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 nβΟmΟβ and q+ΟpΟβ, where m,n, p and q are positive integers. Find m+n+p+q.
Solution:
Because x radians is equivalent to Ο180xβ degrees, the requested special values of x satisfy sinxβ=sinΟ180xββ. It follows from properties of the sine function that either
Ο180xβ=x+360j or 180βΟ180xβ=xβ360k
for some integers j and k. Thus either x=180βΟ360jΟβ or x=180+Ο180(2k+1)Οβ, and the least positive values of x are 180βΟ360Οβ and 180+Ο180Οβ, so m+n+p+q=900β.
The problems on this page are the property of the MAA's American Mathematics Competitions