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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 981.
By an affine transformation to an equilateral triangle, Triangle A2B2C2 is also an equiliateral triangle with sides lengths half of A1B1C1. So S1 = 4*S3And the hexagon is also a regular ones such that area of triangle A1B2C2 = area of triangle OB2C2, symmetrical for the other 3 triangles.So S2 = 2*S3Q.E.D.
Let S(ABC) be the area of ABC. Under the homothetic transformation with center O and scale factor 2, we have A₂B₂C₂→ABC. Hence, S₁=4S₃. Now since B₂C₂ bisects OA₁A₂C₂ bisects OB₁A₂B₂ bisects OC₁Thus S(OB₂C₂)=S(A₁B₂C₂)S(OA₂C₂)=S(B₁A₂C₂)S(OA₂B₂)=S(C₁A₂B₂)Therefore, S₂=2S₃. Hence, S₁=2S₂=4S₃.