# Difference between revisions of "1984 AIME Problems/Problem 12"

## Problem

A function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$. If $x=0$ is a root for $f(x)=0$, what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$?

## Solution 1

If $f(2+x)=f(2-x)$, then substituting $t=2+x$ gives $f(t)=f(4-t)$. Similarly, $f(t)=f(14-t)$. In particular, $$f(t)=f(14-t)=f(14-(4-t))=f(t+10)$$

Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function $$f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}$$ satisfies the conditions and has no other roots.

In the interval $-1000\leq x\leq 1000$, there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$, therefore the minimum number of roots is $\boxed{401}$.

## Solution 2 (non-rigorous)

We notice that the function has reflectional symmetry across both $x=2$ and $x=7$. We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 10$. There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\boxed{401}$. $QED \blacksquare$

Solution by a1b2