# Distance from center of a regular n-sided polygon to its apex

I'm trying to figure out the distance from a regular polygon's apex to its center.

Let $ABP$ be a triangle where $A$ and $B$ are adjacent apexes of the polygon and $P$ is its center. $$\angle APB = \alpha = \frac{2\pi}{n}$$ $$\angle ABP = \angle BAP = \beta = \frac{\pi(n-2)}{2n}$$ Let $a = |AB|$, $r = |AP|$, $h$ be the height of $ABP$. $$\tan \beta = \frac{h}{\frac{a}{2}} \qquad h = \frac{a \tan \beta}{2}$$ $$r^2 = \left(\frac{a}{2}\right)^2+h^2$$ $$r^2 = \frac{a^2}{4} + \frac{a^2 \tan^2 \beta}{4} = \frac{a^2}{4}(1 + \tan^2 \beta)$$ $$r = \frac{a \sec \beta}{2}$$ $$r = \frac{a \csc \frac{\pi}{n}}{2}$$ Is this answer correct? Is there a nicer way to calculate this?

• Did you check whether it gave the right answer for an equilateral triangle, square, and pentagon? If so, it's pretty likely to be correct. Your computation might be ever-so-slightly simpler if you made the radius of the circumcircle of the polygon be 1 (i.e., made $|AP| = |BP| = 1$); the formula for a polygon with radius $r$ the just involves a factor of $r$. – John Hughes May 30 '17 at 15:55

1) Draw a picture of the n-gon with $\triangle ABP$ shown.
2) Split $\triangle ABP$ symmetrically into two right-angled triangles.