Let $\square ABCD$ be a square with sides $r$. Draw a quarter circle starting in each corner, as pictured below. The question is wheter there exists some elementary method to figuring out the value of the padded square in the middle ($z$).

enter image description here

Using integration we can see that

$$ z = 4\int_{r/2}^{\sqrt{3}r/2} \biggl( \sqrt{a^2-x^2} - \frac{a}{2} \biggr) \mathrm{d}r = \frac{r^2}{4}(3 + \pi - 3\sqrt{3}) $$

However, this is far from something a high-school student could come up with. I tried to create a system of equations

$$ \begin{align*} 4x + 4y + z & = r^2 \tag{1} \\ 2x + 3y + z & = \frac{\pi r^2}{4} \tag{2} \end{align*} $$

Alas, I was unable to find a third linear independent equation describing the system.


I think this solution is suitable for high school students: enter image description here

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Imagine that the southwest corner of the square is $(0, 0)$, the square has side $r$, and we coordinatize in the usual way with the $x$-axis along the bottom side of the square and the $y$-axis along its left side.

The area $Area(z)$ can obviously be broken up into four congruent areas, one in each quadrant of the square. Consider the northeastern one. This is that part of the circle of radius $r$ which has $x > r/2, y > r/2$. Call its area $A$; then $Area(z) = 4A$.

This is just a sector of a circle (centered at 0), with two triangles removed. Call the area of the sector $S$ and the area of each triangle $T$; then $z = 4(S-2T) = 4S - 8T$. Now we just need to find $S$ and $T$.

$S$ is a thirty-degree sector of the circle, as you can see because the classical construction of an equilateral triangle is hidden in your figure many times over. So $S = \pi r^2/12$.

One example of $T$ is a triangle with vertices at $(0, 0), (r/2, r/2), (r \sqrt{3}/2, r/2)$; it has base $r(\sqrt{3}-1)/2$ and height $r/2$, so its area is $T = r^2 \times (\sqrt{3}-1)/8$.

Putting it all together we get

$$z = 4S - 8T = 4 \times {\pi r^2 \over 12} - 8 \times {r^2 {\sqrt{3}-1 \over 8}} = r^2 \left( {\pi \over 3} + 1 - \sqrt{3} \right)$$

Although I've expressed this in terms of coordinate geometry there's nothing about the solution that's essentially coordinate-based; I've just chosen to do it this way rather than to draw lots of pictures.

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