Points $P_i$ on an ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$

Question

Consider an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with $O$ as the origin. $n$ points denoted as $P_1,P_2,\cdots$ are taken on the ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ where $i\in(1,n-1)$. Find the value of: $$\sum_{i=1}^n\frac{1}{OP_i^2}$$

I took $P_1$ as $(3,0)$. Let other points be $P_i(3\cos\theta_i,2\sin\theta_i)$.

$$\tan\frac{\pi}{n}=\frac{2}{3}\tan\theta_2$$ $$\tan\frac{2\pi}{n}=\frac{2}{3}\tan\theta_3$$ and so on.

$$OP_1^2=a^2$$ $$OP_2^2=9\cos^2\theta_2+4\sin^2\theta_2=9\cos^2\theta_2\left(1+\tan^2\frac{\pi}{n}\right)$$ $$OP_3^2=9\cos^2\theta_3+4\sin^2\theta_3=9\cos^2\theta_3\left(1+\tan^2\frac{2\pi}{n}\right)$$ and so on.

But I am not able to compute the required sum.

Answer 1

Better to set up a system of two equations: $$ 3\cos\theta_2=r_2\cos(\pi/n) \quad\hbox{and}\quad 2\sin\theta_2=r_2\sin(\pi/n), $$ where $r_2=OP_2$.

From that you get, after eliminating $\theta_2$: $$ {1\over r_2^2}={\cos^2(\pi/n)\over9}+{\sin^2(\pi/n)\over4}, $$ and in general: $$ {1\over r_{k+1}^2}={\cos^2(k\pi/n)\over9}+{\sin^2(k\pi/n)\over4}= {1\over4}-{5\over36}\cos^2{k\pi\over n}. $$ Using $\cos^2\alpha=(1/2)(\cos2\alpha+1)$ we can rewrite that formula as: $$ {1\over r_{k+1}^2}= {13\over72}-{5\over72}\cos{2k\pi\over n}, $$ so that $$ \sum_{k=0}^{n-1}{1\over r_{k+1}^2}= {13\over72}n-{5\over72}\sum_{k=0}^{n-1}\cos{2k\pi\over n}. $$ But the last sum vanishes by symmetry, so the result is $13n/72$.

Answer 2

If we set $P_i=(x_i,y_i)=\left(3\cos\theta_i,2\sin\theta_i\right)$ we have $$\vartheta_0+\frac{\pi(i-1)}{n}=\arctan\frac{y_1}{x_i}=\arctan\left(\frac{2}{3}\tan\theta_i\right)$$ from which: $$ \tan\theta_i = \frac{3}{2}\tan\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) $$ and $$ OP_i^2 = 9\cos^2\theta_i+4\sin^2\theta_i = 4+5\cos^2\theta_i=\frac{9+4 \tan^2\theta_i}{1+\tan^2\theta_i}$$ gives that the wanted sum is: $$ \sum_{i=1}^{n}\frac{1+\tan^2\theta_i}{9+4\tan^2\theta_i}=\sum_{i=1}^{n}\frac{1+\frac{9}{4}\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) }{9+9\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) } $$ or: $$ \frac{1}{36}\sum_{i=1}^{n}\left[4\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) +9\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) \right]$$ that can be computed by exploiting the duplication formulas for $\sin$/$\cos$, De Moivre's formula and geometric series, or simply by noticing that it does not really depend on $\vartheta_0$ by differentiation with respect to $\vartheta_0$: $$\forall\, n,\vartheta_0,\qquad \sum_{i=1}^{n}\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)= \sum_{i=1}^{n}\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)=\color{red}{\frac{n}{2}}\tag{1}$$ gives:

$$ \sum_{i=1}^{n}\frac{1}{OP_i^2} = \color{red}{\frac{13n}{72}}.\tag{2}$$