An Approach to the Triangle

In this Section, we shall create an { infinite series } of { Triangles } :

• starting-off with any { Triangular Shape } , &

• repeatedly applying the { Fermat Point Construction Method } to the { Triangle } , so that :
  • the [ 3 outer-vertices of the 3 { equilateral triangles } ] form a new { triangle } , being the next member of the { infinite series } .

The { infinite series } then converges towards the { equilateral triangle } .

For the series , we start-off with a random triangle of any size & shape , and name this { Triangle A-B-C } .

• We then apply the traditional { Fermat Point Construction Method } to { Triangle A-B-C } , and
  • we treat the triangle formed by the outer-vertices of the 3 { equilateral triangles } as the new triangle { Triangle D-E-F } .

• We then apply the tradtional { Fermat Point Construction Method } to { Triangle D-E-F } , and
  • we treat the triangle formed by the outer-vertices of the 3 { equilateral triangles } as the new triangle { Triangle G-H-I } .

• And so-on-and-so-forth .

Let us now take a look at this example below :

Let us now look at the first 5 { triangles } in the series :

Any guess that this converges towards the { equilateral triangle } is , in fact , correct !

We shall now take an empirical / graphics approach to the solution first , and come back with a more rigorous derivation later in the Section .

Let us now bring back the { Shape Classification System } we have in Section V , and :

• re-draw the { Reference Circle of radius } with the { Equilateral Triangle R-S-T } there-in , & the { Polar North Reference Direction } ,

  for the first 4 { Triangles } :

We then summarize our empirical findings here as follows :

Triangle I.D.	Radius of Reference Circle	Angle	{ Distance O-P }	Ratio of { Distance O-P } : { Radius }
{ Triangle A-B-C }	1.000000	201.62 degrees	0.445400	0.445400
{ Triangle D-E-F }	2.000000	21.62 degrees	0.445400	0.222720
{ Triangle G-H-I }	4.000000	201.62 degrees	0.445400	0.111360
{ Triangle J-K-L }	8.000000	21.62 degrees	0.445400	0.055680

Our findings here then suggest that :

• the { angle } will flip-flop between { 201.62 drgrees } & { 21.62 degrees } , and

• the [ ratio of { Distance O-P } / { Radius } ] will be reduced by [ one-half ] ,

  every time we move-on to the next member of the series , so that :

  • this ratio will eventually come [ very-close-to-zero ] and we reach the [ center ] of the { Reference Circle of radius } ,

    which then yields the { equilateral triangle } .

Let us now look more closely at the { Fermat Point Construction Method } , to come up with a more rigorous derivation .

We then take an alternate approach to constructing the { Fermat Point } , for the purpose of :

• demonstrating some of the key features of the traditional { Fermat Point Construction Method } .

First , we take the same { Triangle D-E-F } and construct the 3 { equilateral triangles } , as before :

• but let us mark-off the centroid of the 3 { equilateral triangles } as { Point R } , { Point S } , & { Point T } respectively , as per the diagram on-the-left , below :

  

  We note immediately that the angles { Angle F-R-D } , { Angle D-S-E } & { Angle E-T-F } are all [ 120 degrees ] , as marked-off in { red-color } in the diagram on-the-left , above .

  Let us now construct 3 circles passing thru the 3 vertices of each of the 3 { equilateral triangles ) , using the { centroid } as the { center of the circle } , as per the diagram in-the-middle , above .

  • And the 3 circles do intersect at one single point , { Point P } , and this is the { Fermat Point } , as per the diagram on-the-right , above .

Let us now take a closer look at the mechanics , leading us to this conclusion :

• 

• For the { diagram on-the-left } , above , we note that :
  • if { Point P } is a point on the { Arc D-P-F } , then the angle { Angle D-P-F } is always equal to [ 120 degrees ] .

    This is because the { 4 angles } of the quadrilateral { Quadrilateral F-R-D-P } always add-up to [ 360 degrees ] :

    • but we already know that { Angle F-R-D } is [ 120 degrees ] , so that :
      • the other { 3 angles } must add-up to [ 240 degrees ] .

    • but because { Triangle D-R-P } & { Triangle P-R-F } are { isosceles triangles } , ( equal-radius as sides ) :
      • { Angle D-P-F } is always equal to one-half of the [ 240 degrees ] , or simply [ 120 degrees ] .

• And we can apply the same logical sequence to the { diagram in-the-middle } & the { diagram on-the-right } , above , to conclude that :
  • if { Point P } is a point on the { Arc D-P-E } , then the angle { Angle D-P-E } is always equal to [ 120 degrees ] ,

  • if { Point P } is a point on the { Arc E-P-F } , then the angle { Angle E-P-F } is always equal to [ 120 degrees ] .

• Thus , if the point ( Point P } is at the intersection of any 2 of the 3 circles , then :
  • the 3 lines { Line P-D } , { Line P-E } & { Line P-F } must intersect one-another at [ 120 degrees ] ,

    • and the intersection point , { Point P } , is then the { Fermat Point } .

  • and it also follows that the remaining 3rd circle will also intersect the first { 2 circles } at this point ,
    • because all the { intersection angles } are always [ 120 degrees ] .

Now , let us take note that { Triangle R-S-T } is an { equilateral triangle } .

• This is because :

  

• For the { diagram on-the-left } :

  • the quadrilateral { Quadrilateral P-R-D-S } is made-up of :
    • the 2 equal-sides { Side P-R } & { Side D-R } , ( equal-radius ) ,
    • the 2 equal-sides { Side D-S } & { Side P-S } , ( equal-radius ) ,

    so that the diagonal { Line R-S } bisects the diagonal { Line D-P } at { right-angles } .

• For the { diagram in-the-middle } :
  • the quadrilateral { Quadrilateral P-S-E-T } is made-up of :
    • the 2 equal-sides { Side P-S } & { Side E-S } , ( equal-radius ) ,
    • the 2 equal-sides { Side E-T } & { Side P-T } , ( equal-radius ) ,

    so that the diagonal { Line S-T } bisects the diagonal { Line E-P } at { right-angles } .

• For the { diagram on-the-right } :
  • the quadrilateral { Quadrilateral P-T-F-R } is made of :
    • the 2 equal-sides { Side P-T } & { Side F-T } , ( equal-radius ) ,
    • the 2 equal-sides { Side F-R } & { Side P-R } , ( equal-radius ) ,

    so that the diagonal { Line T-R } bisects the diagonal { Line F-P } at { right-angles } .

• Since { Point P } is the { Fermat Point } and the 3 lines { Line D-P } , { Line E-P } & { Line F-P } intersects one-anonther at [ 120 degrees ] , we can conclude that :
  • the 3 lines { Line R-S } , { Line S-T } & { Line T-R } intersects one-another at [ 60 degrees ] