Triple Quad Formula
This key result has ramifications and generalizations throughout geometry.
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Suppose that \(A_1,A_2\) and \(A_3\) are points with \(Q_1 \equiv Q(A_2,A_3), Q_2 \equiv Q(A_1,A_3) and Q_3 \equiv Q(A_1,A_2)\). Then
\[ (Q_1+Q_2+Q_3)^2 = 2(Q_1^2+Q_2^2+Q_3^2) \] precisely when \(A_1, A_2 \) and \(A_3\) are collinear.
First verify the important polynomial identity \[ (Q_1 +Q_2 +Q_3)^2 -2(Q_1^2+Q_2^2+Q_3^2) = 4Q_1Q_2 - (Q_1 +Q_2 - Q_3)^2. \]
Assume that \(A_1 \equiv [x_1,y_1], A_2 \equiv [x_2,y_2]\) and \(A_3 \equiv [x_3,y_3]\). Then
\[ Q_1 = (x_3-x_2)^2 + (y_3-y_2)^2 \] \[ Q_2=(x_3-x_1)^2 + (y_3-y_1)^2 \] \[ Q_3 = (x_2-x_1)^2 + (y_2-y_1)^2. \]
Rewrite this in the form \[ Q_1 = a_1^2 +b_1^2 \] \[ Q_2 = a_2^2+b_2^2 \] \[ Q_3 = (a_2-a_1)^2 + (b_2-b_1)^2 \]
where\( a_1 \equiv x_3-x_2, b_1 \equiv y_3 - y_2, q_2 \equiv x_3-x_1\) and \(b_2 \equiv y_3 - y_1\), so that \[ Q_1 +Q_2 - Q_3 = 2(a_1a_2+b_1b_2). \]
Then
\[ 4Q_1Q_2 -(Q_1 +Q_2 - Q_3)^2 \] \[ =4\left( (a_1^2 +b_1^2)(a_2^2 +b_2^2) - (a_1a_2 +b_1b_2)^2\right) \] \[ =4(a_1b_2-a_2b_1)^2 \] \[ =4(x_1y_2-x_1y_3+x_2y_3-x_3y_2+x_3y_1-x_2y_1)^2 \]
where Fibonacci's identity was used to go from the second line to the third. Since \(4\not = 0\), the Collinear points theorem shows that this is zero precisely when the three points \(A_1, A_2\) and \(A_3\) are collinear. \blacksquare