What is the probability that a point randomly taken within a square is at most as far from all four corners of the square as the side of the square: 1[^0]


[^0]:    ${ }^{1}$ The probability that a point $P$ taken within a closed area $S$ is in a closed sub-area $s$ that is completely within $S$, is equal to the fraction $\frac{s}{S}$, regardless of the position of $P$.
From the vertices $A, B, C, D$ of the square, we construct circles with radius $a$; $a$ is the side of the square. These four circles define a curved quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ inside the square. Each arc of the quadrilateral is one-twelfth of the circumference of a circle with radius $a$. (Indeed, the arcs of the circles with radius $a$ that lie inside the square divide each other into three equal parts.) The diagonals of the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, $A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$, intersect at point $O$, the center of the square, and divide the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ into four equal parts.

![](https://cdn.mathpix.com/cropped/2024_05_02_2c1cbe283600ee1d542ag-1.jpg?height=366&width=377&top_left_y=386&top_left_x=858)

The area of one such part, for example, $A^{\prime} O B^{\prime}$, is obtained by subtracting the sum of the areas of triangles $A^{\prime} C O$ and $B^{\prime} C O$ from the area of the sector $A^{\prime} C B^{\prime}$, or twice the area of triangle $A^{\prime} C O$.

The area of the sector $A^{\prime} C B^{\prime}$ is: $\frac{a^{2} \pi}{12}$.

The area of triangle $A^{\prime} C O$ is:

$$
\frac{1}{2} A^{\prime} C \cdot O C \cdot \sin \widehat{A^{\prime} C O} = \frac{1}{2} a \cdot \frac{a \sqrt{2}}{2} \sin \frac{\pi}{12}
$$

The area of the segment $A^{\prime} O B^{\prime}$ is:

$$
\frac{a^{2} \pi}{12} - \frac{a^{2} \sqrt{2}}{2} \sin \frac{\pi}{12}
$$

The area of the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is $4 A^{\prime} O B^{\prime} = a^{2}\left(\frac{\pi}{3} - 2 \sqrt{2} \sin \frac{\pi}{12}\right)$.

The point in question must lie within or on the boundary of this quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. The probability that a randomly chosen point does not lie outside the closed quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is:

$$
v = \frac{A^{\prime} B^{\prime} C^{\prime} D^{\prime}}{A B C D} = \frac{\pi}{3} - 2 \sqrt{2} \sin \frac{\pi}{12}
$$

Now,

$$
\begin{aligned}
\sin \frac{\pi}{12} & = \sqrt{\frac{1 - \cos \frac{\pi}{6}}{2}} = \frac{1}{2} \sqrt{2 - \sqrt{3}} \\
2 \sqrt{2} \sin \frac{\pi}{12} & = 2 \sqrt{2} \cdot \frac{1}{2} \sqrt{2 - \sqrt{3}} = \sqrt{4 - 2 \sqrt{3}} = \sqrt{3 - 2 \sqrt{3} + 1} \\
& = \sqrt{(\sqrt{3} - 1)^{2}} = \sqrt{3} - 1
\end{aligned}
$$

and thus

$$
\begin{gathered}
v = \frac{\pi}{3} - (\sqrt{3} - 1) = \frac{\pi}{3} + 1 - \sqrt{3} = 2.0472 - 1.7321 \\
v = 0.3151
\end{gathered}
$$

István Jankovich (Győr VIII. class, Budapest II.) 

${ }^{1} \widehat{A^{\prime} C O} = \frac{1}{2} \widehat{A^{\prime} C B^{\prime}} = \frac{1}{2} \cdot \frac{2 \pi}{12} = \frac{\pi}{12}$