Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 981.
Saturday, February 15, 2014
Geometry Problem 981. Triangle, Concurrent Cevians, Midpoints, Area, Hexagon
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By an affine transformation to an equilateral triangle, Triangle A2B2C2 is also an equiliateral triangle with sides lengths half of A1B1C1.
ReplyDeleteSo S1 = 4*S3
And 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*S3
Q.E.D.
Let S(ABC) be the area of ABC.
ReplyDeleteUnder the homothetic transformation with center O and scale factor 2,
we have A₂B₂C₂→ABC. Hence, S₁=4S₃.
Now since
B₂C₂ bisects OA₁
A₂C₂ bisects OB₁
A₂B₂ bisects OC₁
Thus
S(OB₂C₂)=S(A₁B₂C₂)
S(OA₂C₂)=S(B₁A₂C₂)
S(OA₂B₂)=S(C₁A₂B₂)
Therefore, S₂=2S₃.
Hence, S₁=2S₂=4S₃.