Treasure hunt using complex numbers & an inequalityby Verdict Tags: complex, hunt, inequality, numbers, treasure 

#1
Feb1913, 01:52 PM

P: 114

1. The problem statement, all variables and given/known data
Question 1: You find an old map revealing a treasure hidden on a small island. The treasure was buried in the following way: the island has one tree and two rocks, one small one and one large one. Walk from the tree to the small rock, turn 90 to the left and walk the same distance in that direction. Mark the point where you end up by A. Next, do the same for the big rock, but here turn 90 to the right. This way you end up at a point that is marked B. The treasure is hidden halfway the line AB. You go to the island to find out that the tree no longer exists. How are you going to find the treasure? Question 2: 2. Relevant equations For the first question, I can't really think of any relevant equations. The question is part of a course in complex analysis, so either using the vector representation of complex numbers, a + ib, or the polar coordinates, re^{i[itex]\theta[/itex]} are bound to be in there somewhere, but how I don't see straight away. For the second question, again from the complex analysis course, the most important formula that I can think of is the triangle inequality, z_{1}+z_{2}[itex]\leq[/itex]z_{1}+z_{2} 3. The attempt at a solution Alright, the first question has me most puzzled. After drawing a few random situations, it seems as if the distance from rock one to the middle of AB is the same length as the distance of rock 2 to that same middle. This is all from just drawing mind you, so in no way proven. It also seems as if you walk to the middle of the two rocks and turn 90 degrees, you could dig a trench either down or up, until you find the treasure. Doesn't sound like a solid plan, either. Apart from this I don't really know where to begin. I've also had a lot of trouble approaching this issue not from a geometric side but from a point where I utilize complex functions instead. Could anyone give me a hint on where to start? The second question I have gotten a bit further, but I somehow feel like either I did something wrong (plausible) or the equality sign in the question is the wrong way around (less plausible, but still possible). Here's what I did: I apologize for the images instead of latex, I should and will learn how to use it some time soon. 



#2
Feb1913, 02:46 PM

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I like question 1 :). It is a puzzle with a surprising result, and complex numbers really help.
Hint: Begin with a coordinate system (for complex numbers). A clever choice of the origin will help to find the coordinates of A and B in a general way. Afterwards, you have to get rid of those coordinates. A general tip for inequalities in complex numbers: Try to avoid them with their components. Usually, this just generates unnecessary complexity. I think I would multiply the equation with those denominators and try to use the triangle equality afterwards. Not sure if that helps, but it would by my first try. Edit: On a second thought, you have to include z1+z2 in any inequality in some way. 



#3
Feb1913, 03:03 PM

P: 114

Hmm. As for coordinate systems, the only ones I know are the Complex (euclidean) plane and the polar representation. I suppose since there isn't a lot of multiplication going on, the euclidean seems like the natural way to go. As for the origin, the only sensible point seems to be the middle between the two rocks. Again, from here on I somehow still start to think in terms of triangles and sides or angles that are equal, which is really not the way to go..
As for the inequality, could you possibly elaborate on that? I could add that they are both smaller than 2 times z1+z2 at the end of the line, but apart from that I don't really see where to go. 



#4
Feb1913, 03:21 PM

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Treasure hunt using complex numbers & an inequalityConcerning the inequality: Hmm, maybe polar coordinates can be useful here. I don't know. 



#5
Feb1913, 03:26 PM

P: 114

Writing it all out in polar coordinates did not do too much for me, as I don't know how to express z1 + z2 in them, other than just replacing them with their polar versions, is there anything more I can do? I suppose picking the treasure as the origin has some merit, as the points a and be will be on opposite sides of a circle with the origin at its center. Which probably isn't the correct way to go either, as that suggests polar again. Meh. 



#6
Feb1913, 04:38 PM

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z1+z2 does not look nice, of course, but the other parts do.
There is another choice for the origin ;). 



#7
Feb1913, 05:24 PM

P: 114

Another one? I really can't think of anything else than just the tree at this point.
For the inequality, the other parts do indeed look nicer. I get that (r_{1}+r_{2}) multiplied with the absolute value of the angles, which is a maximum of 2, is smaller than or equal to z1+z2. Is that what you had in mind? 



#8
Feb1913, 05:34 PM

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I've been playing around with the inequality but haven't found a proof yet. However, I did notice an interesting fact which may prove useful. If you manipulate the left hand side a bit, you can get it equal to this expression:
$$\leftz_1\left(1 + \frac{z_2}{z_1}\right) + z_2 \left(1 + \frac{z_1}{z_2}\right)\right$$ If we put $$r_1 = 1 + \frac{z_2}{z_1}$$ and $$r_2 = 1 + \frac{z_1}{z_2}$$ note that ##r_1## and ##r_2## have the interesting property that ##r_1 r_2 = r_1 + r_2##, or equivalently, $$\frac{1}{r_1} + \frac{1}{r_2} = 1$$ which is reminiscent of conjugate exponents in ##L^p## spaces. 



