Problem:
Equilateral β³ABC 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 β1β parallel to AE, and through E, draw a line β2β parallel to AD. Let F be the intersection of β1β and β2β. Let G be the point on the circle that is collinear with A and F and distinct from A. Given that the area of β³CBG can be expressed in the form pqβ/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 β GCBβ β GAB and β CAGβ β CBG because each pair of angles intercepts the same arc. Also β CAGβ β AFD because AEβ₯DF. Thus β³AFDβΌβ³CBG, and [CBG]=t2[AFD], where t is the similarity ratio. Because mβ ADF=120β,[AFD]=(1/2)ADβ DFβ sin120β=1433β/4. The length of each side of β³ABC is 23β. The Law of Cosines implies that AF2=132+112β2β 13β 11(β1/2)=433, so AF=433β. Therefore t=AFBCβ=433β23ββ, so
Let Ξ±=mβ DAF, and let Ξ²=mβ EAF. Then mβ BCG=Ξ± and mβ CBG=Ξ². Note that [BGC]=(1/2)BCβ CGsinΞ±, and apply the Law of Sines in β³BGC to conclude that sin120βBCβ=sinΞ²CGβ. Then [BGC]=21ββ BCβ sin120βBCsinΞ²βsinΞ±=3βBC2sinΞ±sinΞ²β. Use the Law of Cosines in β³AEF to conclude that AF=433β, and use the Law of Sines to conclude that sin120β433ββ=sinΞ²13β, so sinΞ²=2433β133ββ. The Law of Sines implies that sinΞ±=sinβ AFE=2433β113ββ. Thus