43. It is obvious that any figure with a diameter of 1 can be enclosed within a square with a side of 2: for this, it is sufficient for the center of the square to coincide with any point of the figure. What is the side of the smallest square that can enclose any figure with a diameter of 1?

Note. One can also pose a problem analogous to problems 42 and 43, replacing the square (problem 43) or circle (problem 42) with any other figure. For example, it can be shown that the smallest equilateral triangle that can enclose any figure with a diameter of 1 is a triangle with a side of \(\sqrt{3}\), and the smallest regular hexagon with the same property is a hexagon with a side of \(\frac{\sqrt{3}}{3}\) (see problem 31 in the book by I. M. Yaglom and V. G. Boltyansky, cited on page 10).
43. It is obvious that a circle with a diameter of 1 cannot be enclosed in any square whose side is less than 1. On the other hand, it is almost equally clear that any figure with a diameter of 1 can be enclosed in a square with a side of 1. Indeed, any square that encloses some figure $\Phi$ can always be reduced so that at least one pair of opposite sides of the square "touch" the figure (Fig. 123). But if $A$ and $B$ are points of the figure $\Phi$ with a diameter of 1, lying on opposite sides of the square containing $\Phi$ inside itself, then the distance between these sides of the square (or, equivalently, the side of the square) does not exceed $AB$, and $AB \leqslant 1$.

Thus, the smallest square in which any figure with a diameter of 1 can be enclosed has a side equal to 1.

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-150.jpg?height=394&width=351&top_left_y=928&top_left_x=144)

Fig. 123.

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-150.jpg?height=569&width=575&top_left_y=752&top_left_x=507)

Fig. 124.