Chord Length for Regular Polygon

Theorem

Let $P$ be a regular polygon of $n$ sides.

Let $P$ be inscribed into a circumcircle with radius $r$.

Let $CF$ be a chord of $P$, also a chord of the circumcircle.

$CF$ divides $P$ into two polygons containing $k$ and $n - k$ sides of $P$.

The length of the chord is $2 r \map \sin {\dfrac {k \pi} n}$.

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Proof

Let $\theta = \angle BAC$ be an inscribed angle, subtending one of the sides of $P$, with its vertex coinciding in one of the other vertices of $P$.

Because $P$ is a regular $n$-gon, its sides are all the same length.

We have by hypothesis that $P$ is inscribed into a circle.

Therefore, the central angles subtending a single side all have the same angular measure: $\dfrac {2 \pi} n$.

By the Inscribed Angle Theorem, the angles formed at the vertices of $P$ subtending a single side all have equal angular measure.

That is:

$\theta = \dfrac \pi n$

Let $\angle CAF = \phi$ be an inscribed angle subtending the chord $CF$.

By addition:

$\angle CAF = \angle CAD + \angle DAE + \angle EAF$

Thus, the angle $\phi$ is the sum of $k$ smaller angles each with measure $\theta$.

Hence:

$\phi = k \theta = \dfrac {k \pi} n$

By Length of Chord of Circle:

$CF = 2 r \map \sin {\dfrac {k \pi} n}$

$\blacksquare$

Examples

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For the diagonal of a square inscribed in a circle of radius $r$, we have $k=2$. The central angle subtending one side has measure $\frac {pi} 2$, so the inscribed angle subtending one side has measure $\frac {\pi} 4$. Thus, the length of the diagonal is $2r \map \sin 2 \frac \pi 4 = 2r$.

For the regular hexagon, there are chords enclosing $k=2$ and $k=3$ sides. The latter are diameters of the circumcircle. The central angle subtending one side has measure $\frac \pi 3$, so the corresponding inscribed angle has measure $\frac \pi 6$.

For the case $k=3$ we have $2r \sin 3 \frac \pi 6 = 2r$ and for the case $k=2$ we have $2r \sin 2 \frac \pi 6 = \sqrt{3} r$

For the general case with $n$ sides, the inscribed angle subtending one side has measure $\frac \pi n$ and the length of the chord cutting off $k$ sides is $2r \sin k \frac \pi n$.