Let $A, B, C, D, E$ be the vertices of a regular pentagon inscribed in a circle of radius $r$. Let $F$ be the midpoint of side $AB$. Find the circumradius $AO$ in terms of the side length $x = AB$.
Since the pentagon is regular, $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOE, \triangle EOA$ are congruent isosceles triangles. Their vertex angles add up to 360 degrees, so each vertex angle measures 72 degrees. For $\triangle AOB$: since it is isosceles, the median from the vertex is the same as the angle bisector and the altitude. Thus $OF$ is also the angle bisector, and $\angle AOF = \angle BOF = 36$ degrees. $\triangle AFO$ is a right triangle with right angle at $F$. Using trigonometric functions: $$AO = AF \cdot \csc(\angle AOF) = \frac{x}{2} \csc 36^\circ$$
Let $A, B, C, D, E$ be the vertices of a regular pentagon inscribed in a circle of radius $r$. Let $F$ be the midpoint of side $AB$. Find the circumradius $AO$ in terms of the side length $x = AB$.