mjhjmtu

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 $w_i:$
$$x, x + frac2 pin, x + 2 cdot frac2 pin, cdots, x + (n-1) cdot frac2 pin $$
for some $x$ as shown in the image at the bottom of the post.

Therefore, the sum that we want is
$$ sum_k=0^n-1 cos left( x + k , frac2 pin right)^2.$$

Using Euler's formula $cos(t) = frace^it + e^-it2$, we have

beginalign
sum_k=0^n-1 cos left( x + k , frac2 pin right)^2 &= frac14 sum_k=0^n-1 left(e^2ileft(x + k, frac2pin right) + e^-2ileft(x + k, frac2pin right) + 2right) \
&= fracn2 + (e^2ix + e^-2ix) sum_k=0^n-1 omega^k
endalign
where $omega$ is a non trivial $n$th root of unity. Then we can easily calculate that
$$ sum_k=0^n-1 omega^k = fracomega^n-1omega - 1 = 0$$
which proves the claim for $k = fracn2.$

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:

$$(vcdot w_i)^2=v_x^2cos^2(phi+itheta)+v_y^2sin^2(phi+itheta)+2v_xv_ycos(phi+itheta)sin(phi+itheta) \ =v_x^2left(fraccos(2phi+2itheta)+12right)+v_y^2left(frac1-cos(2phi+2itheta)2right)+v_xv_ysin(2phi+2itheta) \ =fracv_x^2+v_y^22+cos(2phi+2itheta)left(fracv_x^2-v_y^22right)+v_xv_ysin(2phi+2itheta)$$

Now, let's put this into a summation:

$$sum_i=1^n (vcdot w_i)^2=fracn2(v_x^2+v_y^2)+left(fracv_x^2-v_y^22right)left[sum_i=1^ncos(2phi+2itheta)right]+v_xv_yleft[sum_i=1^nsin(2phi+2itheta)right]$$

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

$$sum_i=1^n omega^i=omegafracomega^n-1omega-1=omegafrace^2jntheta-1omega-1=omegafrace^4pi j-1omega-1=omegafrac1-1omega-1=0$$

(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 $2theta=frac4pin$ 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:

$$sum_i=1^n (vcdot w_i)^2=fracn2(v_x^2+v_y^2)+left(fracv_x^2-v_y^22right)[0]+v_xv_y[0]rightarrow sum_i=1^n (vcdot w_i)^2=fracn2|v|^2$$

We can use some vector calculus here. For some constant $k in mathbbR$ define $f : mathbbR^2 to mathbbR$ as $$f(v) = sum_i=1^n (vcdot w_i)^2 - k|v|^2$$

Taking the gradient of this function yields
$$nabla f(v) = sum_i=1^n 2(vcdot w_i) - 2kv = 2left[sum_i=1^n (vcdot w_i)w_i - kvright]$$

because $nabla (v cdot w_i) = w_i$ and $nabla |v|^2 = 2v$.

Now we calculate the sum $sum_i=1^n (vcdot w_i)w_i$. Using the notation from the answer by @Noble Mushtak, we get
beginalign
sum_i=1^n (vcdot w_i)w_i &= sum_i=1^n big(v_xcos(phi+itheta)+v_ysin(phi+itheta)big)beginbmatrixcos(phi + itheta)\ sin(phi + itheta) endbmatrix\
&= beginbmatrixv_x sum_i=1^ncos^2(phi + itheta) + v_y sum_i=1^nsin(phi+itheta)cos(phi + itheta)\ v_x sum_i=1^nsin(phi+itheta)cos(phi + itheta)+ v_y sum_i=1^nsin^2(phi + itheta)endbmatrix\
&= beginbmatrixv_x sum_i=1^nfrac1+cos(2phi + 2itheta)2 + v_y sum_i=1^nfrac12sin(2phi+2itheta)\ v_x sum_i=1^nfrac12sin(2phi+2itheta)+ v_y sum_i=1^nfrac1-cos(2phi + 2itheta)2endbmatrix\
&= fracn2beginbmatrix
v_x \ v_y
endbmatrix\
&= fracn2v
endalign
because $sum_i=1^n cos(2phi + 2itheta) = sum_i=1^n sin(2phi + 2itheta) = 0$.

Therefore, for $k = fracn2$ we have $nabla f equiv 0$ so $f = textconst$. Plugging in $v = 0$ yields $f equiv 0$ so your formula
$$sum_i=1^n (vcdot w_i)^2 = fracn2|v|^2$$
holds.

Here is a solution that is based on expressing the LHS of the given expression under the classical "shape" of a quadratic form by looking for its associated matrix.

Let us begin by notations. Taking one of the "spikes" (vectors) as directing the $x$-axis, we can assume that :
$$W_k:=binomcos(ka)cos(ka) text
associated with w_k:=e^i k ain mathbbC textwhere a:=2 pi/n. tag1$$

We have the following identities :

$$sum |W_k|^2 = sumcos(ka)^2+sumsin(ka)^2=underbrace1+1+cdots +1_n texttimes=n. tag2$$

and, besides, by squaring relationship $sum_k=1^n w_k equiv 0$ :

$$sumcos(ka)^2-sumsin(ka)^2+2i sum sin(ka)cos(ka) = 0. tag3$$

From (2) and (3), we can conclude that :

$$begincases
sum_kcos(ka)^2=sum_ksin(ka)^2=n/2\
sum_k sin(ka)cos(ka) = 0
endcases.tag4$$

Now, we are able to establish the following identity :

$$sum_k=1^n (Vcdot w_k)^2 = fracn2|V|^2 tag5$$

by converting its LHS into the classical shape of a quadratic form :

$$sum_k=1^n (V^TW_k)(W_k^TV)=V^T(sum_k=1^n W_kW_k^T)V = V^TMV$$

(here $W_k$ and $V$ are considered as column vectors)

where the associated matrix is :

$$M=beginpmatrixsum_kcos(ka)^2&sum_k sin(ka)cos(ka)\
sum_k sin(ka)cos(ka)&sum_ksin(ka)^2endpmatrix=beginpmatrixn/2&0\0&n/2endpmatrix$$

(by using relationships (4)).

We can therefore conclude that the LHS of (5) is indeed equal to its RHS.