Problem:
Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B and C, respectively. What fraction of the area of â–³ABC lies outside the circle?
Answer Choices:
A. 2743​π​−31​
B. 23​​−8π​
C. 21​
D. 3​−923​π​
E. 34​−2743​π​
Solution:
Let O be the center of the circle, and without loss of generality, assume that radius OB=1. Because △ABO is a 30−60−90∘ right triangle, AO=2 and AB=BC=3​. Kite ABOC has diagonals of lengths 2 and 3​, so its area is 3​. Because ∠BOC=120∘, the area of the sector cut off by ∠BOC is 31​π. The area of the portion of △ABC lying outside the circle (shaded in the figure) is therefore 3​−31​π. The area of △ABC is 41​3​(3​)2=43​3​, so the requested fraction is
