3.11. The diagonal of a rectangle is equal to $d$ and divides the angle of the rectangle in the ratio $m: n$. Find the perimeter of the rectangle.
3.11. Given that $ABCD$ is a rectangle, $AC = d$, $\frac{\angle ACB}{\angle ACD} = \frac{m}{n}$ (Fig. 3.13). Since $\angle BCD = \frac{\pi}{2}$, then $mx + nx = \frac{\pi}{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-506.jpg?height=273&width=526&top_left_y=106&top_left_x=629)

Fig. 3.13

$x = \frac{\pi}{2(m+n)}$; therefore, $\angle ACB = \frac{m \pi}{2(m+n)}$, $\angle ACD = \frac{n \pi}{2(m+n)}$.

From $\triangle ABC$ and $\triangle ADC$, we find $BC = d \cos \frac{m \pi}{2(m+n)}$,

$$
\begin{aligned}
& DC = d \cos \frac{n \pi}{2(m+n)}. \text{ Therefore, } \\
& P_{ABCD} = 2\left(d \cos \frac{m \pi}{2(m+n)} + d \cos \frac{n \pi}{2(m+n)}\right) = \\
& = 4d \cos \frac{\pi}{4} \cos \frac{\pi(m-n)}{4(m+n)} = 2\sqrt{2}d \cos \frac{\pi(m-n)}{4(m+n)}
\end{aligned}
$$