There’s an old buried treasure puzzle that turns out to be quite easy to solve if you know about complex numbers. So in this video, we’re gonna look at an easy complex way to find some treasure. The feared pirate, Green-and-orange Stripy Beard, buried his hoard of treasure on a small desert island with a pine tree and a bay tree and a nearby gravestone. And then he left some simple instructions to his grandson Nigel to find the spot where the treasure was buried.
Start at the gravestone and walk in a straight line to the bay tree. Turn 90 degrees clockwise and walk the same distance again. Then mark the spot. Return to the gravestone and walk straight to the pine tree. Turn 90 degrees counterclockwise then walk the same distance again and mark the spot.
The treasure is at the midpoint of the two spots you marked. Unfortunately, before Nigel reached the island to retrieve the treasure, some grave robbers literally stole the grave including the gravestone. So how is Nigel going to find his treasure now that he doesn’t have a start point to work from? It is important to point out that this puzzle was set in a time before drone photography and ground-penetrating radar that might have given him some clues.
And he wasn’t able to get any heavy machinery to dig up vast areas at random. He’s got a shovel and he wants to go straight to the treasure and dig it up. And it’s also worth pointing out that Green-and-orange Stripy Beard picked the pine tree and the bay tree because they were uniquely distinctive on the island and would be easy to represent on a simple map. One looks like a triangle and the other looks like a circle.
He didn’t take into account the fact that they both look like a circle when represented in plan view. But let’s not get hung up on small details like that. Pause the video now and have a go at solving the puzzle of how Nigel can find Green-and-orange Stripy Beard’s treasure and then I’ll tell you a way.
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Okay, so our first tactic might be to draw out a simple map, pick a random point for the gravestone, and follow the instructions to see where it leads you. Well here’s the pine tree and here’s the bay tree. Let’s just randomly put the gravestone here and then follow the instructions. Walk in a straight line to the bay tree. Turn 90 degrees clockwise. Walk the same distance again. And then mark that spot.
Then return to the gravestone. Walk straight to the pine tree. Turn 90 degrees counterclockwise. Walk the same distance again and then mark that spot. Then find the midpoint of those two spots and that’s where your treasure is. And when you do this a few times, you notice something interesting. Maybe you’ve already tried this approach. But if you haven’t, then you might want to pause the video again now and have a go yourself before I talk about it.
Let’s do a second demo. This time we put the gravestone here. Walk straight to the bay tree. Turn 90 degrees clockwise. Walk the same distance again. And mark the spot. Return to the gravestone. Walk to the pine tree. Turn 90 degrees counterclockwise. Walk the same distance again. And mark that spot. Then when you find the midpoint of those two points that’s where the treasure is hidden. Interesting, the treasure spot is in the same place.
Is that just a coincidence? Because, if it isn’t, then it doesn’t matter where we start. We’ll always end up in the same spot if we follow the instructions. Obviously, we have to be a bit careful about where we start or we will end up having to mark a spot in the sea, which won’t be easy. But can we come up with a way of proving that wherever the gravestone was, we’ll always find the treasure if we follow the instructions?
Well there are several but let’s use complex numbers to explore one way. Let’s draw a line between the pine tree and the bay tree. Then let’s rotate our view of the island slightly and call it the real axis on a complex plane. Now we can take the midpoint between the trees and call it the origin, then draw in the imaginary axis.
Then if we define our unit as the distance from the origin to the bay tree, then we can represent the location of the bay tree as the complex number one plus zero 𝑖 on that plane and the position of the pine tree as negative one plus zero 𝑖. Now let’s pick an arbitrary point for the gravestone. Let’s call it 𝐺. And while we’re at it, let’s label the point where the trees are as 𝑃 for the pine tree and 𝐵 for the bay tree. And let’s represent the position of 𝐺 with the complex number 𝑎 plus 𝑏𝑖.
It’s got a real component of 𝑎 times the distance from the origin to the bay tree and an imaginary component of 𝑏 times the distance from the origin to the bay tree. The treasure hunt instructions told us to start at 𝐺 and walk to 𝐵 so let’s represent that in vector form on our complex plane. Well to get from 𝐺 to 𝐵, the real component changes from 𝑎 to one. So that component of the vector journey is one minus 𝑎, the difference between one and 𝑎. And the imaginary component changes from 𝑏𝑖 down to zero 𝑖, which can be written zero 𝑖 minus 𝑏𝑖 and simplified just to negative 𝑏𝑖.
So the vector 𝐺𝐵 can be written as one minus 𝑎 plus negative 𝑏𝑖 or even one minus 𝑎 minus 𝑏𝑖. Next, we have to turn 90 degrees clockwise and walk the same distance again. Now on the complex plane we just need to multiply a vector by negative 𝑖 in order to rotate it 90 degrees clockwise. So let’s do that. Negative 𝑖 times 𝐺𝐵 is equal to negative 𝑖 times one minus 𝑎 minus 𝑏𝑖. And when I multiply through by negative 𝑖, I get negative one minus 𝑎𝑖 plus 𝑏𝑖 squared.
