Problem:
Given that (1+sint)(1+cost)=45 and
(1−sint)(1−cost)=nm−k
where k,m, and n are positive integers with m and n relatively prime, find k+m+n.
Solution:
Let s=sint+cost and p=sintcost. It is given that 1+s+p=45; thus p=41−s. It then follows that
1=cos2t+sin2t=s2−2p=s2+2s−21
which leads to s=−1±10/2. Because −2<s<2, however, the only possible value for s is −1+10/2. Then
(1−sint)(1−cost)=1−s+p=45−2s=413−10
so k+m+n=27.
The problems on this page are the property of the MAA's American Mathematics Competitions