Two regular polygons with perimeters $a$ and $b$ are circumscribed about a circle, and a third regular polygon is inscribed in this circle. The second and third polygons have twice as many sides as the first. Find the perimeter of the third polygon.
Let $R$ be the radius of the circle, $x$ be the perimeter of the third polygon, and $n$ be the number of sides of the first polygon. Let $\alpha = \pi / n$. Then

$a = 2 n R \operatorname{tg} 2 \alpha, b = 4 n R \operatorname{tg} \alpha, x = 4 n R \sin \alpha = b \cos \alpha$. Therefore, $\frac{b}{a} = \frac{2 \operatorname{tg} \alpha}{\operatorname{tg} 2 \alpha} = 1 - \operatorname{tg}^{2} \alpha$, hence,

$\operatorname{tg} \alpha = \sqrt{\frac{a - b}{a}}, \cos \alpha = \frac{1}{\sqrt{1 + \operatorname{tg}^{2} \alpha}} = \sqrt{\frac{a}{2 a - b}}$. Consequently, $x = \frac{b \sqrt{a}}{\sqrt{2 a - b}}$.

## Answer

$$
\frac{b \sqrt{a}}{\sqrt{2 a - b}}
$$