# Difference between revisions of "2012 USAJMO Problems/Problem 4"

## Problem

Let $\alpha$ be an irrational number with $0 < \alpha < 1$, and draw a circle in the plane whose circumference has length 1. Given any integer $n \ge 3$, define a sequence of points $P_1$, $P_2$, $\dots$, $P_n$ as follows. First select any point $P_1$ on the circle, and for $2 \le k \le n$ define $P_k$ as the point on the circle for which the length of arc $P_{k - 1} P_k$ is $\alpha$, when travelling counterclockwise around the circle from $P_{k - 1}$ to $P_k$. Supose that $P_a$ and $P_b$ are the nearest adjacent points on either side of $P_n$. Prove that $a + b \le n$.

### Solution

Use mathematical induction. For $n=3$ it is true because one point can't be closest to $P_3$ in both ways, and that $1+2\le 3$. Suppose that for some $n$, the nearest adjacent points $P_a$ and $P_b$ on either side of $P_n$ satisfy $a+b \le n$. Then consider the nearest adjacent points $P_c$ and $P_d$ on either side of $P_{n+1}$. It is by the assumption of the nearness we can see that either $(c,d)=(a+1,b+1)$ still holds, or $P_1$ jumps into the interior of the arc $P_{a+1}P_{n}P_{b+1}$, so that $c$ or $d$ equals to $1$. Let's consider the following two cases.

(i) Suppose $a+b=n$.

Since the length of the arc $P_nP_a$ is $\{(a-n)\alpha\}$ (where $\{x\}$ equals to $x$ subtracted by the greatest integer not exceeding $x$) and length of the arc $P_bP_n$ is $\{(n-b)\alpha\} = \{a\alpha\}$, we now consider a point $P_0$ which is defined by $P_1$ traveling clockwise on the circle such that the length of arc $P_0P_1$ is $\alpha$. We claim that $P_0$ is in the interior of the arc $P_bP_nP_a$. Algebraically, it is equivalent to either $\{0-n\alpha\} < \{a\alpha-n\alpha\}$ or $\{n\alpha -0 \} < \{n\alpha - b\alpha\} = \{a\alpha\}$.

Suppose the latter fails, i.e. $\{n\alpha\} \ge \{a\alpha\}$. Then suppose $n\alpha = m_1 + r_1$ and $a\alpha = m_2 + r_2$, where $m_1$, $m_2$ are integers and $0< r_2\le r_1 <1$ ($r_2$ is not zero because $a\alpha$ is irrational). We now have $$\{0-n\alpha\} = \{-m_1-1 + (1 -r_1)\}=1-r_1$$ and $$\{a\alpha -n\alpha\} = \{m_2-m_1-1 + (1+r_2-r_1)\} = 1+r_2-r_1>1-r_1$$

Therefore $P_0$ is either closer to $P_n$ than $P_a$ on the $P_a$ side, or closer to $P_n$ than $P_b$ on the $P_b$ side. In other words, $P_1$ is the closest adjacent point of $P_{n+1}$ on the $P_{a+1}$ side, or the closest adjacent point of $P_{n+1}$ on the $P_{b+1}$ side. Hence $P_c$ or $P_d$ is $P_1$, therefore $c+d \le n+1$.

(ii) Suppose $a+b\le n-1$ Then either $c+d = (a+1)+(b+1) \le n+1$ when $c=a+1$ and $d=b+1$, or $c+d \le 1+n$ when one of $P_c$ or $P_d$ is $P_1$.

In either case, $c+d\le n+1$ is true.

--Lightest 20:27, 6 May 2012 (EDT)