Given a square with a side length of one unit and 5 equal circular disks. 4 of the disks are fixed in such a way that they internally touch two adjacent sides of the square (there is no overlap). The fifth disk is moved within the part of the square not covered by the stationary disks. What area can the moving disk sweep over in the process?
The radius $r$ of the circular disks cannot be larger than the value of $R$ for which the 5th disk touches all the other 4. In this case, the diagonal of the square - part a) of the diagram -

$$
\sqrt{2}=2 \sqrt{2} R+4 R, \text { from which the diameter: } 2 R=\sqrt{2}-1(\approx 0.414) .
$$

Thus, the 5th disk cannot move, and the rubbed area is $\pi R^{2}$.

The calculation of the rubbed part can proceed in two ways, depending on whether the 5th disk can touch the sides of the square or not. The touch is clearly evident when $r \leq 1 / 6$. In both cases, we calculate the area of the non-rubbed part, always utilizing the symmetries of the square.

![](https://cdn.mathpix.com/cropped/2024_05_02_801392c92a5f78dbe060g-1.jpg?height=689&width=1227&top_left_y=542&top_left_x=444)
I. For $1 / 6 \leq r \leq R$, one quarter of the non-rubbed part is the area bounded by the thick lines in part b) of the diagram, which is the sum of the hatched area, the area of the circle centered at $O_{2}$, and the area of the $B T_{2} T_{3}$ region. The sum of the first two is equal to the area of the pentagon $T_{1} T_{2} O_{2} O_{5} O_{1}$. Indeed, the hatched area is obtained from the pentagon by removing the circular segments $O_{1} T_{1} T_{4}$, $O_{2} T_{2} T_{5}$, and $O_{5} T_{4} T_{5}$, the sum of whose areas is equal to the area of the circle, because the sum of their central angles is obtained from the sum of the angles of the pentagon by removing the angles at $T_{1}$ and $T_{2}$, and thus the remaining is $360^{\circ}$. The pentagon is divided by the diagonal $O_{1} O_{2}$ into a rectangle of area $r(1-2 r)$ and an isosceles triangle $O_{1} O_{2} O_{5}$. The base of this triangle is $1-2 r$, and its height is

$$
\sqrt{O_{1} O_{5}^{2}-O_{1} F^{2}}=\sqrt{4 r^{2}-\left(\frac{1}{2}-r\right)^{2}}=\frac{1}{2} \sqrt{12 r^{2}+4 r-1}
$$

Finally, the area of the $B T_{2} T_{3}$ region is the difference between the area of the square with side $r$ and the area of a quarter circle with radius $r$.

Multiplying all of this by 4, the area of the non-rubbed part is

$$
4 r(1-2 r)+(1-2 r) \sqrt{12 r^{2}+4 r-1}+(4-\pi) r^{2}
$$

and the area of the rubbed part, after appropriate simplification, is

$$
(1-2 r)^{2}+\pi r^{2}-(1-2 r) \sqrt{12 r^{2}+4 r-1}
$$

II. For $r \geq 1 / 6$, the non-rubbed part can be divided into a circle around each corner of the square and 5 regions similar to the above $B T_{2} T_{3}$ region - part c) of the diagram -, so the area of the rubbed part is

$$
1-4\left[\pi r^{2}+5\left(r^{2}-\frac{\pi r^{2}}{4}\right)\right]=1-(20-\pi) r^{2}
$$

For $r=1 / 6$, both expressions give the desired value, and the above triangles become flat line segments.