# 1976 AHSME Problems/Problem 22

## Problem 22

Given an equilateral triangle with side of length $s$, consider the locus of all points $\mathit{P}$ in the plane of the triangle such that the sum of the squares of the distances from $\mathit{P}$ to the vertices of the triangle is a fixed number $a$. This locus

$\textbf{(A) }\text{is a circle if }a>s^2\qquad\\ \textbf{(B) }\text{contains only three points if }a=2s^2\text{ and is a circle if }a>2s^2\qquad\\ \textbf{(C) }\text{is a circle with positive radius only if }s^2

## Solution

We can consider the locus of points $\mathit{P}$ as the set of points satisfying the equation: $[(x_1-x_2)^2+(y_1-y_2)^2+(x_1-x_3)^2+(y_1-y_3)^2+(x_2-x_3)^2+(y_2-y_3)^2=a]$ where $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ are the coordinates of the three vertices of the equilateral triangle.

If we simplify this equation, we get: $[2(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2)-2(x_1x_2+x_1x_3+x_2x_3+y_1y_2+y_1y_3+y_2y_3)=a]$

Since the vertices of the triangle are fixed, the left side of this equation is also fixed. Thus, the locus of points $\mathit{P}$ is a circle centered at the centroid of the triangle with radius $\sqrt{a}$.

We can now determine which of the answer choices is correct:

$\textbf{(A) }\text{is a circle if }a>s^2$ - This is correct, as the radius of the circle will be greater than zero when $a>s^2$.

$\textbf{(B) }\text{contains only three points if }a=2s^2\text{ and is a circle if }a>2s^2$ - This is incorrect, as the locus of points $\mathit{P}$ is always a circle.

$\textbf{(C) }\text{is a circle with positive radius only if }s^2 - This is incorrect, as the locus of points $\mathit{P}$ is always a circle.

$\textbf{(D) }\text{contains only a finite number of points for any value of }a$ - This is incorrect, as the locus of points $\mathit{P}$ is always a circle.

$\textbf{(E) }\text{is none of these}$ - This is incorrect, as the locus of points $\mathit{P}$ is always a circle.

Therefore, the correct answer is $\boxed{\textbf{(A)}}$.