Apostasimetry

In this post I will write about geometry. However, I speak about apostasimetry (Greek: apostasi – “distance”, -metron “measurement”) as I focus on distance.

The concept distance is related to a number that we assign to the gab between two objects, it tells something about “how far apart the two objects are”.

Imagine two rocks on the beach. We measure the distance by putting small pearls between the rocks such that all pearls tough each other. Counting the number of pearls gives a number that we can assign the concept distance. The two rocks are 781 pearls apart. That chain of pearls “connecting” the two rock has a particular shape. We may do the same using another shape and find that the two rocks are 781 pearls apart. This method therefore does not yield a “unique” number and has no practical use. But the question “what is the minimum numbers of pearls required to form a chain between the rocks?” gives a unique number. And THIS is where the concept distance is related to: the minimum number of unit to form a chain between the two objects.

Instead of objects we use ‘points’ denoted by \mathcal P_\jmath. The concept distance between the two points \mathcal{P}_\jmath and \mathcal{P}_k is denoted by D(\mathcal{P}_\jmath,\mathcal{P}_k) and represents a number. Shortly we can write as r_{\jmath k} for the distance between the two points \mathcal{P}_\jmath and \mathcal{P}_k.

But because of the definition of the concept distance, refering to a ‘minimum’ we obtain a particular relation when we consider three points \mathcal{P}_\jmath, \mathcal{P}_k and \mathcal{P}_\ell. That relation is given by:

Distance inequality: r_{\jmath k} + r_{k \ell} \ge r_{\ell \jmath}

The reason is simple, IF r_{\jmath k} + r_{k \ell} < r_{\ell \jmath}, then r_{\ell \jmath} does not represent a distance, for it is not the minimum. Some recognize this inequalicty as the triangular inequality – but it is a fundamental property of the concept distance that follows from definition. I shall refer to it as the “distance inequality”. I shal however write the distance inequality as

Distance inequality: r_{\jmath k} + r_{k \ell} - r_{\ell \jmath} \ge 0

What properties can we derive from the distance inequality?

Distance is allways positive.

The distance inequality is valid for any combination of points, therefore we obtain:

\begin{array}{rcll}  r_{\jmath k} + r_{k \ell} - r_{\ell \jmath} & \ge & 0\\  r_{\ell \jmath} + r_{\jmath k} - r_{k \ell} & \ge & 0\\  \hline  &&& + \\  2 r_{\jmath k} & \ge & 0  \end{array}

So the distance is allways positive. Ok, we may also argue: “as distance is related to counting – it can only be positive!”

Distance is bounded.

We can write

\begin{array}{rcl}  r_{\jmath k} + r_{k \ell} - r_{\ell \jmath} & \ge & 0 \\  r_{k \ell} + r_{\ell \jmath} - r_{\jmath k} & \ge & 0  \end{array}

and therefore

\begin{array}{rcll}  r_{\ell \jmath} - r_{k \ell} & \le & r_{\jmath k}\\  r_{\jmath k} & \le & r_{\ell \jmath} + r_{k \ell}  \end{array}

so

r_{\ell \jmath} - r_{k \ell} \le r_{\jmath k} \le r_{\ell \jmath} + r_{k \ell}.

This may not seem like a usefull inequality – however, when r_{k \ell} = 0 we obtain

r_{\ell \jmath} \le r_{\jmath k} \le r_{\ell \jmath}

and therefore

r_{\ell \jmath} = r_{\jmath k}

When all points are equal – we can only write r_{k k} = \zeta and therefore

r_{k \jmath} - \zeta \le r_{\jmath k} \le r_{k \jmath} + \zeta.

So the distance is bounded, based on the equality of points.

We also see that IF \zeta = 0 then

r_{k \jmath} \le r_{\jmath k} \le r_{k \jmath}

thus

r_{k \jmath} = r_{\jmath k}

so distances are symmetrical.

Metrical distances are symmetric.

A metrical distance is defined as a distance such that r_{k k} = 0 and therefore

r_{k \jmath} = r_{\jmath k}

so metrical distances are symmetrical.

The trinodal identity.

A trinodal is defined as three points.

