The Mathematicians' Party's billboard observers noted that people read the advertisements standing 3 meters away from the center of the half-meter diameter advertising poles. What is the maximum width of the posters the party can produce if they want people to see a complete poster from any direction when looking at the billboard-wrapped poles?
![](https://cdn.mathpix.com/cropped/2024_05_02_050de6cce9d339ba9749g-1.jpg?height=1494&width=811&top_left_y=115&top_left_x=646)

Figure 1

A person standing 3 meters from the center sees the part of the billboard that is bounded by the tangent lines drawn from 3 meters away to the billboard. Clearly, it is sufficient to examine a circle instead of the billboard. Using the notation from Figure 1, the person at point \( M \) sees the arc \( E_{1} E_{2} \). Let the length of this arc be denoted by \( s \), and the angle \( E_{1} O E_{2} \) by \( 2 \varphi \). In the right triangle \( E_{1} O M \), \( \cos \varphi = \frac{1}{4} \), so \( \varphi \approx 75.52^\circ \); the length of the arc is then \( s \approx 0.75 \cdot 2 \pi \cdot \frac{2 \cdot 75.52^\circ}{360^\circ} \approx 1.977 \) meters.

If we denote the maximum width of the posters by \( d \), then \( d \leq \frac{s}{2} \), because if \( d > \frac{s}{2} \), there will always be an arc of length \( s \) that does not contain any complete poster: if the boundary lines of two posters are exactly at the midpoint of the arc, then both posters will extend beyond the arc (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_050de6cce9d339ba9749g-2.jpg?height=731&width=773&top_left_y=117&top_left_x=665)

Since we are looking for posters of maximum width, we can assume that the posters cover the circumference of the column without gaps or overlaps, meaning the circumference of the column is an integer multiple of the width of the posters. If we have \( n \) posters, then \( \frac{0.75 \cdot 2 \pi}{n} \leq \frac{s}{2} \) must hold, from which \( 4.767 < n \) follows.

Therefore, the smallest possible value of \( n \) is 5, and the corresponding poster width is \( \frac{0.75 \cdot 2 \pi}{5} \approx 0.942 \) meters.