284. Circles and Disks. Once at a fair, we saw a man sitting at a table covered with an oilcloth featuring a large red circle in the center. The man offered the public to cover the circle with five thin disks lying nearby, promising a valuable prize to anyone who could do it. All the disks were of the same size, of course, smaller than the red circle (in the illustration, only three disks are shown for clarity).

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-085.jpg?height=494&width=445&top_left_y=1238&top_left_x=811)

The man claimed that the task was very easy and playfully covered the circle with the disks himself. Those who tried to do it after him invariably failed. I forgot to mention one essential condition: once a disk was placed, it could not be moved again, otherwise, completing the task would have been quite simple.

Suppose the diameter of the red circle is 6 dm. What should be the smallest diameter (say, to the nearest $\frac{1}{2}$ dm) of the five disks to be able to cover the circle with them?
284. In the figure, the circle bounding the red circle is shown with a dashed line, and a regular pentagon inscribed in it. The common center of the circle and the pentagon is marked with the letter \(C\). We will find a point \(D\) equidistant from \(A, B\), and \(C\), and draw a circle \(A B C\) with radius \(A D\). Five disks of this size will completely cover the circle if their centers are placed at points \(D, E, F, G\), and \(H\). If the diameter of the large circle is 6 dm, then the diameters of the disks are slightly less than 4 dm (the diameters of the disks are 4 dm "to the nearest \(\frac{1}{2}\) dm"). If you have no secret marks on the circle, it will require some attention and skill to place the disks in the right places without moving them afterward.

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-242.jpg?height=654&width=671&top_left_y=1569&top_left_x=698)

It should be added that the large circle can be covered if the ratio of the diameters exceeds 0.6094185, and it is impossible if the ratio is less than 0.6094180. In our case, when all the disks pass through the center, the ratio is 0.6180340.