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

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===Solution===
 
===Solution===
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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, 2F(x)=F(x-a)+F(x+a).  
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According to the given, <math>f(x-a)+f(x+0.5a)=f(x-0.5a)+f(x)</math>, where <math>x</math> and <math>a</math> are rational. Likewise, <math>f(x-0.5a)+f(x+a)=f(x+0.5a)+f(x)</math>. Hence <math>f(x+a)-f(x)= f(x)-f(x-a)</math>, namely <math>2f(x)=f(x-a)+f(x+a)</math>. Let <math>f(0)=C</math>, then consider <math>F(x)=f(x)-C</math>, where <math>F(0)=0</math> and <math>2F(x)=F(x-a)+F(x+a)</math>. We have:
  
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F(2x)=F(x)+[F(x)-F(0)]=2F(x),
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<cmath>F(2x)=F(x)+[F(x)-F(0)]=2F(x)</cmath>
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F(3x)=F(2x)+[F(2x)-F(x)]=3F(x).
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<cmath>F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)</cmath>
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Easily, by induction, F(nx)=nF(x) for all integers k.
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By induction, <math>F(nx)=nF(x)</math> for all in.tegers <math>k</math>.
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Therefore, for nonzero integer m, (1/m)F(mx)=F(x) , namely F(x/m)=(1/m)F(x)
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Therefore, for nonzero integer <math>m</math>, <math>\frac{1}{m}F(mx)=F(x)</math>, namely <math>F\left(\frac{x}{m}\right)=\left(\frac{1}{m}\right)F(x)</math>.
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Hence F(n/m)=(n/m)F(1). Let 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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Hence <math>F\left(\frac{n}{m}\right)=\left(\frac{n}{m}\right)F(1)</math>. Letting <math>F(1)=k</math>, we obtain <math>F(x)=kx</math>, where <math>k</math> is the slope of the linear functions, and <math>f(x)=kx+C</math>.

Revision as of 17:32, 9 August 2015

Problem

Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ such that\[f(x)+f(t)=f(y)+f(z)\]for all rational numbers $x<y<z<t$ that form an arithmetic progression. ($\mathbb{Q}$ is the set of all rational numbers.)

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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