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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 957.
(∏R2/3-√3 R2/4)+ (√3 R2/4- ∏R2/12)=∏R2/4
Note that 3*yellow area=OA^2*pi-OF^2*pi=pi(OA^2-(OA/2)^2)=3/4*OA^2*pi, so yellow area=OA^2*pi/4=(OA/2)^2*pi=area of circle with radius OF
The circumradius is double of the inradius in an equilateral triangle. Thus the blue area is 1/4 of the circumcircle. Since by symmetry, there are 3 equal yellow area, each of them must also be 1/4 of the circumcirlce.