# Polygon walk

Go on a vector walk and determine which points on the walk are
closest to the origin.

My friends Ulaf and Vicky are each given the vector ${\bf i} = (1, 0)$ and another constant vector ${ \bf u}$ and ${\bf v}$ respectively. Starting at the origin, Ulaf and Vicky take a 2-dimensional 'vector walk' where each step is either ${\bf i}$ or one of their constant vectors, either forwards or backwards.

Thus, Ulaf could reach the points $n{\bf i }+m{\bf u}$ and Vicky could reach the points $n{\bf i }+m{\bf v}$ for any whole numbers $n$ and $m$.

After walking around for a while, Ulaf tells me that the points nearest to the origin he could reach formed the vertices of an equilateral triangle and Vicky tells me that the points nearest to the origin that she could reach formed the vertices of a regular hexagon. Using this information, work out ${ \bf u}$ and ${\bf v}$.

Later that day, Wilber went on a vector walk using the vectors ${\bf i} = (1, 0)$ and ${\bf w}$. He tells me that the points nearest to the origin that he could reach formed a regular pentagon. Why must Wilber be mistaken?

Draw the triangle pointing right such that the rightmost vertex is at $\mathbf{i}$

The coordinates of a regular $n$-gon with a centred on the origin with a vertex at $(1,0)$ are

$$\left(\cos\left(\frac{2m\pi}{n}\right), \sin\left(\frac{2m\pi}{n}\right)\right)\, \text{ where }m=0, \dots, n-1$$

For a pentagon, the coordinates become

$$

(1, 0), \left(\frac{1}{4}\left(\sqrt{5}-1\right), \frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)\right), \left(-\frac{1}{4}\left(\sqrt{5}+1\right), \frac{1}{4}\left(\sqrt{10-2\sqrt{5}}\right)\right)

$$

along with the mirror images in the $x$-axis.

This problem builds on GCSE vector work and provides a foundation for concepts met in the later Core A Level modules.

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.

### Why do this problem?

This problem builds on ideas from the problem Vector Walk.
Students are encouraged to think geometrically about vectors in
order to deduce which vectors could generate particular sets of
points, as well as reasoning why other sets of points could not be
reached in the same way.

### Possible approach

Students could work in small groups to create some vector
walks using the vector ${\bf i} = (1,0)$ and another vector ${\bf
u}$. Encourage them to comment on the similarities
between the points reached when different vectors ${\bf u}$ are
chosen.

After students have a feel for how two vectors can be used to
reach a variety of points on the plane, ask them to sketch possible
arrangements of points around the origin forming an equilateral
triangle, and a hexagon, using points which can be reached using
the vector ${\bf i}$. Some calculation will be needed to work out
the second vector needed to reach the appropriate points.

Finally, set students the last challenge to explain why a
regular pentagon could not be created in the same way. Encourage
them to use both algebra and geometry to justify their
answer.

### Key questions

Can you draw a diagram to show an equilateral triangle or a
hexagon of points around the origin?

Can you show a way to visit all these points using just ${\bf
i}$ and one other vector?

What happens when you try to do the same thing with a
pentagon?