Problem:
Equilateral △ABC is inscribed in a circle of radius 2. Extend AB through B to point D so that AD=13, and extend AC through C to point E so that AE=11. Through D, draw a line ℓ1 parallel to AE, and through E, draw a line ℓ2 parallel to AD. Let F be the intersection of ℓ1 and ℓ2. Let G be the point on the circle that is collinear with A and F and distinct from A. Given that the area of △CBG can be expressed in the form pq/r, where p,q, and r are positive integers, p and r are relatively prime, and q is not divisible by the square of any prime, find p+q+r.
Solution:
Notice that ∠GCB≅∠GAB and ∠CAG≅∠CBG because each pair of angles intercepts the same arc. Also ∠CAG≅∠AFD because AE∥DF. Thus △AFD∼△CBG, and [CBG]=t2[AFD], where t is the similarity ratio. Because m∠ADF=120∘,[AFD]=(1/2)AD⋅DF⋅sin120∘=1433/4. The length of each side of △ABC is 23. The Law of Cosines implies that AF2=132+112−2⋅13⋅11(−1/2)=433, so AF=433. Therefore t=AFBC=43323, so
[CBG]=t2⋅[AFD]=(43323)2⋅41433=4334293]
and p+q+r=865.
OR
Let α=m∠DAF, and let β=m∠EAF. Then m∠BCG=α and m∠CBG=β. Note that [BGC]=(1/2)BC⋅CGsinα, and apply the Law of Sines in △BGC to conclude that sin120∘BC=sinβCG. Then [BGC]=21⋅BC⋅sin120∘BCsinβsinα=3BC2sinαsinβ. Use the Law of Cosines in △AEF to conclude that AF=433, and use the Law of Sines to conclude that sin120∘433=sinβ13, so sinβ=2433133. The Law of Sines implies that sinα=sin∠AFE=2433113. Thus