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 $2$.

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

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

and

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

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

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.