3. Let $E$ be a point on the side $CD$ of the rectangle $ABCD$, such that the angle $\angle AEB$ is a right angle and $3|EA|=2|EC|$. Determine the ratio of the lengths of the sides of the rectangle $ABCD$.


II/3.

![](https://cdn.mathpix.com/cropped/2024_06_07_8c492ac180897ee339acg-21.jpg?height=286&width=511&top_left_y=1679&top_left_x=270)

Let $|A B|=|C D|=a, |B C|=|D A|=b$, and $|E C|=c$. Then $|E A|=\frac{2}{3} c$ and $|E D|=a-c$. By the Pythagorean theorem for triangle $A E D$, we have $b^{2}+(a-c)^{2}=\frac{4}{9} c^{2}$ or $b^{2}=-a^{2}+2 a c-\frac{5}{9} c^{2}$. By the Pythagorean theorem for triangles $B C E$ and $A B E$, we have $c^{2}+b^{2}=|E B|^{2}=a^{2}-\frac{4}{9} c^{2}$ or $b^{2}=a^{2}-\frac{13}{9} c^{2}$. Comparing these two equations, we get $-a^{2}+2 a c-\frac{5}{9} c^{2}=a^{2}-\frac{13}{9} c^{2}$ or $a^{2}-a c-\frac{4}{9} c^{2}=0$. This can be factored as $\left(a-\frac{4}{3} c\right)\left(a+\frac{1}{3} c\right)=0$. Since $a$ and $c$ are positive, it must be that $a=\frac{4}{3} c$. From this, we also get $b^{2}=\frac{1}{3} c^{2}$ or $b=\frac{1}{\sqrt{3}} c$. Therefore, the ratio of the lengths of the sides of the rectangle is $\frac{a}{b}=\frac{4 \sqrt{3}}{3}$.

Pythagorean theorem for triangle $A E D$. 1 point

Pythagorean theorem for triangle $B C E$. 1 point

Pythagorean theorem for triangle $A B E$. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_8c492ac180897ee339acg-21.jpg?height=54&width=1627&top_left_y=2554&top_left_x=220)

Factoring the expression to $\left(a-\frac{4}{3} c\right)\left(a+\frac{1}{3} c\right)=0$.

1 point$\qquad$