(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!
Best Answer
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.