Problem:
Let A,B,C be angles of a triangle with A and C acute and B greater than a right angle satisfying
cos2A+cos2B+2sinAsinBcosC=815 and cos2B+cos2C+2sinBsinCcosA=914.
There are positive integers p,q,r, and s for which
cos2C+cos2A+2sinCsinAcosB=sp−qr,
where p+q and s are relatively prime and r is not divisible by the square of any prime. Find p+q+r+s.
Solution:
Notice that
cos2A+cos2B+2sinAsinBcosC=815
implies
sin2A+sin2B−2sinAsinBcosC=81.
Let R be the circumradius and a,b, and c be the sides opposite A,B, and C, respectively. By the Extended Law of Sines
4R21(a2+b2−2abcosC)=81
and by the Law of Cosines
4R21⋅c2=81
Thus
sin2C=81
Similarly, the second given equation yields sin2A=94, and a similar argument applied to cos2C+cos2A+2sinCsinAcosB shows that it equals 2−sin2B. Because