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!
 A: There probably is an easy, intuitive way to solve part (a), but apparently neither of us has found it. Therefore we try the straightforward but tedious ways: analytic geometry for the intersection of the line and the $x$-axis, and real- and imaginary-parts of the complex numbers for the expression.
Let $p=a+bi$ and $q=c+di$.
Intersection of the line through $p$ and $q$ and the $x$-axis: the slope of the line through $p$ and $q$ is $m=\frac{d-b}{c-a}$. The point-slope form of the line through $p$ with that slope is then
$$y-b=\frac{d-b}{c-a}(x-a)$$
Setting $y=0$ and solving for $x$ we get
$$x=\frac{ad-bc}{d-b}$$
Thus our formula for $z_1$, the complex number at the intersection with the $x$-axis, is
$$z_1=\frac{ad-bc}{d-b}$$
Simplifying the expression: Leaving out some of the tedious simplification steps, we get
$$\begin{align}
z_2 &= \frac{p+q}{pq+1} \\[1em]
 & = \frac{a+bi+c+di}{(a+bi)(c+di)+1} \\[1em]
 & = \frac{(a+c)+(b+d)i}{(ac-bd+1)+(ad+bc)i} \\[1em]
 & = \frac{[(a+c)+(b+d)i]\cdot[(ac-bd+1)-(ad+bc)i]}
    {[(ac-bd+1)+(ad+bc)i]\cdot[(ac-bd+1)-(ad+bc)i]} \\[1em]
 & = \frac{[c(a^2+b^2)+a(c^2+d^2)+a+c]+[-d(a^2+b^2)-b(c^2+d^2)+b+d]i}
    {(a^2+b^2)(c^2+d^2)+2(ac-bd)+1}
\end{align}$$
Since $p$ and $q$ are on the unit circle, $a^2+b^2=1$ and $c^2+d^2=1$. Substituting those we get
$$\begin{align}
z_2 &= \frac{[c+a+a+c]+[-d-b+b+d]i}{1+2(ac-bd)+1} \\[1em]
 &= \frac{a+c}{ac-bd+1}
\end{align}$$
Now we check if those expressions for $z_1$ and $z_2$ are equal.
$$\frac{ad-bc}{d-b} \stackrel{?}{=} \frac{a+c}{ac-bd+1}$$
Cross-multiplying,
$$ad-bc+cd(a^2+b^2)-ab(c^2+d^2) \stackrel{?}{=} ad-bc+cd-ab$$
Again using $a^2+b^2=1$ and $c^2+d^2=1$, we get equality.
Thus the statement in part (a) is proved.

As for part (b), points $P_1$ and $P_{10}$ are opposite vertices of the regular $18$-gon. Therefore segment $\overline{P_1P_{10}}$ is a diameter of the circumcircle of the $18$-gon.
You could choose the particular regular $18$-gon to be inscribed the unit circle, with $P_1$ at unity. $P_{10}$ would then be at $-1$ and the segment between them is on the $x$-axis. If we let $r$ be the first complex $18$th root of $1$, namely $r=e^{i\pi/9}$, then $P_2$ is at $r$, $P_3$ at $r^2$, $P_4$ at $r^3$, and so on.
The diagonal between $P_2$ and $P_{13}$ then meets the requirements of part (a) and you can use the formula to find where that diagonal intersects the $x$-axis, namely at $\frac{r+r^{12}}{r^{13}+1}$. You can do the same for the diagonal between $P_3$ and $P_{15}$ and get $\frac{r^2+r^{14}}{r^{16}+1}$. Show that those two $x$-coordinates are the same, and part (b) is done.
A: let a point on the line segment $pq$ be $$z = tp + (1-t)q, t \text{ a real number.} \tag 1$$  since $p, q$ are on the unit circle, we can write $$p = e^{ia}, q = e^{ib}\tag 2$$
requiring $z$ to be real in $(1),$ gives us $$ t\sin a+(1-t)\sin b = 0 \to t = \frac{\sin b}{\sin b - \sin a}$$ and 
$$\begin{align} z &= \frac{p\sin b - q\sin a }{\sin b - \sin a}\\
&=\frac{p(q-\bar q) -q(p-\bar p)}{q-\bar q -p + \bar p}\\
&=\frac{q \bar p - p \bar q}{q - \bar q - p + \bar p}\\
&=\frac{(q \bar p - p \bar q)pq}{(q - \bar q - p + \bar p)pq}
=\frac{q^2-p^2}{pq^2-p-p^2q+q}\\
&=\frac{(q-p)(q+p)}{(1+pq)(q-p)}\\
&=\frac{p+q}{1+pq}
\end{align}$$ 
for the second part, we have $P_1, P_{10}$ are diametrically opposite. therefore the sum is $0$  and they intersect at the origin. you can do the rest.
A: (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.
A: here is geometric proof of part(a). let us call the points $0, 1, -1, p, q, pq$ on the unit circle by $O, A, B, P, Q, R$  and the  midpoint $\frac 12(p+q)$ by $S.$ we note that $$BR = R - B = pq-(-1) = pq + 1, OS = \frac 12 (p+q)$$
it is easy to verify that $$\angle AOS = \angle OBR = \frac12(arg (p) + arg (q))\quad i.e. BR \parallel OS \tag 1$$ because $\angle QOR = \angle AOP = arg (p).$
let the line cut the real axis at $X.$ we have $$|OS| = \cos\angle POS, |OX| = \frac{cos\angle POS}{\cos\angle AOS},\\
|RB| = 2\cos \angle OBR = 2 \cos\angle AOS$$  therefore $$ |OX| = \frac{2|OS|}{|RB|} = \frac{2OS}{RB} = \frac{p+q}{1+pq} $$
