A square with a side length of 3 units is drawn, and circles with a radius of 3 units are drawn around each vertex of the square. What is the area of the common region enclosed by the four circles?
Let the vertices of the square be denoted by $A, B, C, D$, and the intersection points of the circles inside the square by $E, F, G, H$ (see the figure). The points $A, H, B$ form an equilateral triangle because $A B = A H = B H = 3$. Therefore, $\angle H A B = 60^{\circ}$, so the area of the sector $A B H$ with center $A$ is $\frac{1}{6} \cdot 3^{2} \pi = \frac{3}{2} \pi$. Subtracting the area of triangle $A B H$ from this, we get the area of the segment $B H$: $\frac{3}{2} \pi - \frac{9 \sqrt{3}}{4}$. Clearly, the area of the segment $A H$ is the same. The sector $A B H$ and the segment $A H$ together form the shape $A B H$ bounded by two arcs and one straight line, whose area is thus $\frac{3}{2} \pi + \left(\frac{3}{2} \pi - \frac{9 \sqrt{3}}{4}\right) = 3 \pi - \frac{9 \sqrt{3}}{4}$. Subtracting this from the area of the sector $A B D$, we get the area of the shape $A H D$: $\frac{1}{4} \cdot 3^{2} \cdot \pi - \left(3 \pi - \frac{9 \sqrt{3}}{4}\right) = \frac{1}{4}(9 \sqrt{3} - 3 \pi)$.

The square $A B C D$ is completely covered by the common part of the four circles and the congruent (since they can be mapped onto each other by a $90^{\circ}$ rotation around the center of the square) shapes $A H D, D G C, C F B$, and $B E A$. Therefore, the area of the common part is: $3^{2} - 4 \cdot \frac{1}{4}(9 \sqrt{3} - 3 \pi) = 3 \pi + 9(1 - \sqrt{3}) \approx 2.83$.

Zsolt Katona (Fazekas M. Fóv. Gyak. Gimn., II. o.t.)

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