The upper side of a paper square is white, and the lower side is red. A point F is randomly chosen within the square. Then the square is folded so that one randomly chosen vertex coincides with point F. Find the expected number of sides of the resulting red polygon.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-14.jpg?height=515&width=512&top_left_y=1655&top_left_x=773)
The result depends only on the relative position of point $F$ and the chosen vertex. Therefore, we can assume that the vertex is fixed (let it be vertex $A$), and point $F$ is chosen randomly.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-14.jpg?height=506&width=514&top_left_y=2394&top_left_x=775)

If point $F$ belongs to the orange digon, the result of the addition will be a triangle

(the perpendicular bisector of segment $A F$ will intersect one of the sides $A D$ or $B C$), and if point $F$ is outside the digon, in the gray area, it will be a quadrilateral.

If we assume that the area of the square is 1, then the probabilities $\mathrm{P}_{g}$ and $\mathrm{P}_{o}$ of point $F$ falling into the gray and orange areas, respectively, are equal to the areas of these regions: $\mathrm{P}_{g}=2(1-\pi / 4)=2-\pi / 2, \mathrm{P}_{o}=1-\mathrm{P}_{g}=\pi / 2-1$.

Thus, the expected value is $3 \mathrm{P}_{o}+4 \mathrm{P}_{g}=3(\pi / 2-1)+4(2-\pi / 2)=5-\pi / 2 \approx 3.429$.

## Answer

$5-\pi / 2 \approx 3.429$