Let the side of the square be a unit length, and let one of the circles be $k_{1}$, which touches $k_{2}$ and $k_{3}$. Then $k_{2}$ touches $k_{1}$, and one of $k_{3}$ and $k_{4}$. It cannot be $k_{3}$, because then $k_{1}, k_{2}$, and $k_{3}$ would touch each other pairwise, and $k_{4}$ would touch any of $k_{1}, k_{2}, k_{3}$, thus touching three (and not two) of the other circles. Therefore, $k_{2}$ can only touch $k_{1}$ and $k_{4}$. Similarly, it can be seen that $k_{4}$, apart from $k_{2}$, can only touch $k_{3}$, and $k_{3}$ can only touch $k_{1}$ and $k_{4}$. According to the condition, the radii of the touching circles are equal, so the radii of $k_{1}$ and $k_{4}$ are equal—let's say $r$, and the radii of $k_{2}$ and $k_{3}$ are also equal—let's say $R$.

No circle can touch the opposite sides of the square, since another circle of the same size must fit into the square. Thus, the circles touch two adjacent sides of the square, and their centers are located on the diagonals of the square. It is also clear that none of them can touch the other from the inside, so the four circles sit in the four corners of the square.

First, let's assume that the circles $k_{1}$ and $k_{4}$ with radius $r$ are in adjacent corners of the square. Since $k_{1}$ and $k_{4}$ cannot intersect or even touch each other, $r$ must be less than a quarter of the side of the square, i.e., $1 / 4$. Similarly, the radius of $k_{2}$ and $k_{3}$ must also be less than $1 / 4$, but then $k_{1}$ and $k_{2}$ would not be able to touch each other.

# $1985-03-114-1$. eps 

1. figure

Therefore, $k_{1}$ and $k_{4}$, as well as $k_{2}$ and $k_{3}$, are in opposite corners of the square (1. figure). The radius of $k_{1}$ can be as large as possible so that $k_{1}$ and $k_{4}$ can fit without intersection or touching points, i.e., so that the segment cut from the diagonal by $k_{1}$ is shorter than half the diagonal:

$$
r \sqrt{2} + r < \frac{\sqrt{2}}{2}
$$

which is

$$
r < \frac{\sqrt{2}}{2(\sqrt{2}+1)} = 1 - \frac{\sqrt{2}}{2} = 0.2929
$$

If we choose $r$ to be exactly $(1 - \sqrt{2} / 2)$, then $k_{1}$ and $k_{4}$ will just touch each other at the center of the square. Corresponding $k_{2}$ and $k_{3}$ circles can be found (2. figure).

$$
\begin{gathered}
1985-03-114-2 . \mathrm{eps} \\
\text { 2. figure }
\end{gathered}
$$

If we continuously decrease the radius of $k_{1}$ (and $k_{4}$), the radii of the corresponding $k_{2}$ and $k_{3}$ circles will continuously increase. When $r = 1 / 4$, the radii of all four circles are equal (3. figure). If we further decrease $r$, the radii of $k_{2}$ and $k_{3}$ will increase, and we can decrease $r$ until $k_{2}$ and $k_{3}$ "meet" at the center of the square (4. figure).

$$
\text { 1985-03-115-1.eps }
$$

3. figure

## $1985-03-115-2$. eps

4. figure

In this situation, the radii of $k_{2}$ and $k_{3}$ are the previously calculated $R = 1 - \sqrt{2} / 2$, and from the right-angled triangle drawn with thick lines,

$$
(r + R)^{2} = R^{2} + \left(\frac{\sqrt{2}}{2} - r \sqrt{2}\right)^{2}
$$

which is

$$
r^{2} - 2 r (1 + R) + \frac{1}{2} = 0
$$

From this, considering that $r < 1$, we get

$$
r = 2 - \frac{\sqrt{2}}{2} - \sqrt{4 - 2 \sqrt{2}} = 0.2105
$$

The radius of the circle $k_{1}$ can only be larger than this and smaller than $(1 - \sqrt{2} / 2)$, but it can take any value within this range. The same applies to the circles $k_{2}$ and $k_{3}$. Thus, the answer to the question is: the radii of the circles in question can be any element of the interval

$$
\left(1 - \frac{\sqrt{2}}{2}, 2 - \frac{\sqrt{2}}{2} - \sqrt{4 - 2 \sqrt{2}}\right)
$$

and only these.
A square is inscribed with four circles that do not intersect each other. Each circle touches two sides of the square and exactly two other circles, with the radius of these two circles being equal. What can the radii of the circles be?