Problem source

Evaluate $\displaystyle \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)}$ where $\alpha$ is a fixed angle and $n\ge2$. The fixed point $O$ is a distance $d$ from a fixed line $D$.
For any point $P$, let $s$ be the distance from $P$ to $D$ and let $r$ be the distance from $P$ to $O$. Write down an expression for $s$ in terms of $d$, $r$ and the angle $\theta$, where $\theta$ is as shown in the diagram below.
\begin{center}
\begin{tikzpicture}
    % Clip to the original pspicture dimensions
    \clip (-6.44, -2.23) rectangle (2.5, 6.7);
    % Bézier Curve
    \draw (-5.93, 5.21) .. controls (1.07, 4.02) and (1.05, 0.57) .. (-1.07, -1.78);
    % Vertical line (Line D)
    \draw (2, 6) -- (2, -2);
    % Horizontal line from D to P
    \draw (2, 4) -- (-1.99, 3.99);
    % Diagonal line from P to O
    \draw (-1.99, 3.99) -- (-5, 0);
    % Horizontal line from O to D
    \draw (-5, 0) -- (2, 0);
    % Dimension line with arrows at the bottom
    \draw[<->] (-5.02, -0.25) -- (2, -0.25);
    % Angle arc (theta) at point O (-5, 0)
    % PSTricks param 0.9245 rad is approx 52.97 degrees
    \draw[line width=0.1pt] (-4, 0) arc (0:52.97:1) -- (-5, 0) -- cycle;
    % Labels (using anchor=north west to match rput[tl])
    \node[anchor=north west] at (-3.7, 2.48) {$r$};
    \node[anchor=north west] at (-5.5, 0.05) {$O$};
    \node[anchor=north west] at (-4.5, 0.5) {$\theta$};
    \node[anchor=north west] at (-6.33, 5.61) {$E$};
    \node[anchor=north west] at (-1.7, -0.38) {$d$};
    \node[anchor=north west] at (0.01, 4.45) {$s$};
    \node[anchor=north west] at (-2.15, 4.55) {$P$};
    \node[anchor=north west] at (1.84, 6.66) {$D$};
\end{tikzpicture}
\end{center}
 The curve $E$ shown in the diagram is such that, for any point $P$ on $E$, the relation $r = k s$ holds, where $k$ is a fixed number with $0< k <1$. 
Each of the $n$ lines $L_1$, $\ldots\,$, $L_n$ passes through $O$ and the angle between adjacent lines is $\frac \pi n$. The line $L_j$ ($j=1$,  $\ldots\,$, $n$) intersects $E$ in two points forming a chord of length $l_j$.   Show that, for $n\ge2$, 
\[
\sum_{j=1}^n \frac 1 {l_j} = \frac {(2-k^2)n} {4kd}\,.
\]

Solution source

\begin{align*}
\sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)} &= e^{2i\alpha} \sum_{r=0}^{n-1} \left (\e^{2i\pi/n} \right)^r \\
&= e^{2i\alpha} \frac{1-\left (\e^{2i\pi/n} \right)^n}{1-\e^{2i\pi/n} } \\
&= 0
\end{align*}

$d = s + r \cos \theta$ ie $s = d - r \cos \theta$ 

Therefore $d = \frac{r}{k} + r \cos \theta \Rightarrow r = \frac{kd}{1+k \cos \theta}$. The $l_j$ will come from $r(\alpha + \frac{j \pi}{n} )+r(\alpha + \pi + \frac{j \pi}{n} )$ 

\begin{align*}
&& l_j &= r(\alpha + \frac{(j-1) \pi}{n} )+r(\alpha + \pi + \frac{(j-1) \pi}{n} ) \\
&&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1+k \cos \left ( \alpha+\pi+ \frac{(j-1) \pi}{n}\right)}\\
&&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1-k \cos \left ( \alpha+ \frac{(j-1) \pi}{n}\right)}\\
&&&= \frac{2kd}{1-k^2 \cos^2 \left ( \alpha + \frac{(j-1) \pi}{n}\right)}\\
\Rightarrow && \sum_{j=1}^n \frac 1 {l_j} &=  \sum_{j=0}^{n-1} \frac{1-k^2 \cos^2 \left ( \alpha + \frac{j \pi}{n}\right)}{2kd} \\
&&&=  \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1}  \cos^2 \left ( \alpha + \frac{j \pi}{n}\right) \\
&&&=  \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \frac{1+ \cos \left ( 2\alpha + \frac{2j \pi}{n}\right)}{2} \\
&&&=  \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \sum_{j=0}^{n-1}\cos \left ( 2\alpha + \frac{2j \pi}{n}\right) \\
&&&=  \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \underbrace{\textrm{Re} \left ( \sum_{j=0}^{n-1}e^{  2i(\alpha + \frac{j \pi}{n})} \right)}_{=0} \\
&&&= \frac{n}{2kd} - \frac{nk^2}{4kd} \\
&&&= \frac{n(2-k^2)}{4kd}
\end{align*}