# Sum of Squares in Equilateral Triangle

### Problem

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

Let $ABC\;$ be an equilateral triangle, and $P\;$ a point on its incircle.

Prove that

$AP^{2} + BP^{2} + CP^{2}\;$ is constant.

**Reference**

- C. W. Trigg,
*Mathematical Quickies*, Dover, 1985

### Solution

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.

### Generalization

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

Consider a regular $n\text{-gon}\;$ and a point $P\;$ on its circumcircle (or on any other circle with center at the centroid of the polygon). Denote the vertices of the $n\text{-gon}\;$ as $A_i.\;$ Then,

$\displaystyle \sum_{i=0}^{n-1}A_iP^2\;$ is constant.

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:

Consider a regular $n\text{-gon}\;$ and a point $P\;$ on a sphere whose center coincides with that of the polygon. Denote the vertices of the $n\text{-gon}\;$ as $A_i.\;$ Then,

$\displaystyle \sum_{i=0}^{n-1}A_iP^2\;$ is constant.

### Proof of the Generalization

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

### 2D Problems That Benefit from a 3D Outlook

- Four Travellers, Solution
- Desargues' Theorem
- Soddy Circles and Eppstein's Points
- Symmetries in a Triangle
- Three Circles and Common Chords
- Three Circles and Common Tangents
- Three Equal Circles
- Menelaus from 3D
- Stereographic Projection and Inversion
- Stereographic Projection and Radical Axes
- Sum of Squares in Equilateral Triangle

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny64243827 |