Thursday, April 30, 2015

Geometry Problem 1115: Right Triangle, Angle Trisection, Equilateral

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1115: Right Triangle, Angle Trisection, Equilateral .

3 comments:

  1. (Actually copy from my last solution)

    Since AC₁ = AB₂.
    So AA₁ is the perpendicular bisector of C₁B₂,
    thus AC₁A₁B₂ is a kite, therefore A₁C₁=A₁B₂.

    Similarly, C₁A₁=C₁B₂.
    Hence, ΔB₂A₁C₁ is equilateral.

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  2. Another interesting fact:
    B₁ is actually the center of equilateral ΔB₂A₁C₁

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  3. Reference my proof in the previous problem, A1B2C1 is an isoceles triangle with included angle = 60

    Hence this triangle has to be equilateral

    Sumith Peiris
    Moratuwa
    Sri Lanka

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