Problem:
Equilateral β³ABC has side length 111β. There are four distinct triangles AD1βE1β,AD1βE2β,AD2βE3β, and AD2βE4β, each congruent to β³ABC, with BD1β=BD2β=11β. Find βk=14β(CEkβ)2.
Solution:
Let s=111β and r=11β, let ΞΈ be the common value of β BAD1β and β BAD2β, so that β CAEkβ=ΞΈ,120ββΞΈ,ΞΈ, and 120β+ΞΈ for k=1,2,3, and 4, respectively. Applying the Law of Cosines to each of the triangles ACEkβ gives
CE12βCE22βCE42ββ=CE32β=2s2(1βcosΞΈ),=2s2(1βcos(120ββΞΈ))=2s2(1βcos(240β+ΞΈ)), and =2s2(1βcos(120β+ΞΈ)).β
Because cosΞΈ+cos(120β+ΞΈ)+cos(240β+ΞΈ)=0, the sum βk=14βCEk2β can be simplified to 2s2(4βcosΞΈ). Furthermore, because β BAD1β=β CAE1β, it follows that r2=BD12β=CE12β=2s2(1βcosΞΈ). Thus
