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

Revision as of 02:25, 6 December 2019

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

@@ Line 15: / Line 15: @@
 == See also ==
 {{AIME box|year=2002|n=II|num-b=9|num-a=11}}
+[[Category: Intermediate Trigonometry Problems]]
 {{MAA Notice}}

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

Page

Toolbox

Search

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

Revision as of 02:25, 6 December 2019

Problem

Solution

See also