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If A,B,C are points corresponding to $z_1,z_2,z_3$ respectively such that $\vert z_1\vert=\vert z_2\vert=\vert z_3\vert$ and altitude of trainagle $ABC$ through $A$ meets the circumcircle at $P$, then show that $P$ is represented as -$\frac{z_2z_3}{z_1}$

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Initially, I tried using the property that image of the orthocentre about a side, lies on the circumcircle. But finding the orthocentre takes time. So I used some basic geometry.

Using the above diagram,

  1. $$2(\frac{\pi}{2}-B)=\arg\frac{z}{z_2}$$
  2. $$(\frac{\pi}{2}-B)=\arg\frac{z-z_1}{z_2-z_1}$$
  3. $$2(\frac{\pi}{2}-C)=\arg\frac{z_3}{z}$$
  4. $$(\frac{\pi}{2}-C)=\arg\frac{z_3-z_1}{z-z_1}$$
  5. $$A=\pi-(B+C)=\arg\frac{z_3-z_1}{z_2-z_1}$$

Using equation 1 and 2, $$\frac{z}{z_2}=\frac{(z-z_1)^2}{(z_2-z_1)^2}$$ After simplifying, I got $$z_1^2=z_2z$$ Similarly, using 3 and 4, $$z_1^2=z_3z$$ And using 1,3 and 5, $$z_1^2=z_3z_2$$

But I am not able to prove the given relation.

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  • $\begingroup$ Its an equilateral triangle, suppose $A$ is at $90$, then $B$ is at $210$, and $C$ is at $330$ degrees, $P$ is at $270$. Now $z_2z_3$ will be at $-180$, hence -$\frac{z_2z_3}{z_1}$ will be at $-90$, which is in the direction of $P$. Since their modulii are equal, the division will infact give the point itself. This is the intuitive argument. You can put it down mathematically, $\endgroup$ – Shailesh Jan 2 '16 at 1:48
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Note that $$2C = \arg\left(\frac{z_2}{z_1}\right) $$ Combining this with your observation that $2\left(\frac{\pi}{2}-C\right) = \arg\left(\frac{z_3}{z}\right)$, we have \begin{align} 2C + 2\left(\frac{\pi}{2}-C\right) = \arg\left(\frac{z_2}{z_1}\right)+\arg\left(\frac{z_3}{z}\right)&\implies \pi = \arg\left(\frac{z_2z_3}{zz_1}\right)\\ &\implies \frac{z_2z_3}{zz_1} = -C\\ &\implies z = -\frac{1}{C}\frac{z_2z_3}{z_1} \end{align} for some positive constant $C$. Since $|z| = |z_1| = |z_2| = |z_3|$, it follows that $C=1$.

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