If \(n > 2\) and even, show that \(C = \frac{n}{{2\pi }}\sin \frac{{2\pi }}{n}\).

If \(n > 1\) and odd, it can be shown that \(C = \frac{{n\sin \frac{{2\pi }}{n}}}{{\pi \left( {1 + \cos \frac{\pi }{n}} \right)}}\).

Find the regular polygon with the least number of sides for which the compactness is more than \(0.99\).

If \(n > 1\) and odd, it can be shown that \(C = \frac{{n\sin \frac{{2\pi }}{n}}}{{\pi \left( {1 + \cos \frac{\pi }{n}} \right)}}\).

Comment briefly on whether C is a good measure of compactness.

each triangle has area \(\frac{1}{8}{x^2}\sin \frac{{2\pi }}{n}\;\;\;({\text{use of }}\frac{1}{2}ab\sin C)\)     (M1)

there are \(n\) triangles so \(A = \frac{1}{8}n{x^2}\sin \frac{{2\pi }}{n}\)     A1

\(C = \frac{{4\left( {\frac{1}{8}n{x^2}\sin \frac{{2\pi }}{n}} \right)}}{{\pi {n^2}}}\)     A1

so \(C = \frac{n}{{2\pi }}\sin \frac{{2\pi }}{n}\)     AG

[3 marks]

attempting to find the least value of \(n\) such that \(\frac{n}{{2\pi }}\sin \frac{{2\pi }}{n} > 0.99\)     (M1)

\(n = 26\)     A1

attempting to find the least value of \(n\) such that \(\frac{{n\sin \frac{{2\pi }}{n}}}{{\pi \left( {1 + \cos \frac{\pi }{n}} \right)}} > 0.99\)     (M1)

\(n = 21\) (and so a regular polygon with 21 sides)     A1

Note:     Award (M0)A0(M1)A1 if \(\frac{n}{{2\pi }}\sin \frac{{2\pi }}{n} > 0.99\) is not considered and \(\frac{{n\sin \frac{{2\pi }}{n}}}{{\pi \left( {1 + \cos \frac{\pi }{n}} \right)}} > 0.99\) is correctly considered.

Award (M1)A1(M0)A0 for \(n = 26\).

[4 marks]

EITHER

for even and odd values of n, the value of C seems to increase towards the limiting value of the circle \((C = 1)\) ie as n increases, the polygonal regions get closer and closer to the enclosing circular region     R1

OR

the differences between the odd and even values of n illustrate that this measure of compactness is not a good one.     R1

Most candidates found this a difficult question with a large number of candidates either not attempting it or making little to no progress. In part (a), a number of candidates attempted to show the desired result using specific regular polygons. Some candidates attempted to fudge the result.

In part (b), the overwhelming majority of candidates that obtained either \(n = 21\) or \(n = 26\) or both used either a GDC numerical solve feature or a graphical approach rather than a tabular approach which is more appropriate for a discrete variable such as the number of sides of a regular polygon. Some candidates wasted valuable time by showing that \(C = \frac{{n\sin \frac{{2\pi }}{n}}}{{\pi \left( {1 + \cos \frac{\pi }{n}} \right)}}\) (a given result).

In part (c), the occasional candidate correctly commented that \(C \) was a good measure of compactness either because the value of \(C \) seemed to approach the limiting value of the circle as \(n \) increased or commented that \(C \) was not a good measure because of the disparity in \(C \)-values between even and odd values of \(n \).