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Difference between revisions of "2015 USAMO Problems/Problem 4"

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===Problem===
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===Problem 6===
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Find all functions <math>f:\mathbb{Q}\rightarrow\mathbb{Q}</math> such that<cmath>f(x)+f(t)=f(y)+f(z)</cmath>for all rational numbers <math>x<y<z<t</math> that form an arithmetic progression. (<math>\mathbb{Q}</math> is the set of all rational numbers.)
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Steve is piling <math>m\geq 1</math> indistinguishable stones on the squares of an <math>n\times n</math> grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions <math>(i, k), (i, l), (j, k), (j, l)</math> for some <math>1\leq i, j, k, l\leq n</math>, such that <math>i<j</math> and <math>k<l</math>. A stone move consists of either removing one stone from each of <math>(i, k)</math> and <math>(j, l)</math> and moving them to <math>(i, l)</math> and <math>(j, k)</math> respectively,j or removing one stone from each of <math>(i, l)</math> and <math>(j, k)</math> and moving them to <math>(i, k)</math> and <math>(j, l)</math> respectively.
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Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.
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How many different non-equivalent ways can Steve pile the stones on the grid?
  
 
===Solution===
 
===Solution===

Revision as of 21:57, 9 March 2016

Problem 6

Steve is piling $m\geq 1$ indistinguishable stones on the squares of an $n\times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions $(i, k), (i, l), (j, k), (j, l)$ for some $1\leq i, j, k, l\leq n$, such that $i<j$ and $k<l$. A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively,j or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ respectively.

Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.

How many different non-equivalent ways can Steve pile the stones on the grid?

Solution

According to the given, $f(x-a)+f(x+0.5a)=f(x-0.5a)+f(x)$, where $x$ and $a$ are rational. Likewise, $f(x-0.5a)+f(x+a)=f(x+0.5a)+f(x)$. Hence $f(x+a)-f(x)= f(x)-f(x-a)$, namely $2f(x)=f(x-a)+f(x+a)$. Let $f(0)=C$, then consider $F(x)=f(x)-C$, where $F(0)=0$ and $2F(x)=F(x-a)+F(x+a)$. We have:

\[F(2x)=F(x)+[F(x)-F(0)]=2F(x)\] \[F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)\] By induction, $F(nx)=nF(x)$ for all in.tegers $k$. Therefore, for nonzero integer $m$, $\frac{1}{m}F(mx)=F(x)$, namely $F\left(\frac{x}{m}\right)=\left(\frac{1}{m}\right)F(x)$. Hence $F\left(\frac{n}{m}\right)=\left(\frac{n}{m}\right)F(1)$. Letting $F(1)=k$, we obtain $F(x)=kx$, where $k$ is the slope of the linear functions, and $f(x)=kx+C$.

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