55. How to place three ice cream kiosks on a circular square in the most advantageous way, i.e., so that the greatest distance from points on the square to the nearest kiosk is as small as possible?[^0]
55. Let the radius of the square be $R$, and the maximum distance from a point on the square to the nearest kiosk be $r$. In this case, every point on the square will be at a distance not greater than $r$ from at least one kiosk, i.e., three circles of radius $r$ with centers at the points of the kiosk locations will completely cover the entire square. Conversely, if the entire square is covered by three circles of radius $r$, then by placing the kiosks at the centers of these circles, we can be sure that the distance from each point on the square to the nearest kiosk does not exceed $r$. Therefore, our problem is equivalent to the following: cover a circle of radius $R$ with three equal circles of the smallest possible radius.

Divide the circle of radius $R$ into three equal sectors $OAB$, $OBC$, and $OCA$ with central angles of $120^\circ$ (Fig. 144). The smallest circle covering the sector $OAB$ is the circle with its center at the midpoint $M$ of the chord $AB$ (since the smallest circle covering an obtuse triangle $OAB$ is the circle constructed on the largest side as the diameter; see the solution to problem 18). Construct three such circles covering the sectors $OAB$, $OBC$, and $OCA$; these three circles of radius $\frac{R \sqrt{3}}{2}$ obviously cover the entire circle of radius $R$.

We claim that a circle of radius $R$ cannot be covered by three circles of a smaller radius.

Indeed, let one of the circles of a radius smaller than $\frac{R \sqrt{3}}{2}$ intersect the circumference of the larger circle at points

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-175.jpg?height=391&width=364&top_left_y=584&top_left_x=177)

Fig. 145.

$A'$ and $B'$; then the arc $A'B'$ is $120^\circ$ and, consequently, this arc cannot be covered by a circle with a diameter smaller than $R \sqrt{3}$, which proves our claim.

Thus, the most advantageous placement of the kiosks is such that they are located at the midpoints of the sides of an equilateral triangle inscribed in the contour of the square (in other words, the kiosks are located on three radii of the square, forming angles of $120^\circ$ with each other, at a distance from the center equal to half the radius of the square; see Fig. 144).

Note. We recommend that the reader try to solve a similar problem for the case of $n > 3$ kiosks. For example, for the case of $n=4$ kiosks, the most advantageous placement is at the midpoints of the sides of a square inscribed in the contour of the square (the proof is analogous to the solution of this problem); for the case of $n=7$ kiosks, the most advantageous placement is at the midpoints of the sides of a hexagon inscribed in the contour of the square and at the center of the square (compare with the solution of problem 54), and so on.