On Breaking-down any Triangle into
a pair of Mirror-Imaged Equilateral Triangles

by Frank C. Fung ( 1st published in September, 2021. )

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Section XXV - Summarizing Our Findings for Part IV

Summary for the Section :

• In this Section , we shall summarize our findings for Part IV .

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Explaining the Duality of the Solution Set :

• The question then arises as to why we have 2 sets of solution , namely :
  • The FIRST set being the set { X1 , X2 , X3 } arising from [ W1 ]

    and

  • The SECOND set being the set { X4 , X5 , X6 } arising from [ W2 ] .

• Let us recall from Section XXI the 2 Conditions that gave rise to the solutions :
  • Condition ONE :
    • 

  • Condition TWO :
    • 

  And we see here that [ T ] and [ U ] are indeed interchangible .

  • And that is to say :
    • IF :
      • we were to exchange { [ T ] for [ U ] } and { [ U ] for [ T ] } in any particular solution pair ,

      THEN :

      • the 2 Conditions above would still be fully satisfied and the solution value [ X ] arising therefrom , i.e. :
        • 

        would remain the same .

• Let us now bring in this set of 5 equations from previous Sections in Part IV for the reader's quick review :
  • From Section XXI :
    • 

    • 

    • 

  • From Section XXIII :
    • 

  • From Section XXIV :
    • 

  And these 5 equations then tell us that :

  • we can :
    • either :
      • go with [ W1 ] via :
        • 

      or ,

      • go with [ W2 ] via :
        • 

    and the result will be the same .

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A Graphic Look at the Solution Set { X1 / X2 / X3 } :

• Let us now take a graphic look at the solution set { X1 / X2 / X3 } :

  
  

  And we see here that :

  • [ Vector O-X1 ] = [ Vector O-T1 ] + [ Vector O-U1 ]
  • [ Vector O-X2 ] = [ Vector O-T2 ] + [ Vector O-U2 ]
  • [ Vector O-X3 ] = [ Vector O-T3 ] + [ Vector O-U3 ]

    and

  • { Triangle T1-T2-T3 } is in the formation of an Equilateral Triangle , and

  • { Triangle U1-U2-U3 } is also in the formation of an Equilateral Triangle .

  And indeed we are successful in breaking-down { Triangle X1-X2-X3 } :

  • into its own pair of mirrored-imaged Equilateral Triangles ,
    • resized and re-oriented as may be appropriate .

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Matching { Triangle A-B-C } and { Triangle X1-X2-X3 } :

• We then bring back { Triangle A-B-C } for a quick comparison :

  

  But do know whether { Triangle X1-X2-X3 } will match exactly :

  • { Triangle A-B-C } or { Triangle B-C-A } or { Triangle C-A-B } ?

    or for that matter :

  • { Triangle A-C-B } or { Triangle B-A-C } or { Triangle C-B-A } ?

  Not really :

  • We would need to determine this either via visual contact or alternately via a numerical matching procedure .

• This is because :
  • when we label T1 / T2 / T3 as the cube-roots of [ W1 ] ,
    • 

      there is no specific rule as to which of the 3 cube-roots should be picked as the controlling [ T1 ] ,

      • in our solution procedure as presented in Section XXIII ;

  • when we label T4 / T5 / T6 as the cube-roots of [ W2 ] ,
    • 

      there is also no specific rule as to which of the 3 cube-roots should be picked as the controlling [ T4 ] ,

      • in our solution procedure as presented in Section XXIV ;

  Very simply , the following 3 sets of Complex Numbers are one-and-the-same set :

  • Set { A / B / C } ,

  • Set { X1 / X2 / X3 } , and

  • Set { X4 / X5 / X6 } .

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What's next :

• We shall now move on to Part V of the paper , next .

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go to the next Section : Section XXVI - Outline and Purpose for Part V

go to the last Section : Section XXIV - Solving for [ T ] and [ U ] based on [ W2 ]

return to the Pair of Mirror-Imaged Equilateral Triangles Homepage

Original dated 2021-9-07