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Difference between revisions of "2002 AIME II Problems/Problem 10"

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== Solution ==
 
== Solution ==
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{{solution}}
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Note that <math>x</math> degrees is equal to <math>\frac{\pi x}{180}</math> radians. Also, for <math>\alpha \in \left[0 , \frac{\pi}{2} \right]</math>, the two least positive angles <math>\theta > \alpha</math> such that <math>\sin{\theta} = \sin{\alpha}</math> are <math>\theta = \pi-\alpha</math>, and <math>\theta = 2\pi + \alpha</math>.
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Clearly <math>x > \frac{\pi x}{180}</math> for positive real values of <math>x</math>.
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<math>\theta = \pi-\alpha</math> yields: <math>x = \pi - \frac{\pi x}{180} \Rightarrow x = \frac{180\pi}{180+\pi} \Rightarrow (p,q) = (180,180)</math>.
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<math>\theta = 2\pi + \alpha</math> yields: <math>x = 2\pi + \frac{\pi x}{180} \Rightarrow x = \frac{360\pi}{180-\pi} \Rightarrow (m,n) = (360,180)</math>.
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So, <math>m+n+p+q = \boxed{900}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2002|n=II|num-b=9|num-a=11}}

Revision as of 23:00, 23 June 2008

Problem

While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $x$ for which the sine of $x$ degrees is the same as the sine of $x$ radians are $\frac{m\pi}{n-\pi}$ and $\frac{p\pi}{q+\pi}$, where $m$, $n$, $p$, and $q$ are positive integers. Find $m+n+p+q$.

Solution

Note that $x$ degrees is equal to $\frac{\pi x}{180}$ radians. Also, for $\alpha \in \left[0 , \frac{\pi}{2} \right]$, the two least positive angles $\theta > \alpha$ such that $\sin{\theta} = \sin{\alpha}$ are $\theta = \pi-\alpha$, and $\theta = 2\pi + \alpha$.

Clearly $x > \frac{\pi x}{180}$ for positive real values of $x$.

$\theta = \pi-\alpha$ yields: $x = \pi - \frac{\pi x}{180} \Rightarrow x = \frac{180\pi}{180+\pi} \Rightarrow (p,q) = (180,180)$.

$\theta = 2\pi + \alpha$ yields: $x = 2\pi + \frac{\pi x}{180} \Rightarrow x = \frac{360\pi}{180-\pi} \Rightarrow (m,n) = (360,180)$.

So, $m+n+p+q = \boxed{900}$.

See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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