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 III - Center-of-Mass for the Three-Body Problem with 3 Equal Masses

Summary for the Section :

• In this Section , we shall take a quick look at using :
  • the [ Center-of-Mass ] for the { Three-Body Problem with 3 Equal Masses }
    • as our primary Point-Of-Reference for the { 3-Points Shape } .

Defining the Three-Body Problem with 3 Equal Masses :

• Let [ Mass A ] , [ Mass B ] , and [ Mass C ] be three (3) point-masses of equal mass ,
  • moving around in a 3-Dimensional Space obeying Newton's Law of Gravitation ,

    i.e. :

    • the force of attraction between any 2 masses is given by :
      • 

        where :

        • [ G ] is the Gravitational Constant ,

        • [ M1 ] and [ M2 ] are the mass of the two (2) masses under consideration ,

        • [ R ] is the distance separating the 2 masses .

  The 3 point-masses will , in general , be in the formation of a triangle ,

  • in which case :
    • the [ Center-Of-Mass ] for the 3 Point-Masses

      and

    • the Centroid of { Triangle A-B-C }
      • will coincide and share the exact-same point .

• Since we do not and will not specify any specific { initial conditions } for the { Three-Body Problem with 3 Equal Masses } ,
  • all possibilities for the { 3-Points Shape } formed by the 3 Point-Masses are under consideration throughout this paper .

  And these would include :

  • the general triangle , the case of syzygy , the case of double-collision , and the case of triple-collision .

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The Case of Syzygy :

• The case of Syzygy occurs when the 3 Point-Masses happen to lie on the same straight-line .

  And for our demonstration to follow , we shall assume [ Mass B ] is in-between [ Mass A ] and [ Mass C ] ,

  • as per the diagram on-the-left below .

  

• We can then use a { Straightedge-and-Compass procedure } to find the center-of-mass for [ Mass B ] and [ Mass C ] ,
  • by constructing 2 Circular Arcs with :
    • the radius for both arcs being equal to the length of [ Line Segment B-C ] ,

      and

    • [ Point B ] and [ Point C ] serving as the centers of the 2 Circular Arcs respectively ;

    so that :

    • the 2 arcs will intersect one-another at [ Point P ] and [ Point Q ] ,
      • as per the diagram in-the-middle above .

  We can then use a straightedge to connect [ Point P ] and [ Point Q ] ,

  • so that :
    • [ Line Segment P-Q ] will now intersect [ Line Segment B-C ] at [ Point E ] ;

      and

    • [ Line Segment B-C ] and [ Line Segment P-Q ] do bisect one-another at right-angles ;
      • arising from { B-P-C-Q } being in the formation of a rhombus .

• Suppose we now place 2 masses each equivalent to the equal-mass [ Mass A ] / [ Mass B ] / [ Mass C ] ,
  • one mass at [ Point P ] and one mass at [ Point Q ] .

  Then :

  • the Center-of-Mass for [ Mass B ] and [ Mass C ]

    and

  • the Center-of-Mass for [ Mass P ] and [ Mass Q ]
    • would coincide and share the exact-same point , namely :
      • [ Point E ] as shown in the diagram in-the-middle above ;
        • arising from [ Point E ] being the mid-point of both [ Line Segment B-C ] and [ Line Segment P-Q ] .

• Thus ,
  • the Center-of-Mass for [ Mass A ] , [ Mass B ] and [ Mass C ]

    and

  • the Center-of-Mass for [ Mass A ] , [ Mass P ] and [ Mass Q ]
    • would coincide and share the exact-same position .

  And this point [ Point O ] can be readily found via the centroid of { Triangle A-P-Q } ,

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

• And once we have identified [ Point O ] as the [ Center-of-Mass ] ,
  • we can then create the 3 position-vectors :
    • [ Vector O-A ] , [ Vector O-B ] and [ Vector O-C ] , respectively ;
      • as per the set of 3 diagrams below ;

      to fully define the { 3-Points Shape A-B-C } .

  

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The Case of Double-Collison :

• Let us now take a quick look at the case of Double-Collision ,
  • where 2 of the 3 masses occupy the same point in the { 3-Dimensional Space } .

  For our demonstration to follow , we shall assume :

  • [ Mass B ] and [ Mass C ] share the same point while [ Mass A ] occupies a different point .

• In order to find the [ Center-of-Mass ] , it is necessary for us to create a fourth point , namely [ Point A'] ,
  • via the following procedure :
    • First , we draw-in a circle with [ Point B/C ] as the center and the radius being equal to the length of [ Line Segment A-B/C ] .

    • Then , we extend { Line Segment A-B/C } with a straightedge so that it intersects the circle at [ Point A' ] ,
      • so that :
        • [ Point B/C ] is now the mid-point of [ Line Segment A-A' ] ;
          • as per the diagram in-the-middle below .

  

• We can then construct 2 Circular Arcs with :
  • the radius for both arcs being equal to the length of [ Line Segment A-A' ] ,

    and

  • [ Point A ] and [ Point A' ] serving as the centers of the 2 Circular Arcs respectively ;

  so that :

  • the 2 arcs will intersect one-another at [ Point P ] and [ Point Q ] ,
    • as per the diagram on-the-right above .;

  We can then use a straightedge to connect [ Point P ] and [ Point Q ] , so that :

  • [ Line Segment A-A' ] and [ Line Segmnet P-Q ] will bisect one-another at right-angles ;
    • arising from { A-P-A'-Q } being in the formation of a rhombus .

  As such , [ Line Segment P-Q ] will necessarily pass thru [ Point B/C ] ,

  • [ Point B-C ] being the mid-point of [ Line Segment A-A' ] here .

  

• Suppose we now place 2 masses each equivalent to the equal-mass [ Mass A ] / [ Mass B ] / [ Mass C ] ,
  • one mass at [ Point P ] and one mass at [ Point Q ] .

  Then :

  • the Center-of-Mass for [ Mass B ] and [ Mass C ]

    and

  • the Center-of-Mass for [ Mass P ] and [ Mass Q ]
    • would share the exact-same point , namely :
      • [ Point B/C ] as shown in the diagram in-the-middle above .

  Thus ,

  • the Center-of-Mass for [ Mass A ] , [ Mass B ] and [ Mass C ]

    and

  • the Center-of-Mass for [ Mass A ] , [ Mass P ] and [ Mass Q ]
    • would coincide and share the exact-same position .

  And this point can be readily found via the centroid of { Triangle A-P-Q } ,

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

• And once we have identified [ Point O ] as the [ Center-of-Mass ] ,
  • we can then create the 3 position-vectors :
    • [ Vector O-A ] , [ Vector O-B ] and [ Vector O-C ] , respectively .

    to fully define the { 3-Points Shape A-B-C } .

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The Case of Triple Collision :

• The Case of Triple Collision occurs :
  • when [ Mass A ] , [ Mass B ] and [ Mass C ] all share the exact-same position in the { 3-Dimensional Space } .

  This is a rather simple case as the Center-of-Mass [ Point O ] will also be at the position shared by the 3 Point-Masses ,

  • so that :
    • [ Position Vector O-A ] , [ Position Vector O-B ] and [ Position Vector O-C ] will all be { zero-vectors } .

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What's Next :

• In the next Section , we shall take a quick look at :
  • Mirror-Imaged Triangles and some of the problems arising therefrom .

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go to the next Section : Section V - A First Look at Mirror-Imaged Triangles

go to the last Section : Section III - The Centroid as the Primary Point-Of-Reference

return to the Pair of Mirror-Imaged Equilateral Triangles Homepage

Original dated 2021-9-07