Tuesday, January 28, 2014

Geometry Problem 968: Equilateral Triangle, Rectangle, Common Vertex, Midpoint

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

Click the figure below to see the complete problem 968.

Online Geometry Problem 968: Equilateral Triangle, Rectangle, Common Vertex, Midpoint

3 comments:

  1. Let z(P) be the complex number representing P.

    Let z(A)=0, z(F)=a, z(C)=a-bi
    Then
    z(B)=(1/2 a + √3/2 b)+(√3/2 a - 1/2 b)i
    z(E)=a+(√3/2 a - 1/2 b)i
    z(M)=(1/4 a + √3/4 b)+(√3/4 a - 1/4 b)i

    Easy to check that
    z(F) + (-1/2 + √3/2 i) z(E) + (-1/2 - √3/2 i) z(M) = 0

    Hence, EMF is equilateral triangle.

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  2. Connect CM
    Note that CM ⊥BM => MBEC is cocyclic
    Since M is midpoint of AB => triangle EMF is isosceles
    In circle MBEC , ∠MEC=∠MBC=60 => EMF is equilateral

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  3. CM is perpendicular to AB hence MBEC is con cyclic, so < MEC = < MBC = 60

    Also ACFM is con cyclic so < EFM = < MAC = 60

    So 2 angles of Tr. FEM have been proved to be 60 hence this triangle must be equilateral

    Sumith Peiris
    Moratuwa
    Sri Lanka

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