Wednesday, March 26, 2014

Geometry Problem 998. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Equilateral, Congruence

Geometry Problem.
Post your solution in the comments box below.

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

Click the figure below to see the complete problem 998.

Online Geometry Problem 998. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Equilateral, Congruence

3 comments:

  1. http://s25.postimg.org/8ixkdqthb/pro_998.png

    Connect AE and BC
    Perform rotational transformation, center B, angle of Rot= 120
    Triangle CBD will become EBA
    So AE=DC and ∠ (DMA)=120
    We have FG=1/2AE=1/2DC=GH and ∠ (FGH)=60
    So triangle FGH is equilateral

    ReplyDelete
  2. Let z(P) be the complex number representing P.
    Let z(B)=0, z(D)=a, z(E)=b, denote ω=cis(120°).

    Then
    z(A)=aω, z(C)=bω².
    z(F)=1/2 a(1+ω), z(H)=1/2 b(1+ω²), z(G)=1/2 (a+b)

    z(G) + ω z(F) + ω² z(H)
    = 1/2 (a+b) + 1/2 a(ω+ω²) + 1/2 b(ω²+ω)
    = 0

    Hence, FGH is equilateral.

    ReplyDelete
  3. sort of qualitative justification:

    see http://bleaug.free.fr/gogeometry/998.png

    let's consider I and J as centers of circumscribed triangles ABD and CBE. By construction, IDB and BEJ are equilateral therefore there exist a similarity transformation S that transform IDB into BEJ. Since T=(Id+S)/2 is also a similarity transformation (group + vector space over R), T(IDB)=FGH is equilateral.

    bleaug

    ReplyDelete