# Difference between revisions of "1997 USAMO Problems/Problem 6"

## Problem

Suppose the sequence of nonnegative integers $a_1,a_2,...,a_{1997}$ satisfies

$a_i+a_j \le a_{i+j} \le a_i+a_j+1$

for all $i, j \ge 1$ with $i+j \le 1997$. Show that there exists a real number $x$ such that $a_n=\lfloor{nx}\rfloor$ (the greatest integer $\le nx$) for all $1 \le n \le 1997$.

## Solution

Given nonnegative integers $a_1,a_2,\dots, a_{1997}$ satisfying the given inequalities, let $I_n$ be the set of all $x$ such that $a_n=\lfloor nx\rfloor$. Therefore, $$I_n=\{x\,:\,a_n\le nx We can rewrite this as $\[I_n=\left[\frac{a_n}{n},\frac{a_n+1}{n}\right).\tag{1}$$ So we know that each $I_n$ is an interval. We start by proving that if $n$ and $k$ are positive integers, then $I_n\cap I_{k}\ne \emptyset$. To prove this, we need the following lemma.

Lemma 1: If $n$ and $k$ are positive integers, then $$na_k< k(a_n+1)\tag{2}$$

Proof: We prove this by induction. If $k=1$, then we wish to show that $na_1< a_n+1$. By the given lower bound, we know that $$a_n\ge a_{n-1}+a_1\ge (a_{n-2}+a_1)+a_1\ge \cdots \ge na_1.$$ Therefore, $na_1\le a_n, so the hypothesis is true for $k=1$. If $n=1$, then we wish to prove that $a_k\le k(a_1+1)$. By the given upper bound, we know that $$a_k\le a_{k-1}+(a_1+1)\le (a_{k-2}+(a_1+1))+(a_1+1)\le\cdots\le ka_1+(k-1).$$ Therefore, $a_k, so the hypothesis is true for $n=1$.

Now suppose that (2) holds for all $k,n. If $k=n=j$, then (2) is equivalent to $0, which is obviously true. If $n=j>k$, then by the division algorithm, we can write $n=qk+r$ for nonnegative integers $q$ and $r$ with $0\le r. Thus by applying the lower bound repeatedly (with one of the indices equal to $1$), we find $$k(a_n+1)\ge k(qa_k+a_r+1)=kqa_k+k(a_r+1).\tag{3}$$ But then as $k,r, we can apply the inductive hypothesis with $n=r$ and $k=k$ to find $k(a_r+1)> ra_k$. Substituting this into (3), we find $$k(a_n+1)> kqa_k+ra_k=(kq+r)a_k=na_k.$$ This proves the hypothesis for $n=j>k$.

Suppose that $k=j>n$. Then by the division algorithm, we can write $k=qn+r$ for nonnegative integers $q$ and $r$ with $0\le r. Then by applying the upper bound repeatedly (with one of the indices equal to $1$), we find $$na_k\le n(qa_n+a_r+q)=nqa_n+na_r+nq.\tag{4}$$ But as $n,r, we can apply the inductive hypothesis with $n=n$ and $k=r$, finding that $na_r< r(a_n+1)$. Subsituting this into (4), we find $$na_k< nqa_n+r(a_n+1)+nq=(nq+r)(a_n+1)=k(a_n+1).$$ This proves the hypothesis for $k=j>n$.

Therefore, if the hypothesis is true for all $k,n, then it must be true for all $k,n\le j$. Hence by induction, it must be true for all positive integers $k$ and $n$. $\hfill\ensuremath{\square}$

Now suppose for the sake of contradiction that $I_n$ and $I_k$ are disjoint intervals. Without loss of generality, we may assume that $I_n$ precedes $I_k$ on the number line. Hence the upper bound of $I_n$ is less than or equal to the minimum value of $I_k$, or rather $$\frac{a_{n}+1}{n}\le \frac{a_k}{k}.$$ This simplifies to $na_k\ge k(a_n+1)$. But this contradicts the statement of Lemma 1. Therefore, $I_k\cap I_n\ne \emptyset$.

Now we claim that $I_1\cap I_2\cap \cdots \cap I_{1997}\ne \emptyset$. We prove this by induction. By the above, we know that $I_1\cap I_2\ne \emptyset$. Now suppose that $I_1\cap I_2\cap\cdots\cap I_k\ne \emptyset$. The intersection of two overlapping intervals of the form $[a,b)$ and $[c,d)$ is an interval of the form $[e,f)$, where $e=\max\{a,c\}$ and $f=\min\{b,d\}$. Therefore, by induction, we know that if the intersection of $k$ overlapping intervals is nonempty, then it must also be an interval, say $$I_1\cap I_2\cap\cdots\cap I_k=[x_k,y_k).$$ If $I_{k+1}$ does not intersect $[x_k,y_k)$, then as an interval, it must appear either completely before $x_k$ or completely after $y_k$. If $I_{k+1}$ appears completely before $x_k$, then it has a nonempty intersection with each of $I_1, I_2, \dots, I_k$. But we also know that $x_k$ is a lower bound of one of the intervals, hence $I_{k+1}$ cannot intersect that interval, a contradiction. A similar contradiction arises if $I_{k+1}$ appears completely after $y_k$. Therefore, $$I_1\cap I_2\cap\cdots\cap I_{k+1}\ne \emptyset.$$ Thus by induction, $I_1\cap I_2\cap\cdots\cap I_{1997}\ne \emptyset$. So let $x\in I_1\cap I_2\cap\cdots\cap I_{1997}$. Then by the definition of $I_n$, we know that $a_n=\lfloor nx\rfloor$ for all $1\le n\le 1997$, and we are done.