Question

Consider an equilateral triangle having vertices at the points $A\left(\frac{2}{\sqrt{3}}{e}^{i\pi /2}\right),B\left(\frac{2}{\sqrt{3}}{e}^{-i\pi /6}\right)$ and $C\left(\frac{2}{\sqrt{3}}{e}^{-i5\pi /6}\right).$ Let $P$ be any point on its incircle. then $A{P}^2+B{P}^2+C{P}^2=$

Answer 1

Let,
$z_1=\frac{2}{\sqrt{3}}{e}^{i\pi /2},z_2=\frac{2}{\sqrt{3}}{e}^{-i\pi /6},z_3=\frac{2}{\sqrt{3}}{e}^{-i5\pi /6}$
Clearly, the points lie on the circle $|z|=\frac{2}{\sqrt{3}}$ and $△ABC$ is equilateral triangle.
Here, $z_1+z_2+z_3=0\Rightarrow \overline{z_1}+\overline{z_2}+\overline{z_3}=0$

Since triangle is equilateral, hence
inradius $r=OD=\frac{1}{\sqrt{3}}$ unit

Let $P\left(z\right)$ be any point on the incircle.
Now,
$A{P}^2=|z-z_1{|}^2=|z{|}^2+|z_1{|}^2-(z\overline{z_1}+\overline{z}{z}_1)$
Similarly,
$B{P}^2=|z{|}^2+|z_2{|}^2-(z\overline{z_2}+\overline{z}{z}_2)$
$C{P}^2=|z{|}^2+|z_3{|}^2-(z\overline{z_3}+\overline{z}{z}_3)$

$\therefore A{P}^2+B{P}^2+C{P}^2=3|z{|}^2+|z_1{|}^2+|z_2{|}^2+|z_3{|}^2-z(\overline{z_1}+\overline{z_2}+\overline{z_3})-\overline{z}(z_1+z_2+z_3)$
$=5$