Question

If the complex numbers associated with the vertices $A,B,C$ of $△ABC$ are $e^{i\theta },\omega ,\overline{\omega }$ respectively [where $\omega ,\overline{\omega }$ are the complex cube roots of unity and $cos\theta >Re\left(\omega \right)$], then the complex number of the point where angle bisector of $A$ meets the circumcircle of the triangle is

• A. $e^{i\theta }+e^{-i\theta }$
• B. $e^{-i\theta }$
• C. $\omega \overline{\omega }$
• D. $\omega +\overline{\omega }$

Answer 1

The correct option is D $\omega +\overline{\omega }$

The three vertices are equidistant from the origin. Hence, the circumcentre of the triangle is the origin.
$B$ and $C$ represent $\omega$ and $\overline{\omega }$
Hence, $BC$ subtends a central angle of $\frac{2\pi }{3}$.
Hence, $\angle BAC=\frac{\pi }{3}$
Let $AD$ be the angle bisector of $\angle A$, where D lies on the circumcircle of $\Delta ABC$.
$\angle DAC=\angle DAB=\frac{\pi }{6}$
Consider the angles subtended on the minor arc DB, $\angle DAB=\angle DCB=\frac{\pi }{6}$
Similarly, $\angle DAC=\angle DBC=\frac{\pi }{6}$
Consider $\Delta BCD$,
$\angle DBC=\angle DCB$
Hence, the triangle is isosceles.
Hence, D lies on the perpendicular bisector of $BC$. Hence, D lies on the x-axis.
On the complex plane, the coordinates of D are $(-1,0)$.
Hence, option D is correct.