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 .

Posted by Antonio Gutierrez at 4:36 PM
Labels: , , ,

3 comments:

  1. Jacob HA (EMK2000)April 30, 2015 at 7:03 PM

    (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.

    ReplyDelete
  2. Jacob HA (EMK2000)April 30, 2015 at 7:36 PM

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

    ReplyDelete
  3. Sumith PeirisAugust 6, 2015 at 8:22 AM

    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

    ReplyDelete

Subscribe to: Post Comments (Atom)