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.
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
Let AE,CD cut at X.
ReplyDeleteTr.s ABE and CBD are congruent SAS.
So AE = CD and therefore FG = GH from the midpoint theorem
Also < BDX = BAX and so ADBX is concyclic and < ABX = < AXD = 120 = < AED + < CDE = < FGD + < HGE which gives < FGH = 60
So in Tr. FGH, FG = GH and < FGH = 60 hence the same is equilateral
Sumith Peiris
Moratuwa
Sri Lanka