Triple quad formula proof

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This triple quad formula is a test for collinear points, one of the five basic laws of the rational trigonometry system devised in the early 2000s by Norman J. Wildberger.

It can either be proved by analytic geometry (the preferred means within rational trigonometry) or derived from Heron's formula, using the condition for collinearity that the triangle formed by the three points has zero area.

[edit] The formula

The three points $A, B, C\,$ are collinear precisely when

$(Q(AB) + Q(BC) + Q(AC))^2 = 2(Q(AB)^{2} + Q(BC)^{2} + Q(AC)^{2})\,$

where Q(AB) is the "quadrance", i.e., the square of the distance, between A and B.

[edit] Proof by analytic geometry

Illustration of nomenclature used in the proof.

The line $AB\,$ has the general form:

$ax + by + c = 0\,$

where the (non-unique) parameters $a\,$ , $b\,$ and $c\,$ , can be expressed in terms of the coordinates of points $A\,$ and $B\,$ as:

$a = A_y - B_y\,$

$b = B_x - A_x\,$

$c = A_xB_y - A_yB_x\,$

so that, everywhere on the line:

$(A_y - B_y)x + (B_x - A_x)y + (A_xB_y - A_yB_x) = 0.\,$

But the line can also be specified by two simultaneous equations in a parameter $t\,$ , where $t = 0\,$ at point $A\,$ and $t = 1\,$ at point $B\,$ :

$x = (B_x - A_x)t + A_x\,$ and $y = (B_y - A_y)t + A_y\,$

or, in terms of the original parameters:

$x = bt + A_x\,$ and $y = -at + A_y.\,$

If the point $C\,$ is collinear with points $A\,$ and $B\,$ , there exists some value of $t\,$ (for distinct points, not equal to 0 or 1), call it $\lambda\,$ , for which these two equations are simultaneously satisfied at the coordinates of the point $C\,$ , such that:

$C_x = b\lambda\ + A_x$ and $C_y = -a\lambda\ + A_y.\,$

Now, the quadrances of the three line segments are given by the squared differences of their coordinates, which can be expressed in terms of $\lambda\,$ :

$\begin{matrix}Q(AB) & \equiv & (B_x - A_x)^2 + (B_y - A_y)^2 \\ \ & = & b^2 + (-a)^2 \\ \ & = & a^2 + b^2\end{matrix}$

$\begin{matrix}Q(BC) & \equiv & (C_x - B_x)^2 + (C_y - B_y)^2 \\ \ & = & ((b\lambda\ + A_x) -B_x)^2 + ((-a\lambda\ + A_y) - B_y)^2 \\ \ & = & (b\lambda\ + (A_x -B_x))^2 + (-a\lambda\ + (A_y - B_y))^2\\ \ & = & (b\lambda\ + (-b))^2 + (-a\lambda\ + a)^2\\ \ & = & b^2(\lambda\ - 1)^2 + a^2(-\lambda\ + 1)^2 \\ \ & = & b^2(\lambda\ - 1)^2 + a^2(\lambda\ - 1)^2\\ \ & = & (a^2 + b^2)(\lambda\ - 1)^2\end{matrix}$

$\begin{matrix}Q(AC) & \equiv & (C_x - A_x)^2 + (C_y - A_y)^2 \\ \ & = & ((b\lambda\ + A_x) - A_x)^2 + ((-a\lambda\ + A_y) - A_y)^2 \\ \ & = & (b\lambda\ + A_x - A_x)^2 + (-a\lambda\ + A_y - A_y)^2\\ \ & = & (b\lambda)^2 + (-a\lambda)^2\\ \ & = & b^2\lambda^2 + (-a)^2\lambda^2\\ \ & = & b^2\lambda^2 + a^2\lambda^2\\ \ & = & (a^2 + b^2)\lambda^2\end{matrix}$

where use was made of the fact that $(-\lambda\ + 1)^2 = (\lambda\ - 1)^2$ .

Substituting these quadrances into the equation to be proved:

$(Q(AB) + Q(BC) + Q(AC))^2 = 2(Q(AB)^{2} + Q(BC)^{2} + Q(AC)^{2})\,$

$((a^2 + b^2) + (a^2 + b^2)(\lambda\ - 1)^2 + (a^2 + b^2)\lambda^2)^2 = 2((a^2 + b^2)^2 + ((a^2 + b^2)(\lambda\ - 1)^2)^2 + ((a^2 + b^2)\lambda^2)^2)\,$

$(a^2 + b^2)^2(1 + (\lambda\ - 1)^2 + \lambda^2)^2 = 2(a^2 + b^2)^2(1 + ((\lambda\ - 1)^2)^2 + (\lambda^2)^2)\,$

Now, if $A\,$ and $B\,$ represent distinct points, such that $(a^2 + b^2)\,$ is not zero, we may divide both sides by $Q(AB)^2 = (a^2 + b^2)^2\,$ :

$(1 + \lambda^2 -2\lambda\ + 1 + \lambda^2)^2 = 2(1 + (\lambda^2 -2\lambda\ + 1)^2 + \lambda^4)\,$

$(2\lambda^2 - 2\lambda\ + 2)^2 = 2(1 + \lambda^4 - 2\lambda^3 + \lambda^2 - 2\lambda^3 + 4\lambda^2 - 2\lambda + \lambda^2 - 2\lambda + 1 + \lambda^4)\,$

$4(\lambda^2 - \lambda\ + 1)^2 = 2(2\lambda^4 - 4\lambda^3 + 6\lambda^2 - 4\lambda + 2)\,$

$4(\lambda^4 - \lambda^3 + \lambda^2 - \lambda^3 + \lambda^2 - \lambda + \lambda^2 - \lambda + 1) = 4(\lambda^4 - 2\lambda^3 + 3\lambda^2 - 2\lambda + 1)\,$