2009 IMO Problems/Problem 6

Problem

Let $a_1,a_2,\ldots,a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s=a_1+a_2+\ldots+a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1,a_2,\ldots,a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.

Author: Dmitry Khramtsov, Russia

Solution 1

We will use strong induction on $n$. When $n = 1$, there are no elements in $M$, so the one jump can be made without landing on a point in $M$. When $n = 2$, we consider two cases. If $a_1$ is not in $M$, then the order $a_1, a_2$ will work. If $a_1$ is in $M$, then $a_2$ is not in $M$ and the order $a_2, a_1$ will work. We will assume that the order can be chosen in such a way for all integers $n < k$ for $k \ge 3$.

When $n = k$, we can assume WLOG that $a_1<a_2< \ldots <a_k$. We can also assume that the elements of $M$ are distinct, since if two elements are identical, we can treat them as one and use induction on $n = k-1$. We now consider four cases:


Case 1: $a_k$ is in $M$ but $a_1, a_2, \ldots, a_{k-1}$ are not.

There are at most $k-2$ elements in $M$ greater than $a_k$. Then, from our assumption, there exists an order of jumps starting from $a_k$ of lengths $a_1, a_2, \ldots, a_{k-1}$ where the first jump is $a_i$ such that the grasshopper will not land in any point in $M$. We can then switch the jumps $a_k$ and $a_i$. Since $a_i$ is not in $M$ and no subsequent jumps result in the grasshopper landing in a point in $M$, this sequence is valid.

Case 2: $a_k$ is not in $M$ but at least one of $a_1, a_2, \ldots, a_{k-1}$ is.

There are at most $k-2$ elements in $M$ greater than $a_k$. Then, from our assumption, there exists an order of jumps starting from $a_k$ of lengths $a_1, a_2, \ldots, a_{k-1}$ such that the grasshopper will not land in any point in $M$. Adding $a_k$ at the beginning of this sequence then results in a valid sequence.

Case 3: $a_k$ is in $M$ and at least one of $a_1, a_2, \ldots, a_{k-1}$ is.

Consider the following pairs of distinct integers $(a_1, a_1+a_k), (a_2, a_2+a_k), \ldots, (a_{k-1}, a_{k-1}+a_k)$. Since at most $k-2$ of these points are in $M$, by the pigeonhole principle, there exists an integer $i$ such that $a_i, a_i + a_k$ are both not in $M$. Since there are at most $k-3$ elements greater than $a_k$, at most $k-3$ elements are greater than $a_i + a_k$. Then, from our assumption, there exists an order of jumps starting from $a_i + a_k$ of lengths $a_1, a_2, \ldots, a_{i-1}, a_{i+1}, \ldots, a_{k-1}$ such that the grasshopper will not land in any point in $M$. Adding $a_i, a_k$ at the beginning of this sequence then results in a valid sequence.

Case 4: None of $a_1, a_2, \ldots, a_k$ are in $M$.

Assume that there exists no valid sequence - that any order of jumps will result in the grasshopper landing on at least one element in $M$. We can select any element $m_i$ in $M$ and, from our assumption, construct a sequence beginning at some $a_j$ containing $n-1$ jumps $a_1, a_2, \ldots, a_{j-1}, a_{j+1}, \ldots, a_k$ that will not land on any of the other $n-2$ elements in $M$. Since no valid sequence exists, during this sequence of jumps, the grasshopper must land on $m_i$.

WLOG, we let $m_1 < m_2 < \ldots < m_{k-1}$. We let $i = 1$ and $j = k$ and use the corresponding sequence that only lands on $m_1$ and no other element in $M$. Let the jump immediately after landing on $m_1$ be $a_x$, where $x < n$. We swap jumps $a_x$ and $a_n$. Now, since $a_n > a_x$, all jumps before $a_n$ will land on a integer less than $m_1$. Since $m_1$ is the smallest element in $M$, none of the jumps before $a_n$ will land on a point in $M$. Since no jumps after $a_n$ will land on an element in $M$ by the construction, the grasshopper must then land on an element $m_y > m_1$ in $M$ after making the jump $a_n$. However, this would imply that the original construction would land on another element $m_y$ different from $m_1$, but this is a contradiction. So, a valid sequence must exist.


Since we have an induction on $n$, the statement must be true for all $n$, and we are done.

Solution 2

This solution will use induction rather than strong induction; as a result, it will be less rigorous.

We proceed as the above step to prove that the grasshopper succeeds for $n = 1$ and $n = 2$. We now assume that the grasshopper succeeds for $n$, and we are to prove that it still succeeds for $n + 1$. Let us call the grasshopper's "trajectory" as the path it takes with $n$ elements to never land on a point in $M$. We are now going to add $a_{n + 1}$ the grasshopper can make and another point to $M$ to prove that the grasshopper wins for $n + 1$. There are 4 cases for where this last point to $M$ could be. We will denote this point as $P$.



Case 1: $P$ is not along the trajectory of the grasshopper. In this case, the grasshopper can make the jump of $a_{n + 1}$ after completing its trajectory, since it will not land on $P$. Since $P$ cannot be equal to $s$, we do not have to worry if the $n + 1th$ jump lands on $P$.


Case 2: $P$ is along the trajectory of the grasshopper. Now, we analyze where along this trajectory $P$ is. We will now attempt to prove that it is possible to change the path the grasshopper takes to avoid $P$. Assume $P$ is the point the grasshopper reaches after $k$ jumps. The worst-case scenario would be that all of the points of $M$ block a combination of $k$ jumps. Since no point of $M$ can be at the origin or be equal to $s$, we have that $0 < k < n + 1$. The number of combinations of $k$ jumps would be ${n + 1}\choose{k}$, which always exceeds the number of points in $M$, since the smallest value, where $k = 1$, results in $n + 1$ combinations, but $M$ has only $n$ points and thus $n$ combinations to prevent.


See Also

1959 IMO (Problems) • Resources)
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions
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