We have a rectangle $A B C D$, see the figure. The lengths of sides $A B$ and $B C$ are in the ratio 7 : 5. Inside the rectangle $A B C D$, there are points $X$ and $Y$ such that triangles $A B X$ and $C D Y$ are right isosceles with right angles at vertices $X$ and $Y$. The area common to both triangles is shaded gray and forms a square with an area of $72 \mathrm{~cm}^{2}$. Determine the lengths of sides $A B$ and $B C$.

(L. Šimůnek)

![](https://cdn.mathpix.com/cropped/2024_04_17_27d0b340b16adf4cc1afg-4.jpg?height=477&width=642&top_left_y=538&top_left_x=707)
Let's calculate the length of the diagonal of the gray square, starting from the relationship for calculating the area of a square $S=\frac{1}{2} u^{2}$ :

$$
u=\sqrt{2 S}=\sqrt{2 \cdot 72}=\sqrt{144}=12(\mathrm{~cm}) .
$$

We will denote one unnamed vertex of the gray square as $U$, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_27d0b340b16adf4cc1afg-4.jpg?height=483&width=648&top_left_y=1426&top_left_x=704)

From point $U$, we drop perpendiculars to side $A B$ and then to side $D A$. We denote their feet as $P$ and $Q$. We realize that the angle $X A B$ is the internal angle of the right isosceles triangle $A B X$ and thus has a size of $45^{\circ}$. It follows that the rectangle $A P U Q$ is a square. The figure shows three more squares constructed analogously. We denote the side length of these squares as $s$. From the figure, it is clear that $|A B|=2 s+12(\mathrm{~cm})$ and $|B C|=2 s$. We substitute these expressions into the ratio given in the problem. By manipulating the resulting equation, we obtain $2 s$ :

$$
\begin{aligned}
\frac{|A B|}{|B C|} & =\frac{7}{5}, \\
\frac{2 s+12}{2 s} & =\frac{7}{5}, \\
10 s+60 & =14 s, \\
2 s & =30(\mathrm{~cm}) .
\end{aligned}
$$

The length of side $A B$ is $30+12=42(\mathrm{~cm})$ and the length of side $B C$ is $30 \mathrm{~cm}$.

Another solution. In the given figure, we translate the triangle $C D Y$ so that point $Y$ is mapped to point $X$, see the figure. We translate it by the length of segment $X Y$, and just as in the previous solution, we determine that this length is $12 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_27d0b340b16adf4cc1afg-5.jpg?height=639&width=651&top_left_y=457&top_left_x=702)

Since triangles $A B X$ and $C^{\prime} D^{\prime} Y^{\prime}$ are right isosceles, the resulting rectangle $A B C^{\prime} D^{\prime}$ must be a square. Segment $A B$ represents 7 parts and segment $B C$ represents 5 parts according to the problem. Segment $B C^{\prime}$ is 12 cm longer than $B C$ and also represents 7 parts. 2 parts are thus $12 \mathrm{~cm}$, 1 part is then $6 \mathrm{~cm}$. The length of side $A B$ is $7 \cdot 6=42(\mathrm{~cm})$ and the length of side $B C$ is $5 \cdot 6=30(\mathrm{~cm})$.

Note. The length of the diagonal of the gray square can also be determined as follows. We reflect the intersection of the diagonals according to each side of the square, see the figure. The four resulting points form the vertices of a square that has twice the area of the gray square, i.e., $2 \cdot 72=144\left(\mathrm{~cm}^{2}\right)$. The side length of the resulting square, and thus the length of segment $X Y$, is $\sqrt{144}=12(\mathrm{~cm})$.

![](https://cdn.mathpix.com/cropped/2024_04_17_27d0b340b16adf4cc1afg-5.jpg?height=251&width=182&top_left_y=1719&top_left_x=937)

Evaluation. 1 point for calculating the diagonal of the gray square; 3 points for the insight and justification that the difference in the lengths of sides $A B$ and $B C$ is equal to the length of this diagonal; 2 points for calculating the sides $A B$ and $B C$.