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

(solution added)
(See also)
Line 19: Line 19:
 
  <math>401</math>
 
  <math>401</math>
 
== See also ==
 
== See also ==
* [[1984 AIME Problems/Problem 11 | Previous problem]]
+
{{AIME box|year=1984|num-b=11|num-a=13}}
* [[1984 AIME Problems/Problem 13 | Next problem]]
+
* [[AIME Problems and Solutions]]
* [[1984 AIME Problems]]
+
* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]

Revision as of 14:26, 6 May 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

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS