In-circle, Circumcircle & Regular Polygons

• Inradius: $r = \dfrac{\Delta}{s}$ (area over semi-perimeter).
• Circumradius: $R = \dfrac{abc}{4\Delta} = \dfrac{a}{2\sin A}$.
• Central vs inscribed angle: an arc subtends an angle at the centre that is twice the angle it subtends at any point on the circle: $\angle BOC = 2\angle BAC$.
• Relating the two angles' ratios: with $\alpha = \angle BAC$ and $\beta = \angle BOC = 2\alpha$, the double-angle formulas convert one to the other: $\cos\beta = \cos 2\alpha = \dfrac{1-\tan^2\alpha}{1+\tan^2\alpha} = 1 - 2\sin^2\alpha$, and $\sin\beta = \sin 2\alpha = \dfrac{2\tan\alpha}{1+\tan^2\alpha}$. Option lists mix these up deliberately — $\dfrac{2\tan\alpha}{1+\tan^2\alpha}$ is a sine, not a cosine.
• Chord length: a chord subtending angle $\theta$ at the centre of a circle of radius $R$ has length $2R\sin\dfrac{\theta}{2}$.
• Equal tangents from an external point: the two tangent segments from a point to a circle are equal. For the incircle this means the sides split as $s-a$, $s-b$, $s-c$ at the points of contact — from vertex $A$ the two tangent lengths are both $s-a$ (so $\tan\tfrac A2=\dfrac{r}{s-a}$). A triangle with sides $a,b,c$ whose incircle touches $BC$ at $D$ has $BD=s-b$ and $DC=s-c$; the whole "in-circle" question pair is this fact plus arithmetic.

For a regular polygon of $n$ sides, each of length $s$:

• Interior angle: $\dfrac{(n-2)\,180^\circ}{n}$.
• Inradius (inscribed circle, touching each side): $r = \dfrac{s}{2}\cot\dfrac{\pi}{n}$.
• Circumradius (through the vertices): $R = \dfrac{s}{2}\csc\dfrac{\pi}{n}$.
• The cotangents you will need, since $\pi/n$ is rarely a textbook angle: $\cot 60^\circ = \tfrac{1}{\sqrt3}$ ($n=3$), $\cot 45^\circ = 1$ ($n=4$), $\cot 30^\circ = \sqrt3$ ($n=6$), $\cot 22.5^\circ = 1+\sqrt2$ ($n=8$), $\cot 15^\circ = 2+\sqrt3$ ($n=12$). The last comes from $\tan 15^\circ = \tan(45^\circ-30^\circ) = \dfrac{1-\tfrac{1}{\sqrt3}}{1+\tfrac{1}{\sqrt3}} = 2-\sqrt3$, whose reciprocal is $2+\sqrt3$. So a regular 12-gon of side 1 has inradius $\tfrac12(2+\sqrt3)$ and inscribed-circle diameter $2+\sqrt3$.