# Triple Quad Formula

The **triple quad formula** is a relation invented by Canadian-Australian geometer Norman Wildberger that explains the interaction between the quadrances between any two of three collinear points in the affine \(n\)-space \(\mathbb{A}^n\) over an arbitrary field \(\mathbb{F}\) not of characteristic \(2\).

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## Statement of Theorem

Let an invertible \(n \times n\) matrix \(A\) define a symmetric bilinear form on \(\mathbb{V}^n\), the \(n\)-dimensional vector space associated to \(\mathbb{A}^n\), as follows:

\[ b(u,v) = u \cdot v \equiv uAv^T. \]

Here, we assume vectors in \(\mathbb{V}^n\) to be written in row form, i.e. as \(1 \times n\) matrices.

Suppose that \(A_1, A_2, A_3 \in \mathbb{A}^n\) with

\[ \begin{matrix} Q_1 \equiv Q(A_2,A_3), & Q_2 \equiv Q(A_1,A_3), & Q_3 \equiv Q(A_1,A_2). \end{matrix} \]

Then,

\[ (Q_1+Q_2+Q_3)^2 = 2\big(Q_1^2+Q_2^2+Q_3^2\big) \]

precisely when \(A_1,A_2,A_3\) are collinear.

Without loss of generality, we work in \(\mathbb{A}^1\) and let

\[ \begin{matrix} A_1 \equiv [x_1], & A_2 \equiv [x_2], & A_3 \equiv [x_3]. \end{matrix} \]

Suppose the matrix \(A = \begin{pmatrix} a \end{pmatrix}\) defines a symmetric bilinear form on \(\mathbb{V}^1\) as defined previously, so that

\[ \begin{matrix} Q_1 = a(x_3-x_2)^2, & Q_2 = a(x_3-x_1)^2, & Q_3 = a(x_2-x_1)^2. \end{matrix} \]

We then have that

\[ \begin{split} (Q_1+Q_2+Q_3)^2 &= a^2\left((x_2-x_1)^2+(x_3-x_1)^2+(x_3-x_2)^2\right)^2 \\ &= 4a^2(x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3)^2 \end{split} \]

and

\[ \begin{split} 2\left(Q_1^2+Q_2^2+Q_3^2\right) &= 2a^2\left((x_2-x_1)^4+(x_3-x_1)^4+(x_3-x_2)^4\right) \\ &= 4a^2\left(x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3\right)^2. \end{split} \]

Since the two are equal, our desired result is obtained. \(_\square\)

## Examples

Suppose we are working in \(n\)-dimensional Euclidean space \(\mathbb{E}^n\) over the rational number field \(\mathbb{Q}\), and let \(Q_{01} = 4\) and \(Q_{12} = 9\) be two of the quadrances associated to any two of the three collinear points \(A_0, A_1,\) and \(A_2\). To solve for \(Q_{02}\), we substitute these quantities into the triple quad formula to get

\[ (13+Q_{02})^2 = 2(97+Q_{02}^2). \]

This simplifies to the quadratic equation

\[ Q_{02}^2 - 26Q_{02} + 25 = 0. \]

Factorising the left side into \((Q_{02}-1)(Q_{02}-25)\), we have that \(Q_{02}=1\) or \(Q_{02}=25\). This agrees with our usual notions of "lengths" in Euclidean space; the two answers take into account that we have not imposed an "ordering" of the points on the line.

**Cite as:**Triple Quad Formula.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/triple-quad-formula/