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:

A diagram showing a circle partitioned by three rays emanating from its centre, with three corresponding angles: $2 - 4/n$ right angles (twice) and $2 - 4/p$ right angles.

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.)

A part of the tiling of the plane consisting of regular squares and regular octagons (i.e., the case where $n = 8$ and $p = 4$).

Let’s see what the other integral pairs look like. When \((n,p) = (5,10)\), we have

A regular decagon surrounded by ten regular pentagons (i.e., the case where $n = 5$ and $p = 10$).

This arrangement of tiles around a point cannot be extended to tessellate the plane.

The second is when \((n,p) = (6,6)\); this corresponds to the regular hexagonal tiling of the plane.

A part of the hexagonal tiling of the plane (i.e., the case where $n = p = 6$).

Is there any tiling of the plane by three regular polygons that features the regular decagon?

We could investigate this by generalising the above work to the case of three possibly distinct regular polygons around a vertex.

Which possible triples of regular polygons can meet at a single vertex? Which triples can tile the plane?