Difference between revisions of "2017 USAJMO Problems/Problem 6"
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Suppose I first choose <math>R_1</math> such that <math>B_1</math> is to the right of <math>R_1</math> in the sequence. This implies that <math>\stackrel{\frown}{R_1B_1}</math> does not cross <math>(1,0)</math>. But, neither <math>R_1</math> nor <math>B_1</math> is a commencing <math>B</math> or terminating <math>R</math>. These numbers remain constant, and now after this move we have a sequence of length <math>2n</math>. Hence, by assumption, the total amount of arcs is <math>0+\text{max}(j,k)=\text{max}(j,k)</math>. | Suppose I first choose <math>R_1</math> such that <math>B_1</math> is to the right of <math>R_1</math> in the sequence. This implies that <math>\stackrel{\frown}{R_1B_1}</math> does not cross <math>(1,0)</math>. But, neither <math>R_1</math> nor <math>B_1</math> is a commencing <math>B</math> or terminating <math>R</math>. These numbers remain constant, and now after this move we have a sequence of length <math>2n</math>. Hence, by assumption, the total amount of arcs is <math>0+\text{max}(j,k)=\text{max}(j,k)</math>. | ||
+ | |||
+ | ------------------------- | ||
+ | *Here is a counter-case. <math>BRR_{1}B_{1}RR</math> : j = 1, k = 2 => <math>BRRR</math> : j = 1, k = 3. These numbers may not remain constant. | ||
+ | Thus, this solution probably doesn't work. | ||
+ | ------------------------- | ||
Now suppose that <math>R_1</math> appears to the right of <math>B_1</math>, but <math>B_1</math> is not a commencing <math>B</math>. This implies that there are no commencing <math>B</math>s in the series, because there are no <math>B</math>s to the left of <math>B_1</math>, so <math>j=0</math>. Note that this arc does intersect <math>(1,0)</math>, and <math>R_1</math> must be a terminating <math>R</math>. <math>R_1</math> must be a terminating <math>R</math> because there are no <math>B</math>s to the right of <math>R_1</math>, or else that <math>B</math> would be <math>B_1</math>. The <math>2n</math> length sequence that remains has <math>0</math> commencing <math>B</math>s and <math>k-1</math> terminating <math>R</math>s. Hence, by assumption, the total amount of arcs is <math>1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)</math>. | Now suppose that <math>R_1</math> appears to the right of <math>B_1</math>, but <math>B_1</math> is not a commencing <math>B</math>. This implies that there are no commencing <math>B</math>s in the series, because there are no <math>B</math>s to the left of <math>B_1</math>, so <math>j=0</math>. Note that this arc does intersect <math>(1,0)</math>, and <math>R_1</math> must be a terminating <math>R</math>. <math>R_1</math> must be a terminating <math>R</math> because there are no <math>B</math>s to the right of <math>R_1</math>, or else that <math>B</math> would be <math>B_1</math>. The <math>2n</math> length sequence that remains has <math>0</math> commencing <math>B</math>s and <math>k-1</math> terminating <math>R</math>s. Hence, by assumption, the total amount of arcs is <math>1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)</math>. | ||
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-william122 | -william122 | ||
+ | |||
+ | ==Solution 2== | ||
+ | Lemma: If we switch the ordering of two consecutive <math>R_k</math>, <math>R_{k+1}</math>, the number of arcs crossing <math>(1,0)</math> 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 <math>B_k</math> is to the left of <math>R_{k+1}</math>, 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 <math>(1,0)</math> is the number of arcs containing it will stay invariant. | ||
+ | |||
+ | Lemma 2: We can swap the place of any two points <math>R_i</math> and <math>R_j</math> while keeping the number of crossings invariant. | ||
+ | |||
+ | Proof: By Lemma 1 we can use consecutive arc moves to swap <math>R_i</math> with <math>R_{i+1}</math>, then the new <math>R_{i+1}</math> with <math>R_{i+2}</math> and so on until it swaps with <math>R_j</math>. Now, we can do the reverse swaps to move the new <math>R_{j-1}</math> to <math>R_{j-2}</math>, and so on into the original position of <math>R_i</math>. 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 <math>I</math> be the sequence where <math>R_1</math> through <math>R_n</math> is numbered counterclockwise. We show that any ordering has the same number of crossings as <math>I</math>. We can swap the current place of <math>R_1</math> with the desired position in <math>I</math>. Now we can ignore <math>R_1</math> and swap the currrent place of <math>R_2</math> with the place of <math>R_2</math> in <math>I</math>, and so on until all the points are in the desired position. Thus, all orderings of <math>R_1</math> to <math>R_n</math> have the same number of crossings as <math>I</math>. | ||
+ | |||
+ | -tigershark22 | ||
==See also== | ==See also== | ||
{{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}} | {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}} |
Latest revision as of 15:30, 12 June 2020
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) | ||
Preceded by Problem 5 |
Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |