Question

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 $m + n + p + q.$

Answer

Solution:

An angle of $x$ degrees is $\frac{\pi x}{180}$ radians, so we need $\sin \frac{\pi x}{180} = \sin x.$ Two angles have equal sines exactly when they differ by a multiple of $2\pi$ or sum to $\pi$ plus a multiple of $2\pi.$

The first case gives $x - \frac{\pi x}{180} = 2\pi j,$ so $x = \frac{360 j \pi}{180 - \pi},$ with least positive value $\frac{360\pi}{180 - \pi} \approx 6.4.$ The second gives $x + \frac{\pi x}{180} = (2k + 1)\pi,$ so $x = \frac{180(2k+1)\pi}{180 + \pi},$ with least positive value $\frac{180\pi}{180 + \pi} \approx 3.1.$ These are the two smallest solutions.

Matching $\frac{m\pi}{n - \pi}$ and $\frac{p\pi}{q + \pi}$ gives $m = 360,$ $n = 180,$ $p = 180,$ $q = 180,$ so $m + n + p + q = 900.$

2002 AIME II Problem 10

Solution:

Problem 10 in Other Years