pentagram query

[zhisheng]

Yep, transposing the truth here amounts to taking just the right point of view, wherein the primal nature of the problem stands out and the the irrelevant recedes to the background.

Anyway here is how each line through the plane determines its own unique symmetry relative to pairs of points on a circle. I have drawn a circle and a line L below. The lines in the figures of this post have nothing to do with the line of the previous post.

Note that each pair of points A and B on a circle determines exactly one line of its own, and this line crosses the line L at exactly one point, which I shall label as L(A,B). Another pair of points X and Y on the circle determines its own line, and another point L(X,Y) on the Line L. I will say that two pairs of points on the circle are symmetric relative to the line L if they points they determine on the line L coincide.

Thus, in the figure below, the pair A and B is not symmetric (relative to L)to the pair X and Y because L(A,B) is not L(X,Y), but the pair A and B is symmetric with the pair C and D because L(A,B)=L(C,D).



Note that if you have any pair of points A and B on the circle, and a third point C, there is exactly one line through L(A,B) and C. This line cuts through the circle at exactly one point: we shall denote that point by

[A,R,B]

Thus the pair A and B is symmetric with the pair R and [A,R,B].



Recall tha the major arithmetic operations, like addition and multiplication tend each to have a "neutral" point. For addition, that neutral point is 0: 0+p always equals p. For multiplication, the neutral point is 1, 1 times p always equals p. In other words, a neutral point appearing in an operation does not change the identity of the other point appearing with it. For this reason, a neutral point is often called an identity point.

Once a symmetry is given, choice of a neutral point for it is more or less arbitrary. We will choose a neutral point for our symmetry relative to L by picking a point E at random (see below). Having chosen a neutral point, we can now precisely describe the operation. But first we must give it a name: say #. We don't use + or x because these already have a specific meaning. We define then, for any two points P and Q on the circle,

P#Q=[P,E,Q].

Thus P#Q is the point such that the pair P and Q is symmetric with the pair E and P#Q.



It is important that we keep in mind that the operation # depends both on the line L and the neutral element E, and that if these are changed, then the operation changes too.

In the next posts I will illustrate with some specific examples.

[zhisheng]

For this example I will take for L the vertical line which is tangent to the left side of the circle, and for E the point on the circle horizontally to right of the tangent point.

I have taken two points P and Q more or less at random and shown what their resultant P#Q is.

Now put theoriginal projective line into the picture, and adjust notation so that things mesh. We shall here write + for # and 0 for E:

So, in the figure above, we have taken two numbers P and Q on the number line at random (actually so the action fits on the screen). We want to add the two together.

So we project them back to the circle and get the orange points on the circle P and Q.

The neutral point for addition is 0, and projecting it back to the circle leaves it fixed.

Now we take P#Q where # is the operation relative to the line L and the point E=0. We write # here as +. Using the method described above, we get P+Q. Now we project P+Q back to the number line and get the sum P+Q.

[zhisheng]

For the second example I will take for L the line cutting horizontally through the center of the circle, and for E the point on the very top of the circle. Here the line L cuts through the circle at two points, and so neither of these points can appear in the operation #.

As before, we shall bring in the number line and adjust notation. We shall write x for # and write 1 for E:

So, in the figure above, we have taken two numbers P and Q on the number line at random (actually so the action fits on the screen). We want to multiply the two together.

So we project them back to the circle and get the orange points on the circle P and Q.

The neutral point for multiplication is 1, and projecting it back to the circle brings it to 1.

Now we take P#Q where # is the operation relative to the line L and the point E=1. We write # here as x. Using the method described above, we get PxQ. Now we project PxQ back to the number line and get the product PxQ.

[micpsi]

[zhisheng]

[Raphael]

[Raphael]