A geometric question of proof I have been trying to solve this problem for long time but still unsuccessful. The question is:
A straight line is drawn through the circumcentre of a regular triangle.Prove that sum of square of distances from the vertices of the triangle to that straight line does not depend on the choice of line.
I used Menelaus's theorem,cosine/sine rule and I know that answer would come either in terms of sides of triangle or in terms of circumradius, but these ideas are not working on paper.
I shall be thankful if someone can give me a hint or a complete solution.
 A: Consider the triangle in the complex plane with the imaginary axis chosen along the given line. Then the vertices are given by the complex number $z, \,\omega z, \,\omega^2z$ where:


*

*$|z|=R$ is the radius of the circle circumscribed to the triangle;

*$\omega = e^{i\, 2\pi/3}$ is a primitive cubic root of unity, so that $\omega^3=1$, $\omega \bar\omega=1$ and $1+\omega+\omega^2=0$.
The (signed) distance from $z$ to the imaginary axis is $\operatorname{Re} z = \frac{z+ \bar z}{2}$, so letting $S$ be the sum of the squared distances from the vertices to the given line:
$$
\begin{align}
4 S & = (z+\bar z)^2 + (\omega z + \bar\omega \bar z)^2 + (\omega^2 z + \bar\omega^2 \bar z)^2 \\
 & = z^2(1+\omega^2+\omega^4) + \bar z^2(1+ \bar \omega^2+\bar\omega^4)+ 2 z \bar z(1 + \omega \bar\omega+\omega^2 \bar\omega^2) \\
 & = z^2 \cdot 0 + \bar z^2 \cdot 0 + 2 |z|^2(1+1+1) \\
 & = 6 R^2
\end{align}
$$
Therefore $S = \frac{3}{2} R^2$.  (The proof generalizes easily to show that $S=\frac{n}{2} R^2$ for a regular $n$-gon.)
A: Here is a linear algebra proof.
Let us consider the "standard equilateral triangle" whose vertices' coordinates are the columns of the following matrix:
$$A=\pmatrix{1&-1/2&-1/2\\0&\sqrt{3}/2&-\sqrt{3}/2}$$ 
We are going to consider a rotating triangle and a fixed line. Let us take the $y$ axis as this fixed line  ; thus, we are interested in the sum $s=3/2$ of the squares of abscissas. 
It is not difficult to see that $s=3/2$ is naturally the upper left and lower right coefficient in matrix:
$$AA^T=\pmatrix{s&0\\0&s}=sI_2.$$
Let $R_{\theta}=\pmatrix{\cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta)}$ be the (orthogonal, i.e. $R_{\theta}^T=R_{\theta}^{-1}$) rotation matrix with angle $\theta$.
Now, if we apply $R_{\theta}$ to our 3 points, the new matrix is $R_{\theta}A$, and $AA^T$ is replaced by
$$(R_{\theta}A) . (R_{\theta}A)^T=R_{\theta}AA^TR_{\theta}^T=sR_{\theta}I_2R_{\theta}^T=sI_2$$
Thus, we keep the same matrix, proving the preservation of the sum of the squares of abscissas (or the sum of the squares of ordinates...) whatever the rotation. 
Remark: A very nice cousin problem : [Sum of Squares in Equilateral Triangle] (http://www.cut-the-knot.org/pythagoras/EquiIn3D.shtml)
