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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.This entry contributed by Ajit Athle.Click the figure below to see the complete problem 869.
AB=BDABM=aMBC=60-aDBE=aBM=BE=> ABM~=DBE=> BED=b=AMB=> BMC=180-bDE=AM=MCBF=BCFBE=60-a=> BFE~=BCM=> ABE=180-bABE+BED=180-b+b=180 => DEF collinearEF=MC=AMEF=DE => E midpoint of DF
Let z(P) be the complex number representing P. Let ω=exp(iπ/3). Then z(M) = 1/2[z(A) + z(C)]z(D) = z(B) + ω[z(A) − z(B)] = (1−ω)z(B) + ω z(A)z(E) = z(B) + ω[z(M) − z(B)] = (1−ω)z(B) + ω z(M)z(F) = z(B) + ω[z(C) − z(B)] = (1−ω)z(B) + ω z(C)Obviously z(E) = 1/2[z(D) + z(F)]Thus E is the mid-point of D,F, and so D,E,F collinear.
Consider the rotation anti-clockwise with angle 60° about B. B→BA→DM→EC→FThus BAMC→BDEF, so they are congruent. Since A,M,C collinear with mid-point M, thus D,E,F collinear with mid-point E.
Let us change the statement:Let tr ABC any triangle and let us rotate it 60° counterclockwise with center B obtaining tr A'BC'.Prove that tr MBM' is equilateral, where M is mid point of AC and M' is midpoint of A'C'.This becomes almost trivial, indeed <MBM'=60° (rotation angle is preserved by each line and its image) and MB=M'B.