Wednesday, April 10, 2013

Problem 869: Triangle, Median, Three Equilateral Triangles, Collinear Points, Midpoint

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 869.

Online Geometry Problem 869: Triangle, Median, Three Equilateral Triangles, Collinear Points, Midpoint

4 comments:

  1. AB=BD
    ABM=a
    MBC=60-a
    DBE=a
    BM=BE
    => ABM~=DBE
    => BED=b=AMB
    => BMC=180-b
    DE=AM=MC
    BF=BC
    FBE=60-a
    => BFE~=BCM
    => ABE=180-b
    ABE+BED=180-b+b=180 => DEF collinear
    EF=MC=AM
    EF=DE => E midpoint of DF

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  2. 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.

    ReplyDelete
  3. Consider the rotation anti-clockwise with angle 60° about B.
    B→B
    A→D
    M→E
    C→F

    Thus BAMC→BDEF, so they are congruent.

    Since A,M,C collinear with mid-point M,
    thus D,E,F collinear with mid-point E.

    ReplyDelete
  4. 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.

    ReplyDelete