Review question

How can we arrange regular polygons around a point? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6908

Solution

The sum of the interior angles of a polygon with $n$ sides is $(2n-4)$ right angles. Deduce that each interior angle of a regular polygon is $\left( 2 - \dfrac{4}{n} \right)$ right angles.

A regular polygon with $n$ sides has $n$ interior angles, with all of these angles being equal. Thus, each interior angle is $\frac{1}{n} \left( 2n - 4 \right) = 2 - \frac{4}{n}$ right angles.

Three regular polygons, two having $n$ sides and one having $p$ sides, fit together exactly at a common vertex, as shown:

By using the fact that the sum of the three angles shown is four right angles, prove that $\frac{4}{n} + \frac{2}{p} = 1.$

Since the three angles in the diagram add to four right angles, we have that \begin{align*} \left( 2 - \frac{4}{n} \right) + \left( 2 - \frac{4}{n} \right) + \left( 2 - \frac{4}{p} \right) = 4 &\implies 6 - \frac{8}{n} - \frac{4}{p} = 4 \\ &\implies \frac{8}{n} + \frac{4}{p} = 2 \\ &\implies \frac{4}{n} + \frac{2}{p} = 1 \end{align*}

as required.

Show that this formula can be rearranged into the form $p = \frac{2n}{n-4}.$

We have that \begin{align*} \frac{4}{n} + \frac{2}{p} = 1 &\implies \frac{2}{p} = 1 - \frac{4}{n} \\ &\implies \frac{2}{p} = \frac{n-4}{n} \\ &\implies \frac{p}{2} = \frac{n}{n-4} \\ &\implies p = \frac{2n}{n-4}. \end{align*}

By inserting into this formula various values of $n$, list all the pairs of positive integers $(n,p)$ for which $n < 12$ which satisfy this equation.

As $n$ is a positive integer, in order for $p$ to be positive we require that $n > 4$. By substituting values in, we have the following table.

$n$ $p = \dfrac{2n}{n-4}$
$5$ $10$
$6$ $6$
$7$ $\dfrac{14}{3}$
$8$ $4$
$9$ $\dfrac{18}{5}$
$10$ $\dfrac{20}{6}$
$11$ $\dfrac{22}{7}$

Thus, the suitable pairs $(n,p)$are $(5,10)$, $(6,6)$ and $(8,4)$.

We could note that $(12,3)$ will also work as a pair, and since $p \geq 3$, no other pairs are possible.

By means of a sketch, interpret one of these solutions where $n$ and $p$ are different in terms of three regular polygons meeting at a common vertex.

We will consider the neater of the two cases: $(8,4)$, which is shown below. (In fact, this arrangement will tessellate, that is, tile the entire plane. This is termed the truncated square tiling, and is one of the semi-regular tessellations.)