Someone has proposed to me a problem that I am trying to solve and I invite others to solve it.
Find the locus described by the points such that the sum of the squares of their distances to each of the vertices of a regular polygon of $n$ sides is equal to a fixed square $m^2$.
NB- The integer $m$ obviously must be greater than a certain constant and the solution involves some trigonometric identities that might appear as difficult.
Let the polygon be $p_k =(x_k, y_k) =(\cos(2k\pi/n), \sin(2k\pi/n)) =(\cos(kt), \sin(kt)) $, for $k=0$ to $n-1$, $t = 2\pi/n$.
The distance from a point $p = (x, y)$ to $p_k$ is
$\begin{array}\\ d_k &=\sqrt{|p-p_k|}\\ &=\sqrt{(x-\cos(kt))^2+(y-\sin(kt))^2}\\ &=\sqrt{x^2-2x\cos(kt)+\cos^2(kt)+y^2-2y\sin(kt)+\sin^2(kt)}\\ &=\sqrt{x^2+y^2-2(x\cos(kt)-y\sin(kt))}\\ \end{array} $
If $p = r(\cos(u), \sin(u)) $, then
$\begin{array}\\ d_k &=\sqrt{x^2+y^2-2(x\cos(kt)+y\sin(kt))}\\ &=\sqrt{r^2-2(r\cos(u)\cos(kt)+r\sin(u)\sin(kt))}\\ &=\sqrt{r^2-2r\cos(u-kt)}\\ \text{so}\\ d_k^2 &=r^2-2r\cos(u-kt)\\ \end{array} $
Summing
$\begin{array}\\ D &=\sum_{k=0}^{n-1} d_k\\ &=\sum_{k=0}^{n-1}(r^2-2r\cos(u-kt))\\ &=nr^2-2r\sum_{k=0}^{n-1}\cos(u-kt)\\ &=nr^2-2r\sum_{k=0}^{n-1}Re(e^{i(u-kt)})\\ &=nr^2-2rRe\sum_{k=0}^{n-1}e^{i(u-kt)}\\ &=nr^2-2rRe(e^{iu}\sum_{k=0}^{n-1}e^{-ikt})\\ &=nr^2-2rRe\left(e^{iu}\dfrac{1-e^{-int}}{1-e^{-it}}\right)\\ &=nr^2-2rRe\left(e^{iu}\dfrac{1-\cos(-nt)-i\sin(-nt)}{1-\cos(-t)-i\sin(-t)}\right)\\ &=nr^2-2rRe\left(e^{iu}\dfrac{1-\cos(nt)+i\sin(nt)}{1-\cos(t)+i\sin(t)}\right)\\ &=nr^2-2rRe\left(e^{iu}\dfrac{1-\cos(nt)+i\sin(nt)}{1-\cos(t)+i\sin(t)}\dfrac{1-\cos(t)-i\sin(t)}{1-\cos(t)-i\sin(t)}\right)\\ &=nr^2-2rRe\left(e^{iu} \dfrac{(1-\cos(nt))(1-\cos(t))+\sin(nt)\sin(t)+i(\sin(nt)(1-\cos(t))-\sin(t)(1-\cos(nt)}{(1-\cos(t))^2+\sin^2(t)}\right)\\ &=nr^2-2rRe\left(e^{iu} \dfrac{1-\cos(nt)-\cos(t)+\cos(nt)\cos(t)+\sin(nt)\sin(t)+i(\sin(nt)-\sin(nt)\cos(t))-\sin(t)+\sin(t)\cos(nt)}{2-2\cos(t)}\right)\\ &=nr^2-2rRe\left(e^{iu} \dfrac{a(t)+ib(t)}{2-2\cos(t)}\right) \qquad\text{(getting lazy here)}\\ &=nr^2-2rRe\left( \dfrac{(\cos(u)+i\sin(u))(a(t)+ib(t))}{2-2\cos(t)}\right)\\ &=nr^2-2rRe\left( \dfrac{(\cos(u)a(t)-\sin(u)b(t))+i(\cos(u)b(t)+\sin(u)a(t))}{2-2\cos(t)}\right)\\ &=nr^2-r\left( \dfrac{\cos(u)a(t)-\sin(u)b(t)}{1-\cos(t)}\right)\\ &=nr^2-r\left( \dfrac{\cos(u)v\cos(c)-\sin(u)v\sin(c)}{1-\cos(t)}\right) \qquad\text{if } (a(t), b(t))=v(\cos(c), \sin(c))\\ &=nr^2-rv\left( \dfrac{\cos(u+c)}{2\sin^2(t/2)}\right) \end{array} $
Therefore, we want $m^2 =nr^2-rv\left( \dfrac{\cos(u+c)}{2\sin^2(t/2)}\right) $.
So, find $v$ and $u$ which satisfies this.
Your turn.