14.10. (MS, 83). Find the square of the smallest size in which 5 circles of radius 1 each can be arranged so that no two of them have common interior points.
14.10. Let \(ABCD\) be a square with center \(O\) and side \(a\), containing 5 non-overlapping circles of radius 1. Then the centers of the circles lie within the inner square \(A_1B_1C_1D_1\) with center \(O\) and side \(a-2\) (where \(A_1B_1 \parallel AB\); see Fig. 99). The lines connecting the midpoints of opposite sides of the square \(A_1B_1C_1D_1\) divide it into four smaller squares, one of which (by the pigeonhole principle (Theorem 1)) contains at least two of the centers of the circles. The distance between them, on the one hand, does not exceed the diagonal of the smaller square, and on the other hand, is at least 2. Therefore, we have

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-251.jpg?height=605&width=467&top_left_y=1025&top_left_x=125)

Fig. 100

\[ 2 \leqslant OA_{\mathbf{i}} = \frac{A_1B_1}{2} \cdot \sqrt{2} = \frac{a-2}{2} \sqrt{2}, \]

from which

\[ a \geqslant 2\sqrt{2} + 2 \]

Finally, if \(a = 2\sqrt{2} + 2\) and the centers of the circles lie at points \(O\), \(A_1\), \(B_1\), \(C_1\), \(D_1\), then all the required conditions are satisfied. Thus, the side of the desired square is \(2\sqrt{2} + 2\).