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 XXXV - Setting up the Unit Circles - OLD vs. NEW

Summary for the Section :

• In this Section , we shall set up the { Unit Circles } :
  • under the { OLD System } vs. under the { NEW System } ;

    so that :

    • the distances can be re-calibrated accordingly .

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Setting up the Unit Circles - OLD vs. NEW :

• For our analysis purposes here in Part VI of the paper ,
  • let us now set up { OLD Unit Circle } ,
    • which is the same as the { Unit Circle } we first set up back in Section XIX in Part IV ;
      • as per the diagram on-the-left below .

  

  We then also set up the { NEW Unit Circle } , differently sized ,

  • as per the diagram on-the-right above .

  Consequently , we now set up scalar quantity [ PSI ] :

  • 

    being the ratio of :

    • { OLD Unit Length } : { NEW Unit Length } .

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[ R-sub-A ] / [ R-sub-B ] / [ R-sub-C ] - OLD vs. NEW :

• Let us bring back this set of 4 diagrams below showing :
  • the position vectors [ Vector O-A ] / [ Vector O-B ] / [ Vector O-A ] ,
    • for the random triangle { Triangle A-B-C } set up back in Section XIX .

  

  We then have these relations below relating :

  • the numeric-values for the lengths under the { OLD System }

    and

  • the numeric-values for the lengths under the { NEW System }
    • for above-said { position vectors } ;

    i.e. :

    • 

    • 

    • 

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The assigned { Complex Values A / B / C } - OLD vs. NEW :

• Since the angles [ ALPHA ] / [ BETA ] / [ GAMMA ] do remain unchanged ,
  • the assigned { complex values A / B / C } under { OLD System } and under the { NEW System }
    • would simply be related by the scalar quantity [ PSI ] ,

    i.e. :

    • 

    • 

    • 

  Consequently , we can now write :

  • 

  • 

  • 

    and

  • 

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Recalling [ K2 ] / [ K1 ] / [ K0 ] :

• Let us recall that when we attempted to solve for the 2 Equilateral Triangles in Section XVIII ,
  • we did assume that { A / B / C } are the solution values to a { 3rd-Order Polynomial Equation },

  so that :

  • these 2 equations are deemed to be equivalent throughout our solution procedure :
    • 

    • 

  Consequently , we have :

  • 

  • 

  • 

• Since [ Point O ] is the centroid of { Triangle A-B-C } ,
  • [ Vector O-A ] + [ Vector O-B ] + [ Vector O-C ] = [ zero ]

  It then follows that :

  • 

  Consequently , we have :

  • 

    arising from :

    • 

• Thus , we can re-write our original 3rd-Order Polynomial Equation :
  • 

  as :

  • 

  And on re-arranging terms , we have :

  • 

• Let us now bring in this set of 4 equations from above :
  • 

  • 

  • 

    and

  • 

  Then :

  • For [ K1 ] given by :
    • 

      we can now write :

      • 

  • For [ K0 ] given by :
    • 

      we can now write :

      • 

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What's next :

• And with this in hand , we shall take a closer look at :
  • [ M ] and [ N ] - OLD vs. NEW ;

    next .

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go to the next Section : Section XXXVI - [ M ] and [ N ] - OLD System vs. NEW System

go to the last Section : Section XXXIV - Outline and Purpose for Part VI

return to the Pair of Mirror-Imaged Equilateral Triangles Homepage

Original dated 2021-9-07