12.152. The ratio of the lateral sides of the trapezoid is equal to the ratio of its perimeter to the length of the inscribed circle and is equal to $k$. Find the angles of the trapezoid and the permissible values of $k$.
## Solution.

Let in trapezoid $ABCD$ (Fig. 12.25) $BC \| AD$,

$BC < AD, BE$ and $CK$ - the heights of the trapezoid, $R$ - the radius of the circle inscribed in the trapezoid, $\angle A = \alpha, \angle D = \beta$.

Then $BE = CK = 2R, AB = \frac{2R}{\sin \alpha}, CD = \frac{2R}{\sin \beta}$.

Since $AB + CD = BC + AD$, the perimeter of the trapezoid is

$2(AB + CD), k = \frac{2(AB + CD)}{2\pi R} = \frac{\frac{2R}{\sin \alpha} + \frac{2R}{\sin \beta}}{\pi R} = \frac{2(\sin \alpha + \sin \beta)}{\pi \sin \alpha \sin \beta}$.

By the condition $AB = kCD$. Then $\frac{2R}{\sin \alpha} = \frac{2kR}{\sin \beta}, \sin \beta = k \sin \alpha \Rightarrow$

$\Rightarrow \frac{2(\sin \alpha + \sin \beta)}{\pi \sin \alpha \sin \beta} = k \Leftrightarrow \frac{2(\sin \alpha + k \sin \alpha)}{\pi k \sin^2 \alpha} = k \Leftrightarrow \frac{2(k + 1)}{\pi k \sin \alpha} = k \Leftrightarrow \sin \alpha = \frac{2(k + 1)}{\pi k^2}$.

Then $\sin \beta = k \sin \alpha = \frac{2(k + 1)}{\pi k}$.

Thus, the angles of the trapezoid are $\arcsin \frac{2(k + 1)}{\pi k^2}, \pi - \arcsin \frac{2(k + 1)}{\pi k^2}$, $\arcsin \frac{2(k + 1)}{\pi k}, \pi - \arcsin \frac{2(k + 1)}{\pi k}$.

In this case, $\left\{\begin{array}{l}\frac{2k + 2}{\pi k} \leq 1, \\ \frac{2k + 2}{\pi k^2} \leq 1\end{array} \Leftrightarrow \left\{\begin{array}{l}k(\pi - 2) \geq 2, \\ \pi k^2 - 2k - 2 \geq 0\end{array} \Leftrightarrow \left\{\begin{array}{l}k \geq \frac{2}{\pi - 2}, \\ k \geq \frac{1 + \sqrt{2\pi + 1}}{\pi}\end{array} \Rightarrow k \geq \frac{2}{\pi - 2}\right.\right.\right.$.

Answer: $\arcsin \frac{2(k + 1)}{\pi k^2}, \pi - \arcsin \frac{2(k + 1)}{\pi k^2}, \arcsin \frac{2(k + 1)}{\pi k}$,

$\pi - \arcsin \frac{2(k + 1)}{\pi k}, k \geq \frac{2}{\pi - 2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0911.jpg?height=343&width=444&top_left_y=79&top_left_x=139)

Fig. 12.26

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0911.jpg?height=408&width=525&top_left_y=12&top_left_x=704)

Fig. 12.27