Problem:
Let a,b,c be the three sides of a triangle, and let α,β,γ, respectively, be the angles opposite them. If a2+b2=1989c2, find
cotα+cotβcotγ
Solution:
First note that
cotα+cotβcotγ=sinαsinβcosαsinβ+sinαcosβsinγcosγ=sinγsin(α+β)cosγsinαsinβ=sin2γcosγsinαsinβ
and that
asinα=bsinβ=csinγ
implies
sin2γsinαsinβ=c2ab
Thus
cotα+cotβcotγ=c2abcosγ
Hence, by the Law of Cosines,
cotα+cotβcotγ=2c2a2+b2−c2=2c21989c2−c2=994
The problems on this page are the property of the MAA's American Mathematics Competitions