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

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 $\boxed{401}$ .

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1984 AIME Problems/Problem 12

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