2. Let $M$ be the midpoint of side $B C$ and $N$ the midpoint of side $C D$ of rectangle $A B C D$. Determine the ratio of the lengths of the sides of rectangle $A B C D$ if $A M N$ is a right triangle with hypotenuse $A M$.
I/2. 1. way Let's write $|A B|=2 a$ and $|B C|=2 b$. Since the angle $\Varangle A N M=\frac{\pi}{2}$ is right, $\Varangle D N A+\Varangle M N C=\frac{\pi}{2}$. It follows that $\Varangle C M N=\frac{\pi}{2}-\Varangle N M C=\Varangle D N A$. Therefore, the right triangles $A N D$ and $N M C$ are similar, from which it follows that $|A D|:|D N|=|N C|:|C M|$ or $\frac{2 b}{a}=\frac{a}{b}$. This leads to $a^{2}=2 b^{2}$ and $a: b=\sqrt{2}: 1$.

2. way Let's also denote $|A N|=x,|N M|=y,|A M|=z$. By the Pythagorean theorem,

$$
x^{2}=a^{2}+4 b^{2}, y^{2}=a^{2}+b^{2}, z^{2}=4 a^{2}+b^{2} \text { and } z^{2}=x^{2}+y^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_e3e00bf556075403c4d7g-06.jpg?height=386&width=537&top_left_y=1343&top_left_x=1322)

If we consider the expressions for $x^{2}$ and $y^{2}$ from the first and second equations in the fourth equation, we get $z^{2}=2 a^{2}+5 b^{2}$. Then we equate the right sides of the third and the derived equation. Simplifying, we get $a^{2}=2 b^{2}$, so $a: b=\sqrt{2}: 1$.

1. way Triangles $A N D$ and $N M C$ are similar (angle calculations): 3 points; without angle proof: 1 point. The ratio $|A D|:|D N|=|N C|:|C M|$ (or similar), which follows from the similarity: 2 points. Conclusion $a: b=\sqrt{2}: 1$ (or equivalent result in the form of a ratio $\frac{a}{b}=\sqrt{2}, \frac{b}{a}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$): 2 points; writing in the form $a=b \sqrt{2}: 1$ point (the problem asks for the ratio).
2. way Each correctly written Pythagorean theorem: 1 point (total 4 points). Implicit relationship between $a$ and $b$: 2 points. Calculated ratio $a: b=\sqrt{2}: 1$ (or equivalent answer in the form of a RATIO): 1 point.