59. The area of square $K$ is 1, and the area of the inscribed rectangle $\Pi$ is $s$. What value can $s$ take if

a) $\Pi$ is not a square;

b) $\Pi$ is a square?
59. If \(A_{1} B_{1} C_{1} D_{1}\) is a rectangle inscribed in the square \(A B C D\) with side 1 (Fig. 141), then

\[
\angle B A_{1} B_{1}=90^{\circ}-\angle B B_{1} A_{1}=180^{\circ}-90^{\circ}-\angle B B_{1} A_{1}=\angle C B_{1} C_{1}
\]

and similarly

\[
\angle B A_{1} B_{1}=\angle C B_{1} C_{1}=\angle D C_{1} D_{1}=\angle A D_{1} A_{1}.
\]

Therefore, the right triangles \(B A_{1} B_{1}, C B_{1} C_{1}, D C_{1} D_{1}\) and \(A D_{1} A_{1}\)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-170.jpg?height=375&width=375&top_left_y=1066&top_left_x=210)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-170.jpg?height=390&width=380&top_left_y=1064&top_left_x=658)
b)

Fig. 141.

are similar to each other, and triangles \(B A_{1} B_{1}\) and \(D C_{1} D_{1}\), \(C B_{1} C_{1}\) and \(A D_{1} A_{1}\), are also equal (since \(A_{1} B_{1}=C_{1} D_{1}\) and \(B_{1} C_{1}=D_{1} A_{1}\)). Let the legs of these right triangles be denoted by \(a\), \(b\), \(c\), and \(d\) as shown in Fig. 141. Then

\[
a+d=b+c \quad(=1)
\]

and, due to the similarity of triangles \(B A_{1} B_{1}\) and \(C B_{1} C_{1}\),

\[
\frac{a}{b}=\frac{c}{d}, \text{ i.e., } a d=b c
\]

The sum \(\sigma\) and the product \(p\) of two quantities completely determine these quantities (they are the roots of the quadratic equation \(x^{2}-\sigma x+p=0\)); therefore, the equalities \(a+d=b+c\) and \(a d=b c\) can only hold if \(a=b\) and \(c=d\) (Fig. 141, a) or \(a=c\) and \(b=d\) (Fig. 141, b). In the second case, all four right triangles \(B A_{1} B_{1}, C B_{1} C_{1}\), \(D C_{1} D_{1}\), and \(A D_{1} A_{1}\) are equal to each other, i.e., \(A_{1} B_{1}=B_{1} C_{1}=C_{1} D_{1}=D_{1} A_{1}\), and the rectangle \(A_{1} B_{1} C_{1} D_{1}\) is a square; in the first case, if \(a=b>d\), i.e., points \(A_{1}, B_{1}, C_{1}\), and \(D_{1}\) do not coincide with the midpoints of the sides of the square \(A B C D\), the rectangle \(A_{1} B_{1} C_{1} D_{1}\) is not a square. [If points \(A_{1}, B_{1}, C_{1}, D_{1}\) coincide with the midpoints of the sides of the square \(A B C D\), then \(a=b=c=d\); this arrangement of points \(A_{1}, B_{1}, C_{1}, D_{1}\) belongs to both the first and the second case.]

Now we have:

a) If \(A_{1} B_{1} C_{1} D_{1}\) is not a square (Fig. \(141, a\)), then its area \(s\) is

\[
\begin{aligned}
& s=A_{1} B_{1} \cdot A_{1} D_{1}=\sqrt{2} a \cdot \sqrt{2} d=2 a d=\frac{1}{2}\left[(a+d)^{2}-(a-d)^{2}\right]= \\
& =\frac{1}{2}\left[1-(a-d)^{2}\right] \leqslant \frac{1}{2}
\end{aligned}
\]

more precisely,

\[
0<s<\frac{1}{2}
\]

and a rectangle (not a square!) of any area, as small as desired, or as close to \(1 / 2\) as desired, can be inscribed in the square.

b) If \(A_{1} B_{1} C_{1} D_{1}\) is a square (Fig. \(141, b\)), then its area \(s\) is

\[
\begin{aligned}
& s=A_{1} B_{1}^{2}=a^{2}+b^{2}=a^{2}+d^{2}=\frac{1}{2}\left[(a+d)^{2}+(a-d)^{2}\right]= \\
& \quad=\frac{1}{2}\left[1+(a-d)^{2}\right] \geqslant \frac{1}{2}
\end{aligned}
\]

more precisely,

\[
\frac{1}{2} \leqslant s<1
\]

and a square \(A_{1} B_{1} C_{1} D_{1}\) of area as close to 1 as desired can be inscribed in the square \(A B C D\).

Note. Here is another version of the same problem: a rectangle \(P\) of known area \(s\) is inscribed in the square \(K\) with side 1. Is \(P\) a square? [Answer. \(P\) is not a square (respectively, a square) if \(s<\frac{1}{2}\) (if \(s \geqslant \frac{1}{2}\)).]