Ha megrajzolnánk az összes egyenlő területű téglalapot közös középponttal és párhuzamos oldalakkal, milyen alakú folton feketítenénk be a papirost?

If we were to draw all rectangles of equal area with a common center and parallel sides, what shape would we blacken on the paper?
Let's set up a coordinate system whose origin is the common center $O$ of the rectangles, and whose axes are parallel to the sides of the rectangles. Let's select one of the rectangles with sides $a$ and $b$. The coordinates of the vertex $P$ in the positive quadrant are $a / 2$ and $b / 2$. Since

$$
\frac{a}{2} \cdot \frac{b}{2}=\frac{T}{4}
$$

$P$ lies on one branch of the hyperbola $x y=\frac{T}{4}$, and the vertex opposite to $P$ lies on the other branch. The other two vertices lie on the hyperbola $x y=-\frac{T}{4}$. The spot is located in the infinite region $L$ bounded by the four branches of the hyperbolas.

Since for any point $S$ in $L$, there exists a rectangle whose perimeter includes $S$, the sought spot is the region $L$ itself.