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Difference between revisions of "2010 USAJMO Problems/Problem 5"

(Created page with '==Problem== Two permutations <math>a_1, a_2, \ldots, a_{2010}</math> and <math>b_1, b_2, \ldots, b_{2010}</math> of the numbers <math>1, 2, \ldots, 2010</math> are said to inters…')
 
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For <math>n = 2010</math> we get <math>m = 1006</math>, which is the required special
 
For <math>n = 2010</math> we get <math>m = 1006</math>, which is the required special
 
case of the general result above.
 
case of the general result above.
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 +
== See Also ==
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{{USAJMO newbox|year=2010|num-b=4|num-a=6}}
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[[Category:Olympiad Number Theory Problems]]

Revision as of 20:14, 28 March 2013

Problem

Two permutations $a_1, a_2, \ldots, a_{2010}$ and $b_1, b_2, \ldots, b_{2010}$ of the numbers $1, 2, \ldots, 2010$ are said to intersect if $a_k = b_k$ for some value of $k$ in the range $1 \le k\le 2010$. Show that there exist $1006$ permutations of the numbers $1, 2, \ldots, 2010$ such that any other such permutation is guaranteed to intersect at least one of these $1006$ permutations.

Solution

Let $n$ be a positive integer. Let $m$ be the smallest positive integer with $m > n - m$. Since $n > n - n = 0$, $m \le n$. Let $N = \{1, \ldots, n\}$ be the set of positive integers from $1$ to $n$. Let $M \subset N$, be $M = \{1, \ldots, m\}$.

Let $P_n$ be the set of of permutations of $N$.

Let $C_m$ be the set of cyclic permutations of $M$, there are $m$ cyclic permutations in all, and $C_m$ acts transitively on $M$, i.e. for every pair of elements $a,b \in M$, there is an element of $C_m$ that maps $a$ to $b$.

Let $C'_m \subset P_n$ be the permutations in $P_n$ that leave $N\setminus M$ fixed, and restricted to $M$ yield one of the permutations in $C_m$. There is a natural one-to-one correspondence between $C'_m$ and $C_m$.

We claim that the $m$ permutations $C'_m$ intersect every permutation in $P_n$.

Suppose, to the contrary, that there exists a permutation $p \in P_n$ that does not intersect any permutation in $C'_m$. Since $C'_m$ acts transitively on $M \subset N$ the permutation $p$ cannot send any element of $M$ to any other element of $M$, therefore it must send all the elements of $M$ to $N\setminus M$, but since $N\setminus M$ has $n - m$ elements and $m > n - m$, this is impossible by the pigeonhole principle. Therefore such a permutation cannot exist, and the permutations in $C'_m$ intersect every permutation in $P_n$.

For $n = 2010$ we get $m = 1006$, which is the required special case of the general result above.

See Also

2010 USAJMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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