Consider a polygon that surrounds a convex set, in the simplest case a triangle, such that our points sit on the vertices. Now there are two cases. If one starts with a line that does not intersect the interior then it will stay outside under windmilling. It it goes through the interior, it will allways remain so.

Now take any configuration of points and strip it into an onion of such convex hulls. There is always such an onion. Then the line must be chosen such as to intersect the innermost polygon. If it does so, it will remain so. Furthermore, it will pass through all points in S because in its 360 degree turn, it must touch all of them.

]]>Let set A have this property

Let T be a triangle whose vertexes are in A, and has the remaining points in A in its interior.

Assume T exists for any A (A has no 3 co-linear points, so this should be easy to prove)

A counterexample is easy to find. Let A be the points of a square. There is no T with your desired property.

Also, later, in your proof for Lemma 1, you use C to represent both a circle and a point.

]]>Let set A have this property

Let T be a triangle whose vertexes are in A, and has the remaining points in A in its interior.

Assume T exists for any A (A has no 3 co-linear points, so this should be easy to prove)

Let P be a point outside of T

A union P has this property

Proof of Lemma 1:

Let circle C circumscribe T

Let L be the line that is windmilling through A

Let I be the intersection of C and L

As L turns clockwise, its angle increases, and I moves clockwise

Both the angle and the location of I are continuas.

In order for a cycle to occur in A, L must be at all angles, and intersect C at all points

(Otherwise it would need to jump or move counter-clockwise to reach the original state)

L must intersect P (See Lemma 2)

Let B be the pivot point of L prior to it hitting P.

Let C be the point that L would hit following B, if not for P

The distance between L and C decreases as L pivots around P, until L intersects with C.

L cannot intersect with another point first, or else it would have done so if P were not there as well

The cycle resumes as it would if not for the presence of P, until it has hit all points in A, and repeats.

QED

(I’m not entirely convinced the last two lines of this are necessarily true.)

Lemma 2:

Assume line L intersects circle C at point I.

Assume the angle of L is continuas, increasing, and will hit all angles.

Assume I will be at all points on C, and is continuas.

Assume point P is outside of C

L intersects P

Proof of Lemma 2:

Let C` go through P, and have the same center as C.

Let I` be the intersection of L and C`

Let Z be the distance between I` and P

If I were to move along C at a constant angle, we can see that Z will decrease until it reaches 0;

If the angle of L were to increase, with a constant I, we can see that Z will decrease until it reaches 0;

As both the angle L, and the posistion of I are continuas, when they both increase, Z will decrease until it reaches 0

Main Proof

Let T(A) represent a triangle whose vertices are in set A, and which surrounds all other points in A.

Select a subset A of S, of size 3, such that T(A) does not surround a member of S.

We can see that A has the described property

Add a point P, that is a member of S, to A, such that T(A) does not surround a member of S which is not in A.

A still has the described property.

Repeat until A=S

QED

in the second step is to consider the stereographic projection and see the behaviour of the movement of our line(it will be a line after projection and try to conclude). ]]>

Anyway, It occurs to me that there are two ways to alter the original problem that may be interesting. In the descriptions that follow, “left” and “right” are relative to looking at the line perpendicularly.

First way: points are allowed to be collinear. When the line turns to meet additional points, the new pivot is now the nth from the left if the original was the nth from the right. Any points outside the (new,old) pivot pair are counted as used, while points between the (new,old) pair are not. The new pivot and old pivot need not be distinct.

Second way: a ray is used. When a point intersects with the ray, that point becomes the new pivot. At a pivot change, the ray reverses direction (so if it was infinite to the left, it’s now infinite to the right – and vice-versa).

I believe the original proof still works for the first modification, but it easily fails for the second. However, during my brief thoughts, I have so far failed to think of an example where the second can’t also be solved.

]]>Thus in this case we return to point P each rotation by pi, so the windmill goes through P infinitely often. But, as noted earlier, at any other stage of the windmill with the line going through a point Q we have the very same properties, so the windmill goes through every point visited infinitely often. Appealing to geometry, a point of S can only switch sides of the line if the line has at some point gone through it (okay, this might be another gap but it seems reasonable). So in this case all the points of S have been visited after a rotation through pi. Hence the result follows in the case a=b.

In the other case, we can suppose WLOG that b=a+1 (otherwise again switch orientation by pi). So after the rotation through pi we have a vertical line through some P’ in S with b points on the left and a on the right. This is equivalent to sliding our original line to the right until it hits the first point of S. Now do the rotation through pi again and we have a vertical line through some P’’ with a points of S on the left and b on the right — but then P’’ must be P by the sideways sliding argument again. So in this case we return to P after a rotation through 2pi. Applying similar arguments as in the previous case, the windmill visits every point of S infinitely often as required.

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