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

## Problem

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

## Solution

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

Note that, if, in a move, $B_n$ appears to the left of $R_n$, then $\stackrel{\frown}{R_nB_n}$ intersects $(1,0)$

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

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

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

Suppose I first choose $R_1$ such that $B_1$ is to the right of $R_1$ in the sequence. This implies that $\stackrel{\frown}{R_1B_1}$ does not cross $(1,0)$. But, neither $R_1$ nor $B_1$ is a commencing $B$ or terminating $R$. These numbers remain constant, and now after this move we have a sequence of length $2n$. Hence, by assumption, the total amount of arcs is $0+\text{max}(j,k)=\text{max}(j,k)$.

Now suppose that $R_1$ appears to the right of $B_1$, but $B_1$ is not a commencing $B$. This implies that there are no commencing $B$s in the series, because there are no $B$s to the left of $B_1$, so $j=0$. Note that this arc does intersect $(1,0)$, and $R_1$ must be a terminating $R$. $R_1$ must be a terminating $R$ because there are no $B$s to the right of $R_1$, or else that $B$ would be $B_1$. The $2n$ length sequence that remains has $0$ commencing $B$s and $k-1$ terminating $R$s. Hence, by assumption, the total amount of arcs is $1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)$.

Finally, suppose that $R_1$ appears to the right of $B_1$, and $B_1$ is a commencing $B$. We know that this arc will cross $(1,0)$. Analogous to the previous case, $R_1$ is a terminating $R$, so the $2n$ length sequence which remains has $j-1$ commencing $B$s and $k-1$ terminating $R$s. Hence, by assumption, the total amount of arcs is $1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)$.

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

-william122