#9
Feb1913, 05:40 PM

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Also, note the geometric interpretation:
$$m u_1 + u_2 \leq z_1 + z_2$$ where $$m = \frac{z_1+z_2}{2}$$ is the average of the lengths of ##z_1## and ##z_2##, and ##u_1## and ##u_2## are unitlength versions of ##z_1## and ##z_2##. 



#10
Feb1913, 05:49 PM

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@jbunniii: An interesting way to express the lefthanded side. 



#11
Feb1913, 05:51 PM

P: 114

Hmm, I don't really see the geometric interpretation. That the average of the two lengths is smaller than or equal to the sum of the lengths combined?
I thank you all for your great help, I will go to bed now and continue tomorrow! Edit: Hah, thanks mfb, I guess I wasn't really playing ball with that hint.. So I use Cartesian coordinates (complex though) with the tree as origin. How do I continue from there on though? Do I just try to write the steps you take initially (when the tree is still there) as vectors? 



#12
Feb2013, 05:28 AM

P: 114

The professor has given me the following hint regarding the inequality, which gets me even more confused:
if you take two complex numbers z_1=r_1e^{i\theta} and r_2e^{i\theta}, the l.h.s. is zero, but the r.h.s. is r_1r_2. Now first of all, I don't see how this is general, as you pick the arguments of the z's to be opposite. on top of that, the lhs does not reduce to zero for me, but to (r1+r2)*2cos theta . I don't get the minus sign on the RHS either. It really just made things worse. 



#13
Feb2013, 08:00 AM

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Ah well, let's simplify the inequality a bit. It is invariant with respect to a common phase of z1 and z2: If you multiply both with ##e^{i \theta}##, both sides of the inequality stay the same.
Therefore, without loss of generality, we can assume that the sum of their arguments is zero. The left side will not be zero (in general), but we get closer to a geometric interpretation. 



#14
Feb2013, 09:37 AM

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While playing around with the inequality yesterday, I made the following simplifications:
The goal is to prove that $$(z_1 + z_2) \left\frac{z_1}{z_1} + \frac{z_2}{z_2}\right \leq 2z_1 + z_2$$ The denominators are not even defined if ##z_1 = 0## or ##z_2 = 0##, so we assume that both are intended to be nonzero. Let us write ##z_1 = z_1e^{i\theta_1}## and ##z_2 = z_2e^{i\theta_2}##. Without loss of generality, we may rigidly rotate the complex plane by the angle ##\theta_1##, and then scale it by ##z_1##, to obtain new variables $$u_1 = \frac{z_1}{z_1}e^{i(\theta_1  \theta_1)} = 1$$ and $$u_2 = \frac{z_2}{z_1}e^{i(\theta_2  \theta_1)}$$ Then the stated inequality will be true if and only if this equivalent inequality holds: $$(1 + u_2) \left 1 + \frac{u_2}{u_2}\right \leq 2 1 + u_2$$ We are free to assume either ##u_2 \leq 1## or ##u_2 \geq 1##, whichever makes this work out properly. This corresponds to whether the original unscaled numbers satisfied ##z_2 \leq z_1## or ##z_2 \geq z_1##, respectively. I still didn't manage to prove it, but perhaps you'll find the above useful. 



#15
Feb2013, 09:51 AM

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Can't you just make use of
[tex] \left z_1 + z_2 \right \leq \left z_1 \right + \left z_2 \right [/tex] in the LHS of the inequality? 



#16
Feb2013, 09:51 AM

P: 114

Hmm. The inequality is not really going anywhere for me. The simplifications / changes you make I do understand, but indeed, it does not give the proof just yet.
What I have done for the treasure now, is write out the coordinates of the interesting points, as follows: The origin (0,0) = 0 + 0i = the tree = z1 Rock 1 = z2 = (a,b) = a + ib Rock 2 = z3 = (c, d) = c + id Point A = z4 = z2 plus z2 turned by 90 degrees to the left = (a + ib) + (b + ia) = (ab, b+a) Point B = z5 = z3 plus z3 turned by 90 degrees to the right = (c + id) + (d  ic) = (c+d, dc) The treasure = z6 = in the middle of point A and B = (ab+c+d / 2, b+a+dc /2) = (ab+c+d)/2 + i(b+a+dc)/2 Is this useful, or is it just redundant? I understand that I don't have these coordinates when looking for the treasure, but something about the distances between them or whatever might be extractable from the situation.. 



#17
Feb2013, 09:52 AM

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Since ##u_2/u_2## is a unit vector,
$$\left 1 + \frac{u_2}{u_2} \right$$ cannot be larger than 2. Unfortunately, it is not true that ##(1 + u_2) \leq 1 + u_2##; indeed, the inequality goes in the opposite direction. So obviously we have to combine the two factors on the left side and rewrite them in a clever way if we are going to succeed. I didn't manage to find the trick yet. 


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