Two unit squares are placed in the plane such that the center of one square is a vertex of the other square. What is the area of the common part of the two squares?
I. Solution. Let $O$ be the point which is the center of one square and a vertex of the other, and let $M$ and $N$ be the intersection points of the sides of the two squares, as shown in the first figure.

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\text { 1993-02-071-1.eps }
$$

1. Figure

Then the common part of the two squares is the (possibly degenerate) quadrilateral $O N B M$, whose area is

$$
T_{O N B M}=T_{O B M}+T_{O N B}
$$

The triangle $O N B$ is congruent to the triangle $O M A$, because their angles are right-angled acute angles, and $O A=O B=1 / \sqrt{2}$; therefore, $T_{O M A}=T_{O N B}$. Thus, $T_{O N B M}=T_{O B M}+T_{O M A}=T_{A O B}=1 / 4$.

II. Solution The sides of the other square containing $O$ divide the square centered at $O$ into four congruent parts (2nd figure), since a $90^{\circ}$ rotation around $O$ maps these parts onto each other. Since one of the four congruent shapes obtained is precisely the common part of the two squares, the value of the sought area is $1 / 4$.

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\text { 1993-02-071-2.eps }
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$$
\text { 2. Figure }
$$

Remarks 1. If $M=A$ and $N=B$, then the triangles $O M A$ and $O N B$ are degenerate, in which case the common part of the two squares is exactly the triangle $A O B$.

2. In solving the problem, we used the fact that the square centered at $O$ does not intersect the sides of the other square that do not contain $O$; indeed, if $X$ is a point on the side of the former square, then $O X \leqslant O B \leqslant \frac{1}{\sqrt{2}}<1$, while the sides of the latter square that do not contain $O$ are at a unit distance from $O$.

If the problem is modified so that the square centered at $O$ remains a unit square, while the other square, with vertex $O$, has side length $a$, then the following statements can be easily proven:

If $a \geqslant 1 / \sqrt{2}$, then the common part has an area of $1 / 4$.

If $a \leqslant 1 /(2 \sqrt{2})$, then the square with side $a$ is contained within the other square, so the common part has an area of $a^{2}$.

If $1 /(2 \sqrt{2})<a<1 / \sqrt{2}$, then the area of the common part depends on the position of the two squares; this area is maximal precisely when the sides of the two squares are parallel, and minimal when the sides of the two squares form a $45^{\circ}$ angle with each other.