every proposed “undecasection” was different! We close with the gallery of diagrams accumulated from all of these problems (from which you may be able to surmise what the exact problems may have been).