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

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

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## Saturday, June 20, 2015

### Geometry Problem 1125: Triangle, Four Equilateral Triangles, Centroid, Midpoint

Labels:
centroid,
equilateral,
midpoint,
triangle

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Let ω = cis(2π/3). Let z(P) be the complex number representing P.

ReplyDeleteConsider

z(A₃) + ω z(B₃) + ω² z(C₃)

= 1/2 [z(A) + ω z(B) + ω² z(C)] + 1/2 [z(A₂) + ω z(B₂) + ω² z(C₂)]

= 1/6 {[z(A) + ω z(B) + ω² z(C)] + [z(A₁) + z(B) + z(C)] + ω[z(A) + z(B₁) + z(C)] + ω²[z(A) + z(B) + z(C₁)]}

= 1/6 {[z(A) + ω z(B) + ω² z(C₁)] + [z(A) + ω z(B₁) + ω² z(C)] + [z(A₁) + ω z(B) + ω² z(C)] + [1 + ω + ω²][z(A) + z(B) + z(C)]}

= 1/6 (0 + 0 + 0 + 0)

= 0

The result follows.