B2. Let $E$ be the midpoint of side $A B$ of rectangle $A B C D$ and $F$ the point on diagonal $A C$ such that line $B F$ is perpendicular to diagonal $A C$. Determine the ratio of the sides of rectangle $A B C D$ if segment $E F$ is perpendicular to diagonal $B D$.

(6 points)
B2. Let $\Varangle B A C=\alpha$. Triangle $A B F$ is a right triangle and point $E$ is the midpoint of the hypotenuse, so it is the center of the circumcircle of triangle $A B F$ and we have $|A E|=|B E|=|F E|$. Therefore, $\Varangle A F E=\Varangle E A F=\alpha$, so $\Varangle E F B=\frac{\pi}{2}-\Varangle A F E=\frac{\pi}{2}-\alpha$, from which it follows that $\Varangle F B D=\frac{\pi}{2}-\Varangle E F B=\alpha$. We also have $\Varangle A C B=\frac{\pi}{2}-\alpha$, so $\Varangle C B F=\alpha$. Similarly, $\Varangle D B A=\Varangle B A C=\alpha$. We get $\frac{\pi}{2}=\Varangle C B A=\Varangle D B A+\Varangle F B D+\Varangle C B F=3 \alpha$, from which it follows that $\alpha=\frac{\pi}{6}$. Triangle $A B C$ is half of an equilateral triangle, so $\frac{|A B|}{|B C|}=\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_4050f14f84cf4f327f67g-19.jpg?height=312&width=442&top_left_y=604&top_left_x=861)
It holds that $\Varangle B A C=\Varangle D B A$ ..... 1 point
It holds that $\Varangle B A C=\Varangle C B F$ ..... 1 point
Conclusion $|A E|=|F E|$ or $|B E|=|E F|$ ..... 1 point
Observation $\Varangle E A F=\Varangle A F E$ ..... 1 point
Observation $\Varangle F B D=\Varangle B A C$ ..... 1 point
Calculation $\Varangle B A C=\frac{\pi}{6}$ and conclusion $\frac{|A B|}{|B C|}=\sqrt{3}$ ..... 1 point

2nd method Let $|A B|=a$ and $|B C|=b$, and let $T$ be the intersection of lines $B D$ and $E F$, and $S$ be the intersection of the diagonals. By the Pythagorean theorem, $|A C|=\sqrt{a^{2}+b^{2}}$. Triangles $A B C$ and $A F B$ are similar in size due to the common acute angle and both being right triangles. Therefore, $\frac{|A B|}{|A C|}=\frac{|A F|}{|A B|}$, from which we derive $|A F|=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$. From this, it follows that $|F C|=|A C|-|A F|=\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}$.

By the Pythagorean theorem in triangle $B C F$, we get $|B F|=\sqrt{|B C|^{2}-|C F|^{2}}=\frac{a b}{\sqrt{a^{2}+b^{2}}}$.

Since $\Varangle C A B=\Varangle D B A$, the right triangles $A B C$ and $B T E$ are also similar. Therefore, $\frac{|A B|}{|A C|}=\frac{|B T|}{|B E|}$ and since $|B E|=\frac{|A B|}{2}$, we can express $|B T|=\frac{a^{2}}{2 \sqrt{a^{2}+b^{2}}}$.

By the Pythagorean theorem in triangle $B F T$, we get $|F T|^{2}=|B F|^{2}-|B T|^{2}=\frac{a^{2}\left(4 b^{2}-a^{2}\right)}{4\left(a^{2}+b^{2}\right)}$. The Pythagorean theorem also holds for triangle $S F T$, specifically $|T S|^{2}+|T F|^{2}=|S F|^{2}$. Since $|S F|=$ $|A F|-\frac{|A C|}{2}$ (or $|S F|=\frac{|A C|}{2}-|A F|$, if $F$ is between $A$ and $S$) and $|T S|=\frac{1}{2}|B D|-|B T|$, we derive

$$
\frac{b^{4}}{4\left(a^{2}+b^{2}\right)}+\frac{a^{2}\left(4 b^{2}-a^{2}\right)}{4\left(a^{2}+b^{2}\right)}=\frac{\left(a^{2}-b^{2}\right)^{2}}{4\left(a^{2}+b^{2}\right)}
$$

or $2 a^{4}-6 a^{2} b^{2}=0$. From this, it follows that $2 a^{2}\left(a^{2}-3 b^{2}\right)=0$. Since $a$ and $b$ are positive numbers, the only possibility is $a=\sqrt{3} b$.

![](https://cdn.mathpix.com/cropped/2024_06_07_4050f14f84cf4f327f67g-19.jpg?height=274&width=1585&top_left_y=2399&top_left_x=270)

Derivation of $|A F|=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$ 1 point

Derivation of $|F C|=\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}$ 1 point

Derivation of $|B F|=\frac{a b}{\sqrt{a^{2}+b^{2}}}$ 1 point

Calculation of $|B T|=\frac{a^{2}}{2 \sqrt{a^{2}+b^{2}}}$

Application of the Pythagorean theorem in triangles $B F T$ and $S F T$ 1 point Derivation of the equation $2 a^{4}-6 a^{2} b^{2}=0$ and conclusion $\frac{|A B|}{|B C|}=\sqrt{3}$