A farmer owns a square meadow measuring $33 \mathrm{~m}$ on each side, fenced all around its perimeter. He wishes to divide it into three plots of equal area. Is such a division possible with

1. at most $55 \mathrm{~m}$ of fencing,
2. at most $54 \mathrm{~m}$ of fencing,
With $33 \mathrm{~m}$ of fencing, one can obtain a plot of $11 \mathrm{~m} \times 33 \mathrm{~m}$ and a plot of $22 \mathrm{~m} \times 33 \mathrm{~m}$. With $22 \mathrm{~m}$ of fencing, the second plot can be divided into two equal areas. Instead of dividing the field into three rectangles, it can be divided into two trapezoids of $16.5 \mathrm{~m}$ height (the sum of the parallel sides being $44 \mathrm{~m}$) and a pentagon, as shown in the figure. Thus, instead of using $22 \mathrm{~m} + 2 \times 16.5 \mathrm{~m} = 55 \mathrm{~m}$, it is sufficient to use $20 \mathrm{~m} + 2 \times \sqrt{16.5^{2} + 4^{2}} \mathrm{~m}$, according to the Pythagorean theorem. Now, $16.5^{2} + 16 < 17^{2}$, because $17^{2} - 16.5^{2} = (17 - 16.5)(17 + 16.5) = 16.75$, which is sufficient.
![](https://cdn.mathpix.com/cropped/2024_05_10_88e5f1818c7b44c68275g-174.jpg?height=498&width=1106&top_left_y=648&top_left_x=476)

It is a stroke of luck that the trapezoid with parallel sides of $20 \mathrm{~m}$ and $24 \mathrm{~m}$ works. We could have more generally studied the trapezoid with parallel sides $22 - x$ and $22 + x$, which amounts to studying the function defined by $f(x) = 22 - x + 2 \sqrt{16.5^{2} + (2x)^{2}}$. It is for $5 / 3 < x < 13 / 5$ that $f(x) < 54$, and in particular for $x = 2$, but the minimum $22 + 33 / 4 \sqrt{15} = 53.9521126 \ldots$ is reached at $x = 11 / 5 \sqrt{15} = 2.13014 \ldots$.