10.367 Express the side of a regular decagon in terms of the radius $R$ of the circumscribed circle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-228.jpg?height=312&width=302&top_left_y=128&top_left_x=86)

Let $AB$ be a side of the given regular decagon with center $O$, then $AO = BO = R$, $\angle A = \angle B = 72^{\circ}$. Draw the angle bisector $AK$.

Then $\angle AOK = \angle OAK = 36^{\circ}$, $\angle AKB = \angle ABK = 72^{\circ} \Rightarrow$ $\Rightarrow AK = AB = OK$, $BK = R - AB$.

By Theorem 4,

$\frac{BK}{OK} = \frac{AB}{AO} \Rightarrow \frac{R - AB}{AB} = \frac{AB}{R} \Rightarrow$

$\Rightarrow AB^2 + R \cdot AB - R^2 = 0 \Rightarrow AB = \frac{\sqrt{5} - 1}{2} R$.

Answer: $\frac{\sqrt{5} - 1}{2} R$.