Precalculus unit circle with imaginary axis. (a) Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
(b) Let $P_1 P_2 \dotsb P_{18}$ be a regular 18-gon. Show that $P_1 P_{10}$, $P_2 P_{13}$, and $P_3 P_{15}$ are concurrent.
I have gotten nowhere on this problem, but I have a hint:
One of those three segments is more interesting than the other two. Which one, and why? And how can you use that fact to make part (a) relevant?
Any help is appreciated!

(b) Let $\omega = e^{2 \pi i/18}$, a primitive $18^{\text{th}}$ root of unity. Then $\omega^{18} = 1$. Also, $\omega^{18} - 1 = 0$, which factors as
$(\omega^9 - 1)(\omega^9 + 1) = 0.$Since $\omega^9 = e^{2 \pi i/2} = e^{\pi i} = -1 \neq 1$, $\omega$ must satisfy the equation $\omega^9 + 1 = 0$. We can factor this equation as
$(\omega^3 + 1)(\omega^6 - \omega^3 + 1) = 0.$Since $\omega^3 = e^{2 \pi i/6} = e^{\pi i/3} \neq -1$, $\omega$ must satisfy the equation
$\omega^6 - \omega^3 + 1 = 0.$
We place regular 18-gon $P_1 P_2 \dotsb P_{18}$ in the complex plane so that point $P_1$ goes to 1, point $P_2$ goes to $\omega$, and so on.
To prove that all three lines concur, we are going to find the point where the line joining $\omega$ and $\omega^{12}$ intersects the real axis, and the point where the line joining $\omega^2$ and $\omega^{14}$ intersects the real axis. We will show that these points coincide.
We want to compute the point where the line joining $\omega$ and $\omega^{12}$ intersects the real axis. Since $\omega^9 = -1$, we can write $\omega^{12} = -\omega^3$. Then from our formula in part (a) above, the intersection is
$\frac{\omega - \omega^3}{-\omega^4 + 1}.$
Similarly, the point where the line joining $\omega^2$ and $\omega^{14} = -\omega^5$ intersects the real axis is
$\frac{\omega^2 - \omega^5}{-\omega^7 + 1}.$Hence, we have reduced the problem to showing that
$\frac{\omega - \omega^3}{-\omega^4 + 1} = \frac{\omega^2 - \omega^5}{-\omega^7 + 1}.$Cross-multiplying, we get
$\omega^{10} - \omega^8 - \omega^3 + \omega = \omega^9 - \omega^6 - \omega^5 + \omega^2.$Using the fact that $\omega^9 = -1$, this equation simplifies to
$-\omega - \omega^8 - \omega^3 + \omega = -1 - \omega^6 - \omega^5 + \omega^2,$or
$\omega^8 - \omega^6 - \omega^5 + \omega^3 + \omega^2 - 1 = 0.$We can write this equation as
$(\omega^2 - 1) \omega^6 - (\omega^2 - 1) \omega^3 + (\omega^2 - 1) = 0,$which factors as
$(\omega^2 - 1)(\omega^6 - \omega^3 + 1) = 0.$We know that $\omega^6 - \omega^3 + 1 = 0$. Therefore, our two points of intersection with the real axis coincide, which means that all three lines are concurrent.
