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i know $ |z_1 -z_2 |^2 = |z_1 |^2 + |z_2 |^2 - 2|z_1 | |z_2 | \cos \alpha $ where $ \alpha $ is angle between $z_1$ and $z_2 $. Similarly i get $ |z_1 -z_2 |^2 + |z_2 -z_3 |^2 + |z_3 -z_1 |^2 = 2(4+9+16) - 2(6 \cos \alpha+8 \cos \beta+12\cos \gamma)$ What will be minimum value of $ 6 \cos \alpha+8 \cos \beta+12\cos \gamma $ if $ z_1 , z_2, z_3 $ lie on the 3 given circles

Han de Bruijn
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raj
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  • Can we do this as following:

    $|z_1 -z_2|^2$ is a non negative real.

    We know that $(a+b+c)^2 \ge \sum_{cyc}a^2$ (for non neative reals) Thus, $\sum_{cyc} |z_1 -z_2|^2 \le (\sum_{cyc}|z_1-z_2|)^2 \ge (2\sum_{cyc} |z_1|)^2 = {18}^2= 324 $

     – Ranojoy Dutta May 10 '16 at 09:47
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    The question could also be reworded / translated into geometry: find maximum sum of squared sides of a triangle whose vertices are on the 3 circles. – Heimdall May 10 '16 at 10:20
  • Or look for minimum value of $6\cos\alpha+8\cos \beta+12\cos\gamma$ where $\alpha+\beta+\gamma=0$ (or $2\pi$, or one angle is the sum of other two - these are equivalent conditions for the problem). – Heimdall May 10 '16 at 10:27
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    The maximum is $87$, see answers of this question. – achille hui May 13 '16 at 16:35
  • @achillehui good job this is an exact dupe, should be closed and you should get the bounty, Post yours as the answer. – shai horowitz May 19 '16 at 03:04

1 Answers1

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$$\sum_{cyc}\|z_1-z_2\|^2 = 3\left(\|z_1\|^2+\|z_2\|^2+\|z_3\|^2\right)-\|z_1+z_2+z_3\|^3 \leq \color{red}{87}$$ hence we just need to prove that there is some triangle $ABC$ with centroid $G$ such that $AG=2,BG=3,CG=4$ to prove that the previous inequality is tight.

But $AG^2=4,BG^2=9,CG^2=16$ give the squared lengths of the medians, then the squared lengths of the sides of $ABC$, so here it is our maximal triangle:

enter image description here

Jack D'Aurizio
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