Let $A(Z_1)$,$B(Z_2)$,$C(Z_3)$ are three points on the argand plane such that $|Z_1|=|Z_2|=|Z_3|=4$. The image of $P(\large\frac {-Z_2Z_3}{Z_1})$ about the line BC is ?
This was a question in a test. The best I could do was assume $Z_1 = 1$ , $Z_2 = -1$ and $Z_3 = i$. Then calculate P and find it's image. And to my luck only one of the 4 options satisfied. The answer is $Z_1+Z_2+Z_3$ if that helps.
On
Extend the altitude from $A$ to meet the circumcircle at $P'\equiv z$
We have $\angle POC = 2 \angle PAC = \pi-2C = \theta$.
Also $|z| = |z_3|$
Hence $\dfrac{z_3}{z} = e^{i\theta}$
Similarly $\dfrac{z_2}{z_1} = e^{i2C} = e^{\pi-\theta}$
Hence $\dfrac{z_3}{z} \times \dfrac{z_2}{z_1} = e^{\pi}=-1 \Rightarrow z = -\dfrac{z_2z_3}{z_1}$. Thus given point $P=P'$ i.e. its the point of intersection of the altitude from $A$ with the circumcenter.
By a well known result $P$ and the orthocenter $H$ of $\triangle ABC$ are images of each other across $BC$. Hence reflection of $P$ across $BC$ is the orthocenter.
Finally since circumcenter of $\triangle ABC$ is at origin, the orthocenter corresponds to $z_1+z_2+z_3$
Rotate the coordinate system so that $BC$ is horizontal.Let $Z_1=4e^{i\theta_1}$,$Z_2=4e^{i\theta_2}$. From the condition that $BC$ is horizontal, we have $Z_3=4e^{i(\pi-\theta_2)}=-4e^{-i\theta_2}$. Thus, $$P=\frac{-4e^{i\theta_2}\cdot -4e^{-i\theta_2}}{4e^{i\theta_1}}=4e^{-i\theta_1}$$
Now, to find the image of the point we will shift to Cartesian coordinate system. Here, the line $BC$ is represented by $y=4\sin(\theta_2)$. The $X$ component of $P$ will remain the same upon reflection, whereas the $Y$ component becomes $4\sin(\theta_2)-(4\sin(\theta_1)-4\sin(\theta_2))=4(2\sin(\theta_2)-\sin(\theta_1))$. Thus we have determined the required point.