# Sticks and Stones

Because I’m insufferably vain, I have a search running in my Twitter client for the words “The Maths Behind”, in case someone mentions my book (which is, of course, available wherever good books are sold). On the minus side, it rarely is; on the plus side, the search occasionally throws up tweets like this:

The maths behind this Aboriginal problem of finding something in the desert blows my mind. If you using two fixed objects and right angles you will always find the mark irrespective of distances @drchris_maths @Jason_Loke @MrJohnRowe @AetYates @scibabe pic.twitter.com/LnCLroA2iL

— Matt Jamieson (@STEMJamo) August 6, 2018

Now, I’m always a bit skeptical of claims that a method owes its origins to vaguely-defined groups, so I’ll state for clarity: I don’t know whether this really was a method used by the original inhabitants of what is now Australia. It could easily be, but I have no evidence either way.

In any case, it’s ingenious: you want to bury some treasure. You identify two prominent landmarks (in the example, a rock and a tree) and put a marker in the ground anywhere.

You then walk towards the first landmark, counting your steps; turn through a right angle and walk the same number of steps, placing a second marker in the ground.

Go back to the original marker and walk to the second landmark, again counting your steps. Turn through a right angle and walk the same number of steps before placing a third marker in the ground.

Now bury your treasure midway between the second and third markers.

It turns out that, wherever you put the original marker, you end up burying the treasure in the same place.

### Yes, but why?

There are presumably dozens of ways to prove this, but I’m going to go with a complex numbers approach. Suppose the rock is at a point on the Argand diagram representing the number $R$ and that the tree is at a point representing $T$; we place the first stick at an arbitrary point $z$.

Walking towards the rock, turning a right-angle (anti-clockwise) and walking the same distance forwards is equicalent to rotating the segment from $R$ to $z$ through $+\piby2$ radians. We can achieve the same effect by multiplying $z-R$ by $i$ and adding it to the position of the rock - this gives us $R + i(z-R)$.

Similarly, if we walk to the tree, turn a right-angle clockwise and walk the same distance forwards, it’s the same as multiplying $z-T$ by $-i$ and adding the result to $T$. We get $T - i(z-T)$.

The third marker is midway between the two, so we need to add them up and divide by two, which gives us $\frac{1}{2}\br{R + iz - iR + T - iz + iT}$, which simplifies to $\frac{1}{2}\br{R+T + i(T-R)}$ - which is independent of the initial position of $z$.

### But wait a second…

The complex number that comes out of the calculation turns out to be on the perpendicular bisector of $T$ and $R$, forming a right-angled triangle.

This suggests, among other things, a much simpler method of finding the same spot: simply walk from the rock halfway to the tree, counting your steps; turn a right-angle anti-clockwise and walk the same number of steps forward. The point you arrive at is where the treasure is buried.

I said earlier, there must be a dozen proofs of this. I’d love to hear yours, especially if it’s more intuitive!

* Thanks to @STEMJamo for the puzzle!