2017 USAJMO Problems/Problem 6
Contents
Problem
Let be distinct points on the unit circle other than . Each point is colored either red or blue, with exactly of them red and exactly of them blue. Let be any ordering of the red points. Let be the nearest blue point to traveling counterclockwise around the circle starting from . Then let be the nearest of the remaining blue points to traveling counterclockwise around the circle from , and so on, until we have labeled all the blue points . Show that the number of counterclockwise arcs of the form that contain the point is independent of the way we chose the ordering of the red points.
Solution
I define a sequence to be, starting at and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include , , , , etc. Note that choosing an is equivalent to choosing an in a sequence, and is defined as the closest to when moving rightwards. If no s exist to the right of , start from the far left. For example, if I have the above example , and I define the 2nd to be , then the first will be . Because no or can be named twice, I can simply remove and from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of is:
Note that, if, in a move, appears to the left of , then intersects
Now, I define a commencing to be a which appears to the left of all s, and a terminating to be a which appears to the right of all s. Let the amount of commencing s be , and the amount of terminating s be , I claim that the number of arcs which cross is constant, and it is equal to . I will show this with induction.
Base case is when . In this case, there are only two possible sequences - and . In the first case, does not cross , but both and are , so . In the second example, , , so . crosses since appears to the left of , so there is one arc which intersects. Hence, the base case is proved.
For the inductive step, suppose that for a positive number , the number of arcs which cross is constant, and given by for any configuration. Now, I will show it for .
Suppose I first choose such that is to the right of in the sequence. This implies that does not cross . But, neither nor is a commencing or terminating . These numbers remain constant, and now after this move we have a sequence of length . Hence, by assumption, the total amount of arcs is .
- Here is a counter-case. : j = 1, k = 2 => : j = 1, k = 3. These numbers may not remain constant.
Thus, this solution probably doesn't work.
Now suppose that appears to the right of , but is not a commencing . This implies that there are no commencing s in the series, because there are no s to the left of , so . Note that this arc does intersect , and must be a terminating . must be a terminating because there are no s to the right of , or else that would be . The length sequence that remains has commencing s and terminating s. Hence, by assumption, the total amount of arcs is .
Finally, suppose that appears to the right of , and is a commencing . We know that this arc will cross . Analogous to the previous case, is a terminating , so the length sequence which remains has commencing s and terminating s. Hence, by assumption, the total amount of arcs is .
There are no more possible cases, hence the induction is complete, and the number of arcs which intersect is indeed a constant which is given by .
-william122
Solution 2
Lemma: If we switch the ordering of two consecutive , , the number of arcs crossing stays invariant.
Proof: There are two situations. If the two arcs don't cross this is simple because the actual arcs stay the same, and only the number order of the arcs change. Otherwise, if is to the left of , the two arcs will have some overlap. The order of the points counterclockwise must be RRBB or some rotation of it. Notice how in both the old and the new ordering, the arc between the second R and the first B is counted twice and the arcs between the same colors are only counted once. Thus no matter where is the number of arcs containing it will stay invariant.
Lemma 2: We can swap the place of any two points and while keeping the number of crossings invariant.
Proof: By Lemma 1 we can use consecutive arc moves to swap with , then the new with and so on until it swaps with . Now, we can do the reverse swaps to move the new to , and so on into the original position of . Since the other points were moved once in one direction and once in the opposite, all the other points are kept in place.
Now, let be the sequence where through is numbered counterclockwise. We show that any ordering has the same number of crossings as . We can swap the current place of with the desired position in . Now we can ignore and swap the currrent place of with the place of in , and so on until all the points are in the desired position. Thus, all orderings of to have the same number of crossings as .
-tigershark22
See also
2017 USAJMO (Problems • Resources) | ||
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