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Let ABC be an acute-angled triangle. Let the altitudes from the vertices A, B, C meet the circumcircle at P, Q, R whose corresponding complex numbers are $z_1,z_2$ and $z_3$ respectively. If is $\frac{z_3-z_1}{z_2-z_1}$ is imaginary number then find the value of angle A.

My approach is illustrated below but not able to approach

enter image description here

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  • $\begingroup$ Where have you found the problem? What are your thoughts? $\endgroup$
    – user
    May 18, 2020 at 7:00
  • $\begingroup$ I have added the picture where it is given position of $z_1,z_2,z_3$ we need to find $\frac{z_b-z_a}{z_c-z_a}$ but not able to find the angle.$P=z_1,Q=z_2,R=z_3$ $\endgroup$ May 18, 2020 at 7:05
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    $\begingroup$ $\frac{z_3-z_1}{z_2-z_1}$ being imaginary implies that the points $z_2$ and $z_3$ are diametrically opposite. Drawing the figure like that and some constructions will give $A$ as 45 degrees. $\endgroup$
    – Shooter
    May 18, 2020 at 7:47

2 Answers 2

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Let $A',B',C'$ be the intersection points of the altitudes with the circumscribed circle. Then we have: $$ \angle BAC=\frac{\pi-\angle B'A'C'}2. $$ From $$\Re\frac{z_3-z_1}{z_2-z_1}=0$$ we know $$\angle B'A'C'=\frac\pi2.$$

Can you take it from here?

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From the given, we have $\frac{z_3-z_1}{z_2-z_1} = e^{i\frac\pi2}$, or $Arg \left( \frac{z_3-z_1}{z_2-z_1}\right) =\frac\pi2$

\begin{align} \angle BAC & = \angle BAP + \angle CAP \\ & = \angle BQP + \angle CRP \\ & = Arg \left( \frac{z_2-z_1}{z_2-b} \right) + Arg \left( \frac{z_3-c}{z_3-z_1} \right) \\ & = Arg \left( \frac{z_2-z_1}{z_3-z_1} \frac{z_3-c}{z_2-b} \right) \\ & = Arg \left( \frac{z_2-z_1}{z_3-z_1}\right) +Arg\left( \frac{z_3-c}{z_2-b} \right) \\ & = -\frac\pi2 +\angle RXQ \\ & = - \frac\pi2 +(\pi - \angle BAC) \\ \end{align}

which yields $\angle BAC = \frac\pi4$.

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