Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 5.1 AIME, harder Prove it Italy

Problem:

Let $f$ be a real function of a real variable that satisfies the conditions
(i) $f(10+x)=f(10-x)$
(ii) $f(20+x)=-f(20-x)$
for every real value of $x$ . Prove that $f$ is odd and periodic.
Recall that $f$ is said to be odd if $f(-x)=-f(x)$ for every $x$ ;
$f$ is said to be periodic if there exists $T>0$ such that $f(x+T)=f(x)$ for every $x$ .

Solution

Solution:

We have: $f(20+x)=f(10+(10+x))=f(10-(10+x))=f(-x)$ from (i)

$f(20+x)=-f(20-x)=-f(x)$ from (ii) and the previous one

from which $f(-x)=-f(x)$ for every $x$ and hence $f$ is odd.

Moreover $f(40+x)=f(20+(20+x))=-f(20-(20+x))=-f(-x)=f(x)$
and hence, since $f(40+x)=f(x)$ , it follows that $f$ is periodic.

The following graph shows an example of such a function.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.