The perfume bottle has a cylindrical shape, and a flower-shaped packaging must be constructed to accommodate it. For the production of this packaging, a square sheet with side $n$, four arcs of a circle with centers at the vertices of the square and radius measuring $n$, and a circle tangent to these arcs will be used.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-11.jpg?height=452&width=452&top_left_y=1815&top_left_x=882)

a) What is the measure of the radius of the circle?

b) What is the measure of the lighter area in the figure?

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Solution

a) The radius of the arc measures $n$ and the diagonal of the square measures $n \sqrt{2}$, so the radius of the circle is $\frac{n-(n \sqrt{2}-n)}{2}=\frac{2 n-n \sqrt{2}}{2}$.

b) Dividing the square as shown in the figure, the sum of the areas $A, B$, and $C$ is $\frac{n^{2}}{2}$. Since $B=\frac{\pi \cdot n^{2}}{6}$ (a $30^{\circ}$ circular sector), $C=\frac{n^{2} \sqrt{3}}{4} \cdot \frac{1}{2}=\frac{n^{2} \sqrt{3}}{8}$ (half the area of an equilateral triangle), then $A=\frac{n^{2}}{2}-\frac{\pi n^{2}}{6}-\frac{n^{2} \sqrt{3}}{8}$. Therefore, the area of the lighter region is $8 A=$ $n^{2}\left(4-\frac{4 \pi}{3}-\sqrt{3}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-12.jpg?height=434&width=428&top_left_y=1015&top_left_x=719)