Sum of Squares in Equilateral Triangle

Here's problem #86 from a charming collection by C. W. Trigg, the long term editor of the Problems sections at the Mathematics Magazine.

Reference

1. C. W. Trigg, Mathematical Quickies, Dover, 1985

Trigg credits Leo Moser with the following solution.

Think of the whole diagram as being drawn in three dimensions, so that $A = (1, 0, 0),\;$ $B = (0, 1, 0),\;$ $C = (0, 0, 1),\;$ making the three points lie in the plane $x + y + z = \alpha\;$ (in fact, $\alpha = 1,\;$ in this case but this is convenient to think of it as a generic constant for the sake of a lurking generalization.) The incircle of $\Delta ABC\;$ is the intersection of that plane with the sphere $x^{2} + y^{2} + z^{2} = \beta ,\;$ another constant.

For $P\;$ on the incircle,

$\begin{align} AP^{2} + BP^{2} + CP^{2} &= (1 - x)^{2} + y^{2} + z^{2}\\ &\;\;+x^{2} + (1 - y)^{2} + z\\&\;\;+x^{2} + y^{2} + (1 - z)^{2}\\ &= 3(x^{2} + y^{2} + z^{2}) - 2(x + y + z) + 3\\ &= 3\beta - 2\alpha + 3\\ &= constant. \end{align}$

Observe that the proof goes through for any circle concentric to the incircle. A more general result claims that the sum of squares of the distances from a point to the vertices of a triangle - not necessarily equilateral - is constant for every circle with center at the centroid of the triangle.

Emmanuel José García has observed that a similar result holds for regular polygons, other than triangles, viz.,

Leo Moser's insightful embedding of the problem into 3D naturally extends to polygons with the number of sides beyond $3\;$ and embedding into $\mathbb{R}^n\;$ wherein the polygon vertices are obtained as the intersections of the hyperplane $\displaystyle\sum_{i=0}^{n-1}x_i=\alpha\;$ with the axes. However, the memory of the embedding in 3D suggests a further generalization of Emmanuel José García's statement:

WLOG, assume that the polygon is inscribed into the unit circle and that the vertices are defined as $\displaystyle A_i=\left(\cos\frac{2\pi}{n}i,\sin\frac{2\pi}{n}i\right),\;$ $i=0,\ldots,n-1.\;$ Let point $P\;$ lie on the sphere $x^2+y^2+z^2=\beta.$ Then

$\displaystyle\begin{align} \sum_{i=0}^{n-1}A_iP^{2} &= \sum_{i=0}^{n-1}\left[\left(x-\cos\frac{2\pi}{n}i\right)^2+\left(y-\sin\frac{2\pi}{n}i\right)^2+z^2\right]\\ &=\sum_{i=0}^{n-1}(x^2+y^2+z^2)+\sum_{i=0}^{n-1}\left(\cos^2\frac{2\pi}{n}i+\cos^2\frac{2\pi}{n}i\right)\\ &\;\;\;\;\;-2x\sum_{i=0}^{n-1}\cos\frac{2\pi}{n}i-2y\sum_{i=0}^{n-1}\sin\frac{2\pi}{n}i\\ &=n\beta+n\\ &= constant, \end{align}$

since $\displaystyle\sum_{i=0}^{n-1}\cos\frac{2\pi}{n}i=\sum_{i=0}^{n-1}\sin\frac{2\pi}{n}i=0.$