An elliptical table top, its length is "160 cm", and its width is "1 m". Can the entire table top be completely covered by a rectangular tablecloth measuring $140 \mathrm{~cm} \times 130 \mathrm{~cm}$?
Solution. Consider an arbitrary ellipse. Let the distances from one of its foci to the endpoints of the major axis be $f$ and $d+f$, where $d$ is the distance between the two foci. Draw a rectangle $T$ around the ellipse such that the sides of the rectangle are not parallel to the axes of the ellipse, and then draw lines through the foci parallel to the sides of the rectangle. These lines enclose a smaller rectangle $T^{\prime}$. Let the sides of $T$ be $A$ and $B$, and the corresponding sides of $T^{\prime}$ be $a$ and $b$. Let the distances from the foci to the sides of $T$ be $x$ and $y$ such that $A=a+2 x$ and $B=b+2 y$ (as shown in the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_79113c96a67a28b2d657g-1.jpg?height=564&width=590&top_left_y=431&top_left_x=754)

It is known that if a perpendicular is drawn from one of the foci of the ellipse to its tangent, the foot of the perpendicular lies on the focal circle (the Thales circle of the major axis). From this, it follows that the sides of $T^{\prime}$ intersect the sides of $T$ at the focal circle. Therefore,

$$
x(a+x)=y(b+y)=f(d+f)
$$

since all three products are the power of the focus with respect to the focal circle. By the Pythagorean theorem, we also have $d^{2}=a^{2}+b^{2}$.

The sum of the squares of the sides of the circumscribed rectangle is:

$$
A^{2}+B^{2}=(a+2 x)^{2}+(b+2 y)^{2}=a^{2}+4 x(a+x)+b^{2}+4 y(b+y)=d^{2}+8 f(d+f)
$$

which is a constant depending only on the ellipse. This is equal to the sum of the squares of the ellipse's axes, which is evident in the case of a rectangle with sides parallel to the axes.

From this, it follows that an ellipse can be covered by a rectangle if and only if the sides of the rectangle are not shorter than the minor axis of the ellipse, and the sum of the squares of the sides is at least as large as this constant of the ellipse. (Indeed, if the latter conditions are met, either the rectangle is longer than the major axis of the ellipse, making the covering trivial, or the ellipse has two parallel tangents whose distance is the length of one side of the rectangle, in which case the distance between the perpendicular tangents is at most the length of the other side of the rectangle, so the ellipse can be covered. The converse is also easily provable.)

In our case (considering the decimeter as the unit of length)

$$
14^{2}+13^{2} \geq 16^{2}+10^{2}
$$

and the sides are longer than the minor axis, so the conditions are met, and the table can be covered with the given tablecloth.