Now remember, 𝑖 squared is equal to negative one. So that last term just becomes minus 𝑏. Now let’s put the real component first and the imaginary component second to give me negative 𝑏 minus one minus 𝑎 𝑖. And in fact I don’t like all these kind of negative signs floating around here. So I’m just going to write that as negative 𝑏 plus 𝑎 minus one 𝑖. But that vector just represents a movement in the complex plane, a direction which is 90-degrees- clockwise rotated from the vector 𝐺𝐵 and the same magnitude or distance as vector 𝐺𝐵.
We need to lay down that vector so that its start point or initial point is at the bay tree. And then we can find its end or terminal point to tell us where to mark a point on the ground. Let’s call it 𝐵 dash. Now our vector, negative 𝑏 plus 𝑎 minus one 𝑖, can be called vector 𝐵𝐵 dash or this vector here. So to find the position vector of 𝐵 dash, we need to start at the origin and specify the vector that takes us to point 𝐵 dash.
Now to do that, let’s make the journey from 𝑂 to 𝐵 and then from 𝐵 to 𝐵 dash. So vector 𝑂𝐵 dash is equal to vector 𝑂𝐵 plus vector 𝐵𝐵 dash. And remember, we just worked out an expression for vector 𝐵𝐵 dash. So vector 𝑂𝐵 was one plus zero 𝑖 and vector 𝐵𝐵 dash was negative 𝑏 plus 𝑎 minus one 𝑖. Then pulling out the real components, we’ve got one take away 𝑏 and the imaginary components zero 𝑖 and 𝑎 minus one 𝑖.
So the position vector for point 𝐵 dash simplifies down to one minus 𝑏 plus 𝑎 minus one 𝑖. So let’s make a note of that over here. And now we can do a similar process to find out where point 𝑃 dash is after starting from 𝐺, walking in a straight line to 𝑃, turning 90 degrees counterclockwise this time, and then walking the same distance again in that direction. So let’s go back to the gravestone and walk in a direct line towards a point 𝑃.
And 𝐺𝑃’s real component is the difference between negative one and 𝑎, which is negative one minus 𝑎 or simply negative 𝑎 plus one. And its imaginary component is the difference between zero 𝑖 and 𝑏𝑖, which is zero 𝑖 minus 𝑏𝑖 or simply negative 𝑏𝑖. So we’ve got this expression for vector 𝐺𝑃: negative 𝑎 plus one plus negative 𝑏𝑖 or simply negative 𝑎 plus one minus 𝑏𝑖. Then to perform a 90-degree counterclockwise rotation, we just need to multiply that by 𝑖, which gives us 𝑖 times 𝐺𝑃 is equal to 𝑖 times negative 𝑎 plus one take away 𝑏𝑖.
And multiplying through by 𝑖 gives us negative 𝑎 plus one 𝑖 minus 𝑏𝑖 squared. Well again, 𝑖 squared is negative one. So we’ve got negative 𝑏 times negative one, which is positive 𝑏. Then swapping those round to put the real component, first we’ve got 𝑖 times 𝐺𝑃 is 𝑏 minus 𝑎 plus one 𝑖. And just like with vector 𝐵𝐵 dash, this represents the direction and magnitude of vector 𝑃𝑃 dash. And when we draw that on our diagram, it does march straight through our working out. But it tells us that 𝑃 dash is down here.
And just like before, to work out the position vector of 𝑃 dash I’m going go from 𝑂 to 𝑃 and then from 𝑃 to 𝑃 dash. So the position vector of 𝑃 dash is equal to the position vector of the pine tree plus the vector 𝑃𝑃 dash. So that’s negative one plus zero 𝑖, the position of the pine tree, plus 𝑏 minus 𝑎 plus one 𝑖. Taking the real components, we’ve got negative one plus 𝑏 and the imaginary components zero plus negative 𝑎 plus one 𝑖. So rather than writing negative one plus 𝑏, I’m gonna write that as 𝑏 minus one.
So this is the position vector for 𝑃 dash. And let’s make a note of that over here. Now we only need to find the midpoint between 𝐵 dash and 𝑃 dash to find the treasure. And to do that, we just need to find the mean of these two position vectors. Just add them together and divide by two. And that just means we’ve got to simplify this: one minus 𝑏 plus 𝑎 minus one 𝑖 plus 𝑏 minus one minus 𝑎 plus one 𝑖 all over two. So thinking about the real components first, we’ve just got one minus 𝑏 plus 𝑏 minus one. Well one take away one is nothing and negative 𝑏 plus 𝑏 is nothing. So that simplifies to zero.
And taking the imaginary components second, taking care to remember that this is negative 𝑎 take away one, we’ve got 𝑎 take away 𝑎 is nothing. But this time we’ve got negative one take away another one, which is negative two. So this can be simplified to zero in the real part plus negative two over two 𝑖. And of course negative two over two simplifies to negative one. So that’s zero plus negative 𝑖, which of course is zero minus 𝑖 or just negative 𝑖. So on our map, the treasure is at zero minus 𝑖.
So that’s one unit in the negative direction on the imaginary axis, which means that this distance here is the same as this distance here. But more importantly, it’s completely independent of 𝑎 and 𝑏. It doesn’t matter where we started our little journey. We’ll always end up at zero minus 𝑖. So after a bit of mathematical analysis of the problem, we come up with a new easier method for finding the treasure. It’s on the perpendicular bisector of the line segment between the two trees at a distance of half of that between the trees from the line segment, with the bay tree just to the right as you look at it. Happy Treasure Hunting!