Brief Introduction

First, a regular polygon has all sides equal and all interior angles equal. These two conditions are equivalent for convex polygons, and there are many strategies using these such as angle chasing, and computing areas.

The key is having the right formulas memorized and knowing when to exploit symmetry rather than brute-forcing coordinates.

Angles

Interior Angle

The sum of interior angles of any $n$-gon is $(n-2) \cdot 180°$. Since all angles in a regular $n$-gon are equal, we find that each interior angle is

Exterior Angle

At each vertex, the interior and exterior angles are supplementary to each other. The exterior angle of a regular $n$-gon is

An important fact to understand is that all exterior angles of any convex polygon always sum to exactly $360°$. hence why the exterior angle formula is $360°/n$. (Because all $n$ exterior angles are equal and they sum to $360°$, each must be $360°/n$.)

Table of Polygons

Side Length, Circumradius, and Inradius

We have a regular $n$-gon with center $O$, side length $s$, and circumradius $R$ (center to vertex), and inradius $r$ (center to midpoint of a side, also called the apothem).

Then, we connect the center to two adjacent vertices creates an isoceles triangle with two sides of length $R$ and a base of length $s$, and the central angle is $360°/n$. From this triangle,

Furthermore, the apothem is the height from the center to a side:

The second identity follows from the Pythagorean theorem applied to the right triangle formed by $R$, $r$, and $s/2$.

Area of Polygons

The area formula comes from dividing the polygon into $n$ congruent isoceles triangles from the center. Each triangle has base $s$ and height $r$ (the apothem), so therefore,

where $P = ns$ is the perimeter. A useful form of the formula is that the area equals half the perimeter times the apothem.

Hence in terms of side length alone, we find that

In terms of the circumradius,

Some special cases of is for an equilateral triangle with side $s$, where

For a square with side $s$,

For a regular hexagon with side $s$,

For hexagons, $R = s$, where the circumradius equals the side length.

Diagonals

Another important concept when talking about regular $n$-gon is that it has $\binom{n}{2} - n = \dfrac{n(n-3)}{2}$ diagonals.

The length of a diagonal connecting vertices $k$ positions apart (where $1 \leq k \leq \lfloor n/2 \rfloor$) is

We can prove this identity because the central angle subtended is $2\pi k/n$, and from the chord length formula, we get $d_k = 2R\sin(k\pi/n)$.

Symmetry

A regular $n$-gon has:

• $n$ lines of symmetry (through each vertex and the midpoint of the opposite side for even $n$), in each vertex and midpoint of the opposite side for odd $n$).
• We also see a rotational symmetry of order $n$, where there are rotations by $360°k/n$ for $k=0,1,\ldots,n-1$ all map the polygon to itself.
• Dihedral group $D_n$ of order $2n$ as its symmetry group. The concept of dihedral groups can be found in the USAMO Series.

Examples

Example 1: Finding Exterior Angles

Problem. A regular polygon has an interior angle of $150°$. How many sides does it have?

Solution. Exterior angle $= 180° - 150° = 30°$. Since exterior angles sum to $360°$: $n = 360°/30° = \mathbf{12}$ sides.

Example 2: Finding Diagonals of a Hexagon

Problem. A regular hexagon has side length $2$. What is the area of the triangle formed by every other vertex?

Solution. The vertices of a regular hexagon at distance $R = 2$ from the center form two equilateral triangles when alternated. Each such triangle connects vertices $2$ apart, therefore its side length is $d_2 = 2R\sin(2\pi/6) = 2(2)\sin(60°) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$.

Hence the area of equilateral triangle with side $2\sqrt{3}$:

Example 3: Area via Apothem

Problem. A regular octagon is inscribed in a circle of radius $5$. Find the area of the octagon.

Solution. Use $A = \frac{1}{2}nR^2\sin(2\pi/n)$ with $n=8$, $R=5$:

Example 4: Using Pentagons and its Golden Ratio

Problem. In a regular pentagon with side length $1$, find the length of a diagonal.

Solution. First, we label the vertices $A,B,C,D,E$. The diagonal $AC$ and side $AB$ form the base of an isoceles triangle $ABC$ with $\angle ABC = 108°$ and base angles $36°$ each. By the sine rule,

Since $\sin 108° = \sin 72° = \cos 18°$ and using the identity $\frac{\sin 72°}{\sin 36°} = 2\cos 36° = 1 + \frac{1}{\phi}$.

We let $d$ be the diagonal. The triangle $ABC$ has $AB = 1$ and $AC = d$. By the similarity of the golden gnomon: $d/1 = 1/(d-1)$, giving $d^2 - d - 1 = 0$, so $d = \frac{1+\sqrt{5}}{2} = \phi \approx \mathbf{1.618}$.

Practice Problems