Question

If the lines xsin^2a + are concurrent where a b c are angles of a triangle

Answer 1

Answer:

Step-by-step explanation:

Let α, β and γ be the angles of a triangle; then it holds

sin2α+sin2β+sin2γ=2+2cosαcosβcosγ

If this equals 2, we conclude

cosαcosβcosγ=0

so one of the angles is a right angle.

Proof of the claim

Let's use that γ=π−α−β, so cosγ=−cos(α+β). Then

cos2α+cos2β+cos2γ=cos2α+cos2β+cos2(α+β)=cos2α+cos2β+cos2αcos2β+sin2αsin2β−2sinαsinβcosαcosβ=cos2α+cos2β+1−cos2α−cos2β+2cos2αcos2β−2sinαsinβcosαcosβ=1+2cosαcosβ(cosαcosβ−sinαsinβ)=1−2cosαcosβcosγ

giving the final relation

cos2α+cos2β+cos2γ=1−2cosαcosβcosγ

Now

sin2α+sin2β+sin2γ=3−cos2α−cos2β−cos2γ=2+2cosαcosβcosγ

as claimed at the beginning.