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

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m (Solution: extra parens)
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<cmath>f(t)=f(14-t)=f(14-(4-t))=f(t+10)</cmath>
 
<cmath>f(t)=f(14-t)=f(14-(4-t))=f(t+10)</cmath>
  
Since <math>0</math> is a root, all multiples of <math>10</math> are roots, and anything congruent to <math>4\pmod{10}</math>) are also roots. To see that these may be the only integer roots, observe that the function
+
Since <math>0</math> is a root, all multiples of <math>10</math> are roots, and anything congruent to <math>4\pmod{10}</math> are also roots. To see that these may be the only integer roots, observe that the function
 
<cmath>f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}</cmath>
 
<cmath>f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}</cmath>
 
satisfies the conditions and has no other roots.
 
satisfies the conditions and has no other roots.
  
 
In the interval <math>-1000\leq x\leq 1000</math>, there are <math>201</math> multiples of <math>10</math> and <math>200</math> numbers that are congruent to <math>4 \pmod{10}</math>, therefore the minimum number of roots is <math>\boxed{401}</math>.
 
In the interval <math>-1000\leq x\leq 1000</math>, there are <math>201</math> multiples of <math>10</math> and <math>200</math> numbers that are congruent to <math>4 \pmod{10}</math>, therefore the minimum number of roots is <math>\boxed{401}</math>.
 +
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=11|num-a=13}}
 
{{AIME box|year=1984|num-b=11|num-a=13}}

Revision as of 18:30, 30 June 2013

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 $\boxed{401}$.

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
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