Maclaurin 2015 Q6

Let the exterior angle in each polygon be \(\alpha^\circ\), so that \(\alpha = \dfrac{360}{n}\), and let \(P\), \(Q\), \(R\), \(S\) and \(T\) be five consecutive vertices of one of the polygons, as shown, where \(QRS\) is part of the perimeter of the inner region.

From the symmetry, the point \(O\) where \(PQ\) and \(TS\) meet is the centre of the inner region and hence

Now the sum of the angles in quadrilateral \(OSRQ\) is 360°, so we have

and therefore

Multiplying each term by \(mn\), we obtain

so that

Now \(m > 2\) for the ring to exist. Hence \((m - 2)\,(n - 6)\) is a product of two positive integers, so that \(n - 6\) is a positive factor of 12. The only possibilities for \(n - 6\) are therefore 1, 2, 3, 4, 6 and 12, and thus the only possibilities for \(n\) are 7, 8, 9, 10, 12 and 18. The following figures show the corresponding rings.

Thus a ring of polygons of the required form is possible for six different values of \(n\).