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

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== Solution ==
 
== Solution ==
{{solution}}
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If <math>\displaystyle f(2+x)=f(2-x)</math>, then substituting <math>\displaystyle t=2+x</math> gives <math>\displaystyle f(t)=f(4-t)</math>. Similarly, <math>\displaystyle f(t)=f(14-t)</math>. In particular,
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<math>\displaystyle f(t)=f(14-t)=f(14-(4-t))=f(t+10)</math>
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Since 0 is a root, all multiples of 10 are roots, and anything of the form "4 minus a multiple of 10" (that is, anything congruent to 4 modulo 10) are also roots. To see that these may be the only integer roots, observe that the function
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<math>\sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}</math>
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satisfies the conditions and has no other roots.
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In the interval <math>-1000\leq x\leq 1000</math>, there are 201 multiples of 10 and 200 numbers that are congruent to 4 modulo 10, therefore the minimum number of roots is
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<math>401</math>
 
== See also ==
 
== See also ==
 
* [[1984 AIME Problems/Problem 11 | Previous problem]]
 
* [[1984 AIME Problems/Problem 11 | Previous problem]]
 
* [[1984 AIME Problems/Problem 13 | Next problem]]
 
* [[1984 AIME Problems/Problem 13 | Next problem]]
 
* [[1984 AIME Problems]]
 
* [[1984 AIME Problems]]

Revision as of 18:58, 26 March 2007

Problem

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

Solution

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

$\displaystyle f(t)=f(14-t)=f(14-(4-t))=f(t+10)$


Since 0 is a root, all multiples of 10 are roots, and anything of the form "4 minus a multiple of 10" (that is, anything congruent to 4 modulo 10) are also roots. To see that these may be the only integer roots, observe that the function

$\sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}$

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 modulo 10, therefore the minimum number of roots is

$401$

See also

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