# Difference between revisions of "2017 USAMO Problems/Problem 4"

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==Problem== | ==Problem== | ||

Let <math>P_1</math>, <math>P_2</math>, <math>\dots</math>, <math>P_{2n}</math> be <math>2n</math> distinct points on the unit circle <math>x^2+y^2=1</math>, other than <math>(1,0)</math>. Each point is colored either red or blue, with exactly <math>n</math> red points and <math>n</math> blue points. Let <math>R_1</math>, <math>R_2</math>, <math>\dots</math>, <math>R_n</math> be any ordering of the red points. Let <math>B_1</math> be the nearest blue point to <math>R_1</math> traveling counterclockwise around the circle starting from <math>R_1</math>. Then let <math>B_2</math> be the nearest of the remaining blue points to <math>R_2</math> travelling counterclockwise around the circle from <math>R_2</math>, and so on, until we have labeled all of the blue points <math>B_1, \dots, B_n</math>. Show that the number of counterclockwise arcs of the form <math>R_i \to B_i</math> that contain the point <math>(1,0)</math> is independent of the way we chose the ordering <math>R_1, \dots, R_n</math> of the red points. | Let <math>P_1</math>, <math>P_2</math>, <math>\dots</math>, <math>P_{2n}</math> be <math>2n</math> distinct points on the unit circle <math>x^2+y^2=1</math>, other than <math>(1,0)</math>. Each point is colored either red or blue, with exactly <math>n</math> red points and <math>n</math> blue points. Let <math>R_1</math>, <math>R_2</math>, <math>\dots</math>, <math>R_n</math> be any ordering of the red points. Let <math>B_1</math> be the nearest blue point to <math>R_1</math> traveling counterclockwise around the circle starting from <math>R_1</math>. Then let <math>B_2</math> be the nearest of the remaining blue points to <math>R_2</math> travelling counterclockwise around the circle from <math>R_2</math>, and so on, until we have labeled all of the blue points <math>B_1, \dots, B_n</math>. Show that the number of counterclockwise arcs of the form <math>R_i \to B_i</math> that contain the point <math>(1,0)</math> is independent of the way we chose the ordering <math>R_1, \dots, R_n</math> of the red points. | ||

+ | |||

+ | ==Solution== | ||

+ | I define a sequence to be, starting at <math>(1,0)</math> and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include <math>RB</math>, <math>RBBR</math>, <math>BBRRRB</math>, <math>BRBRRBBR</math>, etc. Note that choosing an <math>R_1</math> is equivalent to choosing an <math>R</math> in a sequence, and <math>B_1</math> is defined as the <math>B</math> closest to <math>R_1</math> when moving rightwards. If no <math>B</math>s exist to the right of <math>R_1</math>, start from the far left. For example, if I have the above example <math>RBBR</math>, and I define the 2nd <math>R</math> to be <math>R_1</math>, then the first <math>B</math> will be <math>B_1</math>. Because no <math>R</math> or <math>B</math> can be named twice, I can simply remove <math>R_1</math> and <math>B_1</math> from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of <math>BBRRRB</math> is: <math>BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3</math> | ||

+ | |||

+ | Note that, if, in a move, <math>B_n</math> appears to the left of <math>R_n</math>, then <math>\stackrel{\frown}{R_nB_n}</math> intersects <math>(1,0)</math> | ||

+ | |||

+ | Now, I define a commencing <math>B</math> to be a <math>B</math> which appears to the left of all <math>R</math>s, and a terminating <math>R</math> to be a <math>R</math> which appears to the right of all <math>B</math>s. Let the amount of commencing <math>B</math>s be <math>j</math>, and the amount of terminating <math>R</math>s be <math>k</math>, I claim that the number of arcs which cross <math>(1,0)</math> is constant, and it is equal to <math>\text{max}(j,k)</math>. I will show this with induction. | ||

+ | |||

+ | Base case is when <math>n=1</math>. In this case, there are only two possible sequences - <math>RB</math> and <math>BR</math>. In the first case, <math>\stackrel{\frown}{R_1B_1}</math> does not cross <math>(1, 0)</math>, but both <math>j</math> and <math>k</math> are <math>0</math>, so <math>\text{max}(j,k)=0</math>. In the second example, <math>j=1</math>, <math>k=1</math>, so <math>\text{max}(j,k)=1</math>. <math>\stackrel{\frown}{R_1B_1}</math> crosses <math>(1,0)</math> since <math>B_1</math> appears to the left of <math>R_1</math>, so there is one arc which intersects. Hence, the base case is proved. | ||

+ | |||

+ | For the inductive step, suppose that for a positive number <math>n</math>, the number of arcs which cross <math>(1,0)</math> is constant, and given by <math>\text{max}(j, k)</math> for any configuration. Now, I will show it for <math>n+1</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>. | ||

+ | |||

+ | 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>. | ||

+ | |||

+ | Finally, suppose that <math>R_1</math> appears to the right of <math>B_1</math>, and <math>B_1</math> is a commencing <math>B</math>. We know that this arc will cross <math>(1,0)</math>. Analogous to the previous case, <math>R_1</math> is a terminating <math>R</math>, so the <math>2n</math> length sequence which remains has <math>j-1</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}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)</math>. | ||

+ | |||

+ | There are no more possible cases, hence the induction is complete, and the number of arcs which intersect <math>(1,0)</math> is indeed a constant which is given by <math>\text{max}(j,k)</math>. | ||

+ | |||

+ | -william122 |

## Revision as of 15:02, 22 December 2017

## Problem

Let , , , be distinct points on the unit circle , other than . Each point is colored either red or blue, with exactly red points and blue points. 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 travelling counterclockwise around the circle from , and so on, until we have labeled all of 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 .

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