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Treasure hunt using complex numbers & an inequality 
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#37
Feb2013, 04:32 PM

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\lefte^{i\theta_1} + e^{i\theta_2}\right^2 &= \lefte^{i\theta_2} (1 + e^{i(\theta_1  \theta_2)})\right^2 \\ &= \left1 + e^{i(\theta_1  \theta_2)}\right^2 \\ &= \lefte^{i(\theta_1  \theta_2)/2} (e^{i(\theta_1  \theta_2)/2} + e^{i(\theta_1  \theta_2)/2})\right^2 \\ &= \left (e^{i(\theta_1  \theta_2)/2} + e^{i(\theta_1  \theta_2)/2})\right^2 \\ &= \left 2 \cos((\theta_1  \theta_2)/2) \right^2 \\ &= 4 \cos^2((\theta_1  \theta_2)/2) \\ &= 4 \left(\frac{1}{2} + \frac{1}{2} \cos(\theta_1  \theta_2)\right) \\ &= 2 + 2 \cos(\theta_1  \theta_2)\\ \end{align}$$ Probably you can find a way to consolidate a few of the steps to make it shorter. 


#38
Feb2013, 04:41 PM

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Hi Verdict!
It would be a lot easier to put the two rocks at ±1, and the tree at a general point z 


#39
Feb2013, 04:51 PM

P: 114

I wrote out all the multiplications, and you are left with two terms on the left that are smaller than or equal to two terms on the right, which is nice, but a third term on the left that is bigger than or equal to the term on the right. So that isn't as nice as I had hoped it to be, and it is of course not the same as what you reduced it to. Do you use more identities to get there? @tinytim Hey, thanks! I did manage to work it out in the end (it's somewhere in this heap of posts, haha), and yeah it is a lot harder than it could have been, sadly. Ah well, if it works, it works. The digging twice part is a nuisance, so it is indeed a lesser answer, but the question didn't technically require me to only dig once, so I guess it's not that bad. 


#40
Feb2013, 05:08 PM

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#41
Feb2013, 05:16 PM

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$$\begin{align} e^{i\theta_1} + e^{i\theta_2}^2 &= \overline{(e^{i\theta_1} + e^{i\theta_2})} (e^{i\theta_1} + e^{i\theta_2}) \\ &= (e^{i\theta_1} + e^{i\theta_2}) (e^{i\theta_1} + e^{i\theta_2}) \\ &= 2 + e^{i(\theta_1  \theta_2)} + e^{i(\theta_1  \theta_2)} \\ &= 2 + 2\cos(\theta_1  \theta_2) \end{align}$$ 


#42
Feb2013, 05:42 PM

P: 114

Alright, so now I have
4r_{1}r_{2} + 2r_{1}^{2}cos(θ_{1}θ_{2}) + 2r_{2}^{2}cos(θ_{1}θ_{2}) is smaller than or equal to 2r_{2}^{2} + 2r_{1}^{2} +4 r_{1}r_{2}cos(θ_{1}θ_{2}) So I see that on the LHS the two terms with the cosine are, smaller than or equal to the two corresponding ones on the RHS. However, the cosine on the RHS is smaller than or equal to the term on the left side. Oh, and one small question: When you do z1+z2², you write (z1+z2)(z1*+z2*). But isn't that just z1+z2, without the square? 


#43
Feb2013, 06:15 PM

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$$\begin{align} 4r_1 r_2 + 2 r_1^2 \cos(\theta_1  \theta_2) + 2r_2^2\cos(\theta_1  \theta_2) \leq 2r_2^2 + 2r_1^2 + 4r_1 r_2 \cos(\theta_1  \theta_2)\\ \end{align}$$ Moving everything to one side of the inequality, we get $$\begin{align} 0 &\leq 2r_1^2 (1  \cos(\theta_1  \theta_2)) + 2r_2^2 (1  \cos(\theta_1  \theta_2)) + 4r_1 r_2 ( \cos(\theta_1  \theta_2)  1) \\ &\leq (2r_1^2 + 2r_2^2  4r_1 r_2) (1  \cos(\theta_1  \theta_2)) \\ \end{align}$$ Now it should be pretty easy to argue that both factors on the right hand side are nonnegative. 


#44
Feb2013, 06:27 PM

P: 114

Alright, so the part with the cosine is obvious indeed: a cosine has a maximal value of 1, so at most, the term is zero, never negative. Inequality holds.
For the first part though.. Of course it makes sense, that if you take something squared, plus something smaller squared, that that is bigger than 2 times something times the smaller one.. It sounds like there is a theorem/inequality for this type of thing, I'll look for it. Edit: Ok, that was one of the easier things I have missed so far. Got it, inequality proven, thanks so much! You guys are really great. Also, I made a sneaky edit in my last post: When you do z1+z2², you write (z1+z2)(z1*+z2*). But isn't that just z1+z2, without the square? 


#45
Feb2013, 06:42 PM

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#46
Feb2013, 06:47 PM

P: 114

Again, thanks so much, for all the time and effort! 


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