Let us consider three points and denote the distances as x, y and z. These values are all positive. We obtain the three inequalities:

\begin{array}{rcl}  x + y + z & \ge & 0\\  x + y - z & \ge & 0 \\  z + x - y & \ge & 0 \\  y + z - x & \ge & 0  \end{array}

and therefore

\begin{array}{rcl}  \Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) & \ge & 0  \end{array}

However, is the “reverse” also valid?

When we look to the signs we have several possibilities:

\begin{array}{rcl}  \Big( + \Big) \Big( + \Big) \Big( + \Big) \Big( + \Big) & \ge & 0\\\\  \Big( + \Big) \Big( + \Big) \Big( - \Big) \Big( - \Big) & \ge & 0\\\\  \Big( + \Big) \Big( - \Big) \Big( + \Big) \Big( - \Big) & \ge & 0\\\\  \Big( + \Big) \Big( - \Big) \Big( - \Big) \Big( + \Big) & \ge & 0  \end{array}

But due to the symmetry we can permute the symbols such that

x \ge y \ge z

thus

\begin{array}{rcl}  y & = & z + p\\  x & = & z + p + q  \end{array}

where both p and q are positive. Therefore we obtain

\begin{array}{rcl}  \Big( 3z + p + q \Big) \Big( 2p + q \Big) \Big( z + q \Big) \Big( y + z - x \Big) & \ge & 0  \end{array}

Two of the three inequalities are always satisfied. Therefore we can write

\Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) \ge 0 \Longleftrightarrow \left[\begin{array}{rcl}  x + y + z & \ge & 0\\  x + y - z & \ge & 0 \\  z + x - y & \ge & 0 \\  y + z - x & \ge & 0  \end{array}\right.

Let us therefore define

H(x,y,z) = \sqrt{ \Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) }

It is clear that

\begin{array}{rcl}  H(x,y,z) & = & \sqrt{ \Big( x + y + z \Big) \Big( x + y - z \Big) \Big( z + x - y \Big) \Big( y + z - x \Big) }\\  & = & \sqrt{ \Big( [x + y] + z \Big) \Big( [x + y] - z \Big) \Big( z + [x - y] \Big) \Big( z - [x - y] \Big) }\\  & = & \sqrt{ \Big( [x + y]^2 - z^2 \Big) \Big( z^2 - [x - y]^2 \Big) }\\  & = & \sqrt{ \Big( 2 x y + \left[ x^2 + y^2 - z^2 \right] \Big) \Big( 2 x y - \left[ x^2 + y^2 - z^2 \right] \Big) }\\  & = & \sqrt{ \Big( 4 x^2 y^2 - \left[ x^2 + y^2 - z^2 \right]^2 \Big) }\\  & = & \sqrt{ \Big( 2 x^2 y^2 + 2 y^2 z^2 + 2 z^2 x^2 - \left[ x^4 + y^4 + z^4 \right] \Big) }\\  & = & \sqrt{ \Big( \left[ x^2 + y^2 + z^2 \right]^2 - 2 \left[ x^4 + y^4 + z^4 \right] \Big) }\\  \end{array}

The function H(x,y,z) is named after Hero of Alexandria – as the Heron’s ‘surface’ of a triangle is given by \frac{1}{4} H(x,y,z). Note that we have not defined a concept like ‘surface’ – but that ‘form’ of ‘surface’ is embedded in the function H(x,y,z) and is due to properties of distance.

Every trinodal satisfies

H(x,y,z) \ge 0

that I shall call the trinodal inequality.

The linear trinodal

As we deal with inequalities in the form \cdots \ge 0 there is a special case associate with the equality \cdots = 0.

I shall call a trinodal linear IF H(x,y,z) = 0. A linear trinodal has the property that H(x,y,z) = 0. Note that this is just a definition.

Going back to the definition of H(x,y,z) we find for a linear trinodal:

H(x,y,z) = 0 \Longrightarrow x = y + z \vee y = z + x \vee z = x + y \vee x = y = z = 0

And this is what we basically have for a line as we use it normal – but now we have it defined from the distance inequality.

The link with normal geometry is that the surface of a line (or a point) is zero and therefore H(x,y,z) = 0.

So-far this post – I hope you liked it!

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