Equidistant points on a unit circle (part 2)

Define $y_k$ per the diagram; we have

$x_k^2+y_k^2=2^2$

Now note that by symmetry, $y_k=x_{14-k}$ ; this justifies the first line below:

$\begin{aligned} 2\sum_{k=0}^{14}x_k^2 &= \sum_{k=0}^{14}x_k^2 + \sum_{k=0}^{14}y_k^2 \\ &= \sum_{k=0}^{14} \left( x_k^2 +y_k^2 \right) \\ &= \sum_{k=0}^{14} 4 \\ &= 60 \\ \\ \sum_{k=0}^{14}x_k^2 &= 30 \end{aligned}$

and since $x_0=0$ , this is in fact the required sum.

Relevant wiki: Cosine Rule (Law of Cosines)

By cosine rule, $x_k^2 = 1^2 + 1^2 - 2\cos \dfrac {k\pi}{14}$ . Then,

$\begin{aligned} S & = \sum_{k=1}^{14} x_k^2 \\ & = \sum_{k=1}^{14} \left(2 - 2\cos \frac {k\pi}{14} \right) \\ & = \sum_{k=1}^{14} 2 - 2 \sum_{k=1}^{14} \cos \frac {k\pi}{14} \\ & = 28 - 2\left(\sum_{k=1}^6 \cos \frac {k\pi}{14} + \cos \frac {7\pi}{14} + \sum_{k=8}^{13} \cos \frac {k\pi}{14} + \cos \frac {14\pi}{14} \right) \\ & = 28 - 2\left(\sum_{k=1}^6 \cos \frac {k\pi}{14} + \cos \frac \pi 2\ {\color{#3D99F6} + \sum_{k=1}^6 \cos \left(\pi - \frac {k\pi}{14}\right)} + \cos \pi \right) & \small \color{#3D99F6} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ & = 28 - 2\left(\cancel{\sum_{k=1}^6 \cos \frac {k\pi}{14}} + 0\ \cancel{{\color{#3D99F6} - \sum_{k=1}^6 \cos \frac {k\pi}{14}}} -1 \right) \\ & = 28 - 2(-1) = 30 \end{aligned}$

Therefore, $\lfloor 100S\rfloor = \boxed{3000}$ .

Let $A=(0,0)$ and the 28 points on the unit circle, each of these is a roots of $x^{28}=1$ . This means that each point can be represented by a complex number $w_k = \cos\frac{k\pi}{14} + i\sin\frac{k\pi}{14}$ , where $k=0,1,2, \ldots, 27$ . The sum of roots is $\displaystyle\sum_{k=0}^{27} w_k=0$ .

We may use the idea of vector. Now $x_k =\left|1-w_k\right|$ and hence $x_k^2 = \left|1-w_k\right|^2 = \left(1-w_k\right) \cdot \left(1-w_k\right) = \left(1\cdot 1\right) -2\left(1\cdot w_k\right) + \left(w_k\cdot w_k\right) = 2 - 2\left(1\cdot w_k\right)$

Note that $\sum_{k=0}^{27} x_k^2 = \sum_{k=0}^{27} \left(2 - 2\left(1\cdot w_k\right)\right) = 56 - 2 \left( 1\cdot \sum_{k=0}^{27} w_k \right) = 56$

By symmetry, $\displaystyle S=\sum_{k=1}^{13} x_k^2= \sum_{k=15}^{27} x_k^2$ . It is clear that $x_{14}^2 = 2^2=4$ as $x_{14}$ is the length of a diameter. Now $\displaystyle 56=\sum_{k=0}^{27} x_k^2 =\sum_{k=1}^{27} x_k^2 = \left(\sum_{k=1}^{13} x_k^2 \right)+x_{14}^2 + \left(\sum_{k=15}^{27} x_k^2 \right) = 2S+4$ which implies that $S=26$ .

So $\displaystyle \sum_{k=1}^{14} x_k^2 = 26+4=30$ and hence $\displaystyle\left\lfloor 100\sum_{k=1}^{14} x_k^2 \right\rfloor =3000$ .