# 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

## Solution 3

Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \pm 5, \pm 10, \pm 15... \pm 1000$ so the answer is 400 + 1 = $\boxed{401}$

## Solution 4

Let $z$ be an arbitrary zero. If $z=2-x$, then $x=2-z$ and $2+x=4-z$. Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$. From $0$, we get $4$ and $14$. Now note that applying either of these twice will return $z$, so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$, respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\cdot200+1=\boxed{401}$

## See also

 1984 AIME (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Invalid username
Login to AoPS