Difference between revisions of "2015 USAMO Problems/Problem 4"
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| + | ===Problem=== | ||
| + | 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.) | ||
| + | |||
| + | ===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, 2F(x)=F(x-a)+F(x+a). | 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). | ||
Revision as of 02:00, 12 May 2015
Problem
Find all functions 
such that![\[f(x)+f(t)=f(y)+f(z)\]](http://latex.artofproblemsolving.com/7/e/c/7ecb395c1e56e6d9726fc45f1bc7a66d98333da2.png)
for all rational numbers 
that form an arithmetic progression. (
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, 2F(x)=F(x-a)+F(x+a).
F(2x)=F(x)+[F(x)-F(0)]=2F(x), F(3x)=F(2x)+[F(2x)-F(x)]=3F(x). Easily, by induction, F(nx)=nF(x) for all integers k. Therefore, for nonzero integer m, (1/m)F(mx)=F(x) , namely F(x/m)=(1/m)F(x) 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.
