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## 'Polygon Walk' printed from http://nrich.maths.org/

After drawing a couple of regular hexagons such that there is a
vertex at the top, and joining up those vertices, it became clear
that 2 finite lines can reach any vertex without stopping at an
internediate location nearer to the centre, and those 2 lines are
like the vectors $\mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix}$,
and $\mathbf{u} =
\begin{pmatrix}\frac{\sqrt{3}}{2}\\\frac{1}{2}\end{pmatrix}$.

But we wanted one of the vectors to be $\mathbf{i}$, so we can
rotate the whole problem by 90$^{\circ}$ (swap two components and
multiply either by -1) so we have $\mathbf{i} =
\begin{pmatrix}1\\0\end{pmatrix}$, and $\mathbf{u} =
\begin{pmatrix}-\frac{1}{2}\\\frac{\sqrt{3}}{2}\end{pmatrix}$.

The vertices of the equilateral triangle coincide with vertices of
the hexagon, so the vectors are the same for the triangle, so
$\mathbf{v} = \mathbf{u}$. But Ulaf must detour, Vicky need
not.

For the pentagon, draw one in some orientation, say with a central
top vertex. If we say that the vertices sit on a unit circle, we
can see that a multiple of the y-coordinate of $\mathbf{w}$ must be
-1, and another multiple must be $\cos(36^{\circ})$. Having a look
at $\cos(36^{\circ})$ in the calculator, I am pretty sure it is
irrational, and thus we cannot have the multiples described.

I first converted 36$^{\circ}$ to radians, to get $\frac{\pi}{5}$
radians. I first looked at using the cosine rule, but then realised
that I could use the fact that $\cos(5 \theta) = -1$, expand
$\cos(5\theta)$ in terms of $\cos(\theta)$, and solve to get an
exact answer which I could show was irrational. However this
expends to a quintic which is hard to solve. So I decided to try
solving for $\sin(5\theta) = 0$, then use the identity to convert
the answer to $\cos(\theta)$. Expanding $\sin(5\theta) = 0$ in
terms of $\sin(\theta)$, using identities, we get $16s^4 - 20s^2 +
5 = 0 \therefore s^2 = \frac{5\pm \sqrt{5}}{8}$, checking with the
calculator, the one we want is $s = \sqrt{\frac{5- \sqrt{5}}{8}}$,
then we find that $\cos(36^{\circ}) = \sqrt{\frac{3+\sqrt{5}}{8}}$,
and since all square roots of numbers that are not themselves
squares are irrational, $\cos (36^{\circ})$ is irrational. Thus
Wilber was mistaken.