Geometry Problem.

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

Click the figure below to see the complete problem 998.

## Wednesday, March 26, 2014

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

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

ReplyDeleteConnect 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

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

ReplyDeleteLet 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.

sort of qualitative justification:

ReplyDeletesee 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