Circumradius:Inradius

The inradius of a regular polygon is its apothem , i.e. the line from the center of the polygon to the middle of one of its sides. An adjacent inradius and circumradius will be two sides of a right triangle as shown below on the left.

We have that $\cos \theta = \dfrac{r}{R} = \dfrac{1}{\sqrt{5} - 1} = \dfrac{\sqrt{5} + 1}{4}$ , so $\theta = 36^{\circ}$ (see note below). Then $n = \dfrac{360^{\circ}}{2 \theta} = \boxed{5}$

Note: To see that $\cos 36^{\circ} = \dfrac{\sqrt{5} + 1}{4}$ we can use a regular pentagon with side lengths $1$ as shown above on the right.

Since the internal angle of a regular pentagon is $108^{\circ}$ , we have that $\angle BCG = 54^{\circ}$ , so $\angle CBG = 36^{\circ}$ .

Also, since the external angle of a regular pentagon is $72^{\circ}$ , we have that $\angle BAF = 18^{\circ}$ .

Then calculating the length $BD$ two different ways we get

$2 \cos 36^{\circ} = 2 \sin 18^{\circ} + 1$

Using a double-angle identity, we know that $\cos36^{\circ} = 1 - 2 \sin^2 36^{\circ}$ , so setting $\alpha = \sin 36^{\circ}$ gives us

$\begin{array}{rl} 4\alpha^2 +2\alpha - 1 &= \ 0 \\ \alpha &= \ \dfrac{\text{-}1 \pm \sqrt{5}}{4} \end{array}$

We take the positive root as $\alpha = \sin 36^{\circ} > 0$ ; and since we know from the original equation that $2 \cos 36^{\circ} = 2 \alpha + 1$ , we get that $\cos 36^{\circ} = \dfrac{\sqrt{5} + 1}{4}$ .