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Let $z_1, z_2, z_3$ be complex numbers satisfying the condition $\left|z\right|^2\leq 4\left|z+\overline{z}\right|+33$. Find the maximum value of $$P=\left|z_1-z_2\right|+\left|z_2-z_3\right|+\left|z_3-z_1\right|$$


If $z=a+bi$ then $a^2+b^2\leq 8\left|a\right|+33 \Leftrightarrow \left(\left|a\right|-4\right)^2+b^2\leq 49$. This follows that $z_i$ belongs to the union of two circles $\left(x-4\right)^2+y^2\leq 49$ (when $Re z_i>0$) and $\left(x+4\right)^2+y^2\leq 49$ (when $Re z_i<0$). Let $A,B,C$ be the points responding to $z_1,z_2,z_3$. Then $P=AB+BC+CA$. I have no idea to find the maximum value of $P$. But when I choose $z_1=-8+\sqrt{33}i, z_2=-8-\sqrt{33}i, z_3=11$, I can see $P=2\sqrt{33}+2\sqrt{394}$. I cannot find any other triple $\left(z_1,z_2,z_3\right)$ so that $P$ is greater. Help me some ideas. Thank you.

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  • $\begingroup$ Why $P=AB+BC+CA$? $\endgroup$
    – user
    Commented Apr 14, 2021 at 17:59
  • $\begingroup$ @user Presumably the notation $AB$ is the geometric notation for distance between $A$ and $B$, which is precisely $|z_1-z_2|$. $\endgroup$ Commented Apr 14, 2021 at 18:01
  • $\begingroup$ @user because $|z_1-z_2|$=dist(A,B)="AB". $\endgroup$
    – Jean Marie
    Commented Apr 14, 2021 at 18:01
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    $\begingroup$ @JeanMarie I see. I thought $AB $ stays for product of $A $ and $B $. $\endgroup$
    – user
    Commented Apr 14, 2021 at 18:33
  • $\begingroup$ @JeanMarie yes, I mean $z_2=-8-\sqrt{33}i$. Thank you $\endgroup$
    – Chivul
    Commented Apr 15, 2021 at 0:19

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use $$ z_1=q+\bigg[\sqrt{49-(q+4)^2}\bigg]i, \\z_2=q-\bigg[\sqrt{49-(q+4)^2}\bigg]i,\\ z_3=11 $$ you should find $q_{max} = -8$

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    $\begingroup$ My guess is this is what they did to find their candidate for the maximum. It makes intuitive sense that this should be the maximum, but as I understand it, they're asking for a proof that the maximum perimeter occurs on this 1-dimensional parametrization of triangles. $\endgroup$ Commented Apr 14, 2021 at 18:19

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