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Question Statement:- Given that $\displaystyle\sum_{i=1}^{4}{b_i}=0$ and $\displaystyle\sum_{i=1}^{4}{b_iz_i}=0$, where $b_i$'s are non zero real numbers, no three $z_i$'s form a straight line, prove that the $z_i$'s are concyclic if $$b_1b_2|z_1-z_2|^2=b_3b_4|z_3-z_4|^2$$

Attempt at a solution:-

I was not able to do much to it, so here is my attempt at the solution.

$$\begin{aligned} &\displaystyle\sum_{i=1}^{4}{b_iz_i}=0 \implies b_1z_1+b_2z_2+b_3z_3+b_4z_4=0 \\ &\implies b_1z_1+b_3z_3=-(b_2z_2+b_4z_4) \hspace{5cm} \ldots (1) \end{aligned}$$

Also, we have the following relation:- $$\sum_{i=1}^{4}{b_i}=b_1+b_2+b_3+b_4=0 \implies b_1+b_3=-(b_4+b_5) \tag{2}$$

On dividing $(1)$ and $(2)$, we get $$\dfrac{b_1z_1+b_3z_3}{b_1+b_3}=\dfrac{b_2z_2+b_4z_4}{b_2+b_4}$$

This denotes that the point that divides the line joining $z_1$ and $z_3$ in the ratio $b_3:b_1$ and the point that divides $z_2$ and $z_4$ in the ratio $b_4:b_3$ are same but I am not able to conclude anything more.

Also other approaches are welcome.

user350331
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1 Answers1

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The same argument you used also gives \begin{align*} \dfrac{b_1z_1+b_2z_2}{b_1+b_2}=\dfrac{b_3z_3+b_4z_4}{b_3+b_4} \end{align*} Let this point be $P(z)$. This is the point of intersection of line joining $z_1, z_2$ and $z_3, z_4$. We have \begin{align*} z- z_1 = \dfrac{b_1z_1+b_2z_2}{b_1+b_2} - z_1 &= \dfrac{b_2(z_2-z_1)}{b_1+b_2}\\ z- z_2 = \dfrac{b_1z_1+b_2z_2}{b_1+b_2} - z_2 &= \dfrac{b_1(z_1-z_2)}{b_1+b_2}\\ z- z_3 = \dfrac{b_3z_3+b_4z_4}{b_3+b_4} - z_3 &= \dfrac{b_4(z_4-z_3)}{b_3+b_4}\\ z- z_4 = \dfrac{b_3z_3+b_4z_4}{b_3+b_4} - z_4 &= \dfrac{b_3(z_3-z_4)}{b_3+b_4} \end{align*} Thus \begin{align*} (z-z_1)(z-z_2) &= -\dfrac{b_1b_2(z_1-z_2)^2}{(b_1+b_2)^2} \\ (z-z_3)(z-z_4) &= -\dfrac{b_3b_4(z_3-z_4)^2}{(b_3+b_4)^2} \end{align*} From the condition $b_1b_2|z_1-z_2|^2=b_3b_4|z_3-z_4|^2$, it follows that $b_1b_2$ and $b_3b_4$ have the same sign and hence we have \begin{align*} |b_1b_2||z_1-z_2|^2=|b_3b_4||z_3-z_4|^2 \end{align*} We thus have \begin{align*} |z-z_1||z-z_2| &= \left|\dfrac{b_1b_2(z_1-z_2)^2}{(b_1+b_2)^2}\right| \\ &= \left|\dfrac{b_3b_4(z_3-z_4)^2}{(b_3+b_4)^2}\right| \\ &= |z-z_3||z-z_4| \end{align*} If the given points are $A,B,C,D$, then we have $PA \cdot PB = PC\cdot PD$ and hence $A,B,C,D$ are concyclic.