Problem:
An angle x is chosen at random from the interval 0β<x<90β. Let p be the probability that the numbers sin2x,cos2x, and sinxcosx are not the lengths of the sides of a triangle. Given that p=d/n, where d is the number of degrees in arctanm and m and n are positive integers with m+n<1000, find m+n.
Solution:
Because cos(90ββx)=sinx and sin(90ββx)=cosx, it suffices to consider x in the interval 0β<xβ€45β. For such x,
cos2xβ₯sinxcosxβ₯sin2x
so the three numbers are not the lengths of the sides of a triangle if and only if
cos2xβ₯sin2x+sinxcosx
which is equivalent to cos2xβ₯21βsin2x, or tan2xβ€2. Because the tangent function is increasing in the interval 0ββ€xβ€45β, this inequality is equivalent to xβ€21β(arctan2)β. It follows that
p=45β21β(arctan2)ββ=90β(arctan2)ββ
so m+n=92β.
The problems on this page are the property of the MAA's American Mathematics Competitions