12.404 The ratio of the perimeter of a parallelogram to its larger diagonal is $k$. Find the angles of the parallelogram, given that the larger diagonal divides one of the angles of the parallelogram in the ratio 1:2.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-289.jpg?height=184&width=314&top_left_y=1292&top_left_x=163)

Let $AC$ be the longer diagonal of the given parallelogram $ABCD$, $\angle CAD = x$, and $\angle BAC = 2x$. Extend side $AD$ by a segment $KD$ equal to $CD$.

$$
\angle AKC = \angle DCK = \frac{\angle ADC}{2} = \frac{180^\circ - 3x}{2} \Rightarrow
$$

$$
\Rightarrow \angle ACK = \frac{180^\circ + x}{2}. \text{ By the Law of Sines for } \triangle ACK
$$

$$
\frac{AK}{AC} = \frac{AD + DC}{AC} = \frac{k}{2} = \frac{\sin \angle ACK}{\sin \angle AKC} = \frac{\cos \frac{x}{2}}{\cos \frac{3x}{2}} = \frac{1}{4 \cos^2 \frac{x}{2} - 3} \Rightarrow 4 \cos^2 \frac{x}{2} = \frac{2}{k} + 3 \Rightarrow
$$

$$
\Rightarrow 2 \cos x = \frac{2}{k} + 1 \Rightarrow x = \arccos \frac{2 + k}{2k}, \angle A = 3x, \angle B = \pi - 3x
$$

Answer: $3 \arccos \frac{2 + k}{2k}, \pi - 3 \arccos \frac{2 + k}{2k}$.