In a square piece of paper, a point $O$ is chosen at random. The square is then folded so that each vertex coincides with the point $O$. The figure below shows one of the possible folding schemes. Find the expected number of sides of the resulting polygon.

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-02.jpg?height=452&width=460&top_left_y=1209&top_left_x=799)
A small geometric investigation shows that the result can be a hexagon, a pentagon, and a quadrilateral. A quadrilateral is obtained only if point O falls exactly at the center or at a vertex of the square. The probability of this is zero.

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-02.jpg?height=472&width=504&top_left_y=1902&top_left_x=777)

A hexagon is formed if point $O$ is inside the figure bounded by four semicircles (the orange four-petaled flower in the diagram). If point $O$ is in any other place (the gray area), a pentagon is formed.

If the area of the square is considered to be 1, then the probability that point $O$ falls into the orange area is equal to the area of this region:

$P_{o}=S_{o}=8 \cdot(\pi / 16-1 / 8)=\pi / 2-1$.

The probability $P_{g}$ that point $O$ falls into the gray area is $1-P_{o}=2-\pi / 2$.

Thus, the expected number of sides of the polygon is $5(2-\pi / 2)+6(\pi / 2-1)=4+\pi / 2 \approx$ $5.571$.

## Answer

$4+\pi / 2 \approx 5.571$.

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