Comments on: Minipolymath3 project: 2011 IMO

By: Christian Wieczerkowski

Christian Wieczerkowski — Wed, 17 Oct 2012 15:25:14 +0000

Not sure where this blog really is but it seems to me that the solution should make use of convex hulls.

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.

By: twio

twio — Mon, 02 Apr 2012 13:20:37 +0000

brandon wrote:
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.

By: brandon

brandon — Mon, 02 Apr 2012 05:21:14 +0000

Lemma 1:
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

By: mkhida

mkhida — Wed, 28 Mar 2012 14:09:14 +0000

sorry for my English,I have just a remark on this problem,I just tried to see this problem in dimension three (in the space), so I remarked that all points on a sphere verify the condition :each three point are not collinear,it means that there is a plane passing through each set of three points.Now if we consider starting point P and a plane witch contain it .then we can do the same procedure and try to meet all remaining points.the problem is to define the movement of our plane such that:
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).

By: twio

twio — Tue, 14 Feb 2012 13:24:24 +0000

Just encountered this site, and this problem, it’s wonderfully distracting.

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.

By: Anonymous

Anonymous — Thu, 19 Jan 2012 20:05:10 +0000

(apologies for doing my latex wrong). I just wanted to clarify, though, that a line advances if it passes over the sector with center and boundary equal to the starting point of and angle . Once the whole sector is on the other side of , then I believe we will have hit

By: Anonymous

Anonymous — Thu, 19 Jan 2012 19:59:07 +0000

I hope this hasn’t been said already, but this seems close to me. Consider one convex hull $$\{ p1, p2, …\}$$ inside another $$\{q1, …\}$$, and suppose that a line pivoting on $$p_i$$ needs to sweep out some angle $$\alpha$$ to reach $$p_{i+1}$$. Then I think that even if there are points $$q_{j_1}, q_{j_2}, …$$ between $$p_i$$ and $$p_{i+1}$$, the line keeps advancing until it starts to pivot on $$p_{i+1}$$. If this could be extended to many hulls then we may be done.

By: David Holland

David Holland — Mon, 16 Jan 2012 10:16:55 +0000

To deal with the second gap in the proof, where it is assumed that every point is visited after a rotation by pi, first consider the case when a=b. If a point is not visited by the windmill, WLOG it is on the left of the line initially (otherwise change orientation by pi). But then at each stage of the windmill, when a new pivot is visited, this point remains on the left as we consider the frame of reference of the line (imagine rotating the plane anticlockwise by an equal amount as the line has rotated clockwise to keep the line vertical). But we know that after the rotation by pi, every point on the left of the line initially is now on the right — again considering the frame of reference of the line. The unvisited point can’t be both on the left and on the right. This contradiction shows that every point of S is visited in this case. Now consider the remaining case, WLOG that b=a+1. After the rotation by pi the vertical line goes through P’, the next point to the right of P in S. All a points on the left of the line initially are on the right of the line after the rotation by pi, so by the same argument as before they must all have been visited by the windmill in between. All points but P’ on the right initially end up on the left so they also must have been visited. But P and P’ were also visited, so it is still true that the windmill visits every point of S after a rotation by pi.

By: David Holland

David Holland — Sat, 14 Jan 2012 15:55:57 +0000

Okay, I think I can fill in the gaps. Let us first suppose that we choose the orientation so the line l is vertical and also assume that it only goes through the single point P of S. Suppose we count a points of S on the left and b on the right of the line. WLOG we can suppose as we rotate the line clockwise the first point we come to Q in S is on the right (otherwise switch the orientation by pi). At the moment the line goes through both P and Q there are are still a points of S on the left but b-1 on the right as Q has been removed. Now rotate a small enough angle clockwise about Q so the line now only goes through the one point Q of S. This can be done as S is finite. Now we have again a points of S on the left and b on the right of the line. If we now imagine rotating the plane anticlockwise so that the line through Q is vertical, we see that the number of points on either side of the line is preserved at every stage of a windmill. As has been pointed out, we can choose our original line so that either a=b or |a-b|=1. In the simpler case where a=b, we can see after a rotation through pi we have got a vertical line through some P’ in S with a points of S on either side. Hence P’=P, since we can also achieve the new vertical line by sliding the old one to the left or the right (or not moving it) until we hit P’, but if P’ is not P we would have changed the number of points on both sides by at least one.

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.

By: David Holland

David Holland — Sat, 14 Jan 2012 12:14:07 +0000

IMO comment 14 is not a complete solution. It may well be correct, but there is a gap. There is an assumption, perhaps using spatial intuition, that when the line has rotated by pi it will have returned to the original, central pivot. This may be true but it requires a proof. The pivot point changes and can move back and forth to points on both sides as well as the central point. How do we know that, after the rotation by pi, it hasn’t moved so far to the left or right that the central point can’t be the pivot? I have struggled to find a proof, or at least some idea pointing in that direction, but I haven’t managed it.