Problem:
Equilateral △ABC has side length 111. There are four distinct triangles AD1E1,AD1E2,AD2E3, and AD2E4, each congruent to △ABC, with BD1=BD2=11. Find ∑k=14(CEk)2.
Solution:
Let s=111 and r=11, let θ be the common value of ∠BAD1 and ∠BAD2, so that ∠CAEk=θ,120∘−θ,θ, and 120∘+θ for k=1,2,3, and 4, respectively. Applying the Law of Cosines to each of the triangles ACEk gives
CE12CE22CE42=CE32=2s2(1−cosθ),=2s2(1−cos(120∘−θ))=2s2(1−cos(240∘+θ)), and =2s2(1−cos(120∘+θ)).
Because cosθ+cos(120∘+θ)+cos(240∘+θ)=0, the sum ∑k=14CEk2 can be simplified to 2s2(4−cosθ). Furthermore, because ∠BAD1=∠CAE1, it follows that r2=BD12=CE12=2s2(1−cosθ). Thus
