How many identical squares can be placed in the coordinate system so that no two have a common interior point, and each square has a point on both coordinate axes on its perimeter?
Let the side length of the squares be $2a$. Since the longest segments connecting a square's center to its other points are the diagonals, which are $\sqrt{2}a$ in length, the centers of the squares can be no more than this distance from the axes, meaning they cannot fall outside the square centered at the origin with side length $2\sqrt{2}a$ and sides parallel to the axes (Figure 1).

$$
\text{1986-01-011-2.eps}
$$

Figure 1

On the other hand, no two squares can share an interior point, so the centers of the squares cannot be closer to each other than $2a$.

The square shown in Figure 1 is divided by the axes into four smaller squares, each with a diagonal of length $2a$. Since the maximum distance between two points in a square is the length of its diagonal, no square with side length $\sqrt{2}a$ can contain more than one center, and it can contain two only if these centers are at opposite vertices of the square.

If we now want to place more than four $2a$ side length squares in the specified manner, then at least one of the four smaller squares with side length $\sqrt{2}a$ must contain at least two centers. These centers, therefore, must be at the endpoints of a diagonal of the $\sqrt{2}a$ side length square.

If one endpoint of this diagonal is the origin, then we can place the additional centers only at the vertices of the $2a$ side length square shown in Figure 1, making the total number of centers no more than 5. If, however, the diagonal in question is the other one, then we can place only two additional centers at the mirror images of the previous two points with respect to the origin.

# 1986-01-011-3.eps 

Figure 2

Thus, we have found that the maximum number of squares that can be placed in the specified manner is 5, and this is only possible if one of the squares has its center at the origin, and the other four have their centers at the vertices of the $2\sqrt{2}a$ side length square shown in Figure 1 (Figure 2).