Two congruent rectangles have a common center, and their sides are rotated $45^{\circ}$ relative to each other. Together, they cover $q$ times the area of one of them. What is the maximum possible value of $q$? Does $q$ determine the ratio of the sides of the rectangle?
I. Let the two rectangles be $T$ and $T_{1}$, with a common center $O$, and side lengths $a$ and $b$, where $a \geq b$. Each side of $T_{1}$ is inclined at a $45^{\circ}$ angle to each side of $T$, so the midpoints (axes of symmetry) of the $a$-length sides $C D$ and $C_{1} D_{1}$ also form a $45^{\circ}$ angle ($\angle C O C_{1} = 45^{\circ}$, $\angle C_{1} O D = 135^{\circ}$). The shape $X$ formed by $T$ and $T_{1}$ is symmetric with respect to the angle bisectors $t_{1}$ and $t_{2}$ of these angles. 

![](https://cdn.mathpix.com/cropped/2024_05_02_713b12229a1949203587g-1.jpg?height=460&width=1270&top_left_y=344&top_left_x=424)

Reflecting $T$ and $T_{1}$ over these bisectors, $T$ and $T_{1}$ map onto each other, with $C D$ mapping to $C_{1} D_{1}$ or $D_{1} C_{1}$. Let $A$ be the vertex of $T$ in the sector $C O C_{1}$ (on the line $O C_{1}$ if $a = b$). Thus, the common part of $T$ and $T_{1}$ is an octagon $N$ or a rhombus $R$ depending on whether $C A$ is greater than $C_{1} A$ (or equal, in which case $A$ is a vertex of the rhombus); we can also say: the common part is an octagon if $\angle A O C > 45^{\circ} / 2 = 22.5^{\circ}$, otherwise it is a rhombus. The area covered by $X$ (once) is $2 a b - N$, or $2 a b - R$, where $N$ and $R$ denote the areas of the respective shapes.

II. In the first case, the side of $T$ of length $b$ starting from $A$ intersects the side of $T_{1}$ of length $b$ at point $E$ on $t_{1}$, and the side of length $a$ at points $F$ and $G$ on $t_{2}$. Then $X$ is obtained by adding two pairs of isosceles right triangles with hypotenuses $G E$ and $G F$ to $T$ (since $N$ is equal to the difference between the area of $T$ and the areas of the four triangles).

The lengths of the two hypotenuses can be calculated from the following system of equations:

\[
\begin{aligned}
& \frac{1}{\sqrt{2}} \cdot G E + \left(1 + \frac{1}{\sqrt{2}}\right) G F = a \\
& \left(\frac{1}{\sqrt{2}} + 1\right) G E + \frac{1}{\sqrt{2}} \cdot G F = b
\end{aligned}
\]

($a$ and $b$ each contain one leg of each type of triangle, plus one of the hypotenuses), and we get

\[
G E = \frac{1}{\sqrt{2}}(b - (\sqrt{2} - 1) a), \quad G F = \frac{1}{\sqrt{2}}(a - (\sqrt{2} - 1) b)
\]

from which the sum of the areas of the four triangles is

\[
2\left(\frac{G E^{2}}{4} + \frac{G F^{2}}{4}\right) = \left(1 - \frac{1}{\sqrt{2}}\right)(a^{2} + b^{2} - \sqrt{2} a b)
\]

and adding this to $a b$ gives the area of the shape $X$

\[
X_{1} = \left(1 - \frac{1}{\sqrt{2}}\right)(a + b)^{2}
\]

This result is valid as long as $G E \geq 0$, i.e., $b / a \geq \sqrt{2} - 1$ (and of course $b / a \leq 1$). This limit is exactly $\tan(45^{\circ} / 2)$, which requires that the angle $\gamma$ between the diagonal and the axis of symmetry of $T$ not be less than $22.5^{\circ}$, as we can also read from the diagram. ($G F$ is always positive.)

In the second case, the side of $R$ is equal to the hypotenuse of an isosceles right triangle with leg $b$, and its height is $b$, so the area of the shape is

\[
X_{2} = 2 a b - \sqrt{2} b^{2}
\]

provided

\[
\frac{b}{a} \leq \sqrt{2} - 1 = \tan 22.5^{\circ}, \quad \text{or if} \quad \angle A O C \leq 22.5^{\circ}.
\]

III. Let's examine the ratio $q$ of the area of $X$ to the area of $T$ as a function of the ratio $b / a = r$ while $r$ decreases from 1 to $\sqrt{2} - 1$; then further to 0 (but $r > 0$), i.e., while reducing the width of $T$ and $T_{1}$ and keeping their lengths constant.

In the first interval,

\[
q = \frac{X_{1}}{a b} = \left(1 - \frac{1}{\sqrt{2}}\right)\left(\frac{b}{a} + 2 + \frac{a}{b}\right) = \left(1 - \frac{1}{\sqrt{2}}\right)\left(r + \frac{1}{r} + 2\right)
\]

and the variable part of the second factor increases monotonically as $r$ decreases. If indeed $\sqrt{2} - 1 \leq r_{2} < r_{1} \leq 1$, then

\[
r_{2} + \frac{1}{r_{2}} - r_{1} - \frac{1}{r_{1}} = \frac{(r_{2} - r_{1})(r_{1} r_{2} - 1)}{r_{1} r_{2}} > 0
\]

since both factors are positive. Thus, for our first question, we only need the value of $q = \sqrt{2}$ corresponding to $r = \sqrt{2} - 1$. For the second question, the value corresponding to $r = 1$ (the smallest possible value) is $q = 4 - 2 \sqrt{2} ( = 1.172)$.

For the second interval, if $0 < r < \sqrt{2} - 1$, then

\[
2 - \sqrt{2} < q < 2
\]

and if

\[
\sqrt{2} \leq q < 2
\]

then from (2),

\[
r = \frac{2 - q}{\sqrt{2}}
\]

György Turán (Budapest, Fazekas M. Gyak. Gimn., II. o. t.) András Kovalcsik (Balassagyarmat, Balassi B. Gimn., II. o. t.)