We covered a square with a regular hexagon so that their centers coincide. What is the maximum percentage of the square's area relative to the hexagon's area?
When a square is rotated $90^{\circ}$ around its center, we get the same square. Therefore, the $90^{\circ}$ rotated version of the hexagon must also contain the square. However, if both the hexagon and its $90^{\circ}$ rotated version contain the square, then the intersection of these two hexagons, which is a regular 12-sided polygon - denoted as $A_{1} A_{2} \ldots A_{12}$ on the diagram - must also contain the entire square.

The area of a square is larger the longer its diagonal. Since the center of the 12-sided polygon and the square coincide, the diagonal of the square can be at most as long as the diameter of the circumscribed circle of the 12-sided polygon. The sides of the 12-sided polygon are seen from the center of the circumscribed circle at an angle of $\frac{360^{\circ}}{12}=30^{\circ}$. Therefore, the diagonal $A_{1} A_{4}$ of the 12-sided polygon is seen from the center at an angle of $3 \cdot 30^{\circ}=90^{\circ}$, and the diagonals $A_{4} A_{7}$, $A_{7} A_{10}$, and $A_{10} A_{1}$ are also seen at an angle of $90^{\circ}$. This implies that $A_{1} A_{4} A_{7} A_{10}$ is the square with the largest possible diagonal. Thus, this is a square with the maximum area that can be inscribed.

Let the radius of the circumscribed circle of the original regular hexagon - denoted as $A B C D E F$ on the diagram - be $r$. Then the area of the hexagon is equal to the area of 6 equilateral triangles with side length $r$, which is $6 r^{2} \frac{\sqrt{3}}{4}$. Using the notation on the diagram: $O P = r \cdot \frac{\sqrt{3}}{2}$ (where $O P$ is the height of the equilateral triangle $A B O$), $P O A_{1} \angle = \frac{1}{2} P O A \angle = 15^{\circ}$, so $O A_{1} = \frac{O P}{\cos 15^{\circ}}$. Therefore, the area of the square is $2 \cdot O A_{1}^{2} = \frac{6 r^{2}}{4 \cos ^{2} 15^{\circ}}$. The ratio of the two areas is

$$
\frac{6 r^{2}}{4 \cos ^{2} 15^{\circ}} : \left(6 r^{2} \cdot \frac{\sqrt{3}}{4}\right) = \frac{1}{\sqrt{3} \cdot \cos ^{2} 15^{\circ}} \approx 0.62.
$$

Thus, the area of the square is at most $62\%$ of the area of the hexagon.

Based on the work of Mátrai Tamás Fazekas M. Főv. Gyak. Gimn., II. o.t.

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