Csdrhrt

Let $v$ be any vector in the plane, and ${w_i}$ be $n>2$ vectors evenly spaced around the unit circle. Then it seems true that
$$sum_{i=1}^n (vcdot w_i)^2 = k |v|^2$$
where $k$ is a constant independent of $v$. My intuition is that the left-hand side, as a function of the argument $theta$ of $v$, has "too much symmetry" for its low frequency.

Is there some slick proof of this identity, using e.g. the algebraic structure of the $n$ roots of unity?

This is similar to Noble's answer but I wanted to highlight the geometric aspect of the problem a little more. First we note that
$$v cdot w_i = || v || cos(theta) $$
where $theta$ is the angle between $v$ and $w_i$. Since $||v||^2$ is on both sides of the equation, we can assume $||v|| = 1$.

Now consider a unit vector $v$ and a regular polygon with $n$ sides centered at the origin. Connect the vertices of the polygon to the center. $v$ lies between two of the radii. Now going counter clockwise, $v$ makes the following angles with the $w_i:$
$$x, x + frac{2 pi}n, x + 2 cdot frac{2 pi}n, cdots, x + (n-1) cdot frac{2 pi}n $$
for some $x$ as shown in the image at the bottom of the post.

Therefore, the sum that we want is
$$ sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2.$$

Using Euler's formula $cos(t) = frac{e^{it} + e^{-it}}2$, we have

begin{align}
sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2 &= frac{1}4 sum_{k=0}^{n-1} left(e^{2ileft(x + k, frac{2pi}n right)} + e^{-2ileft(x + k, frac{2pi}n right)} + 2right) \
&= frac{n}2 + (e^{2ix} + e^{-2ix}) sum_{k=0}^{n-1} omega^k
end{align}
where $omega$ is a non trivial $n$th root of unity. Then we can easily calculate that
$$ sum_{k=0}^{n-1} omega^k = frac{omega^n-1}{omega - 1} = 0$$
which proves the claim for $k = frac{n}2.$

Since the $w_i$ are evenly spaced, we can say:

$$w_i=cos(phi+itheta)hat i+sin(phi+itheta)hat j$$

Also, since they are evenly spaced around the full circle, the distances between each $w_i$, which is $theta$ radians, times the number of vectors, which is $n$, must be a full circle: That is, $ntheta=2pi$

Now, $vcdot w_i=v_xcos(phi+itheta)+v_ysin(phi+itheta)$, so:

$$(vcdot w_i)^2=v_x^2cos^2(phi+itheta)+v_y^2sin^2(phi+itheta)+2v_xv_ycos(phi+itheta)sin(phi+itheta) \ =v_x^2left(frac{cos(2phi+2itheta)+1}{2}right)+v_y^2left(frac{1-cos(2phi+2itheta)}{2}right)+v_xv_ysin(2phi+2itheta) \ =frac{v_x^2+v_y^2}{2}+cos(2phi+2itheta)left(frac{v_x^2-v_y^2}{2}right)+v_xv_ysin(2phi+2itheta)$$

Now, let's put this into a summation:

$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)left[sum_{i=1}^ncos(2phi+2itheta)right]+v_xv_yleft[sum_{i=1}^nsin(2phi+2itheta)right]$$

Now, I will use $j=sqrt{-1}$. The first summation and second summation are related: They are the real and imaginary part, respectively, of $sum_{i=1}^n e^{2jphi+2ijtheta}=e^{2jphi}sum_{i=1}^n e^{2ijtheta}$. If we let $omega=e^{2jtheta}$, the sum becomes $e^{2jphi}sum_{i=1}^n omega^i$, which is a geometric series:

$$sum_{i=1}^n omega^i=omegafrac{omega^n-1}{omega-1}=omegafrac{e^{2jntheta}-1}{omega-1}=omegafrac{e^{4pi j}-1}{omega-1}=omegafrac{1-1}{omega-1}=0$$

(Note that I used $ntheta=2pi$ and $e^{4pi j}=cos(4pi)+jsin(4pi)=1$ in the above derivation.)

(Also, notice that this proof fails when $omega=e^{2jtheta}=1$ because of a division-by-zero error from above. More specifically, this occurs when $2theta=frac{4pi}{n}$ is a multiple of $2pi$, which is the case when there are $n=1$ or $n=2$ points.)

Thus, this summation is $0$, so both its real and imaginary parts are 0. This gives us:

$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)[0]+v_xv_y[0]rightarrow sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}|v|^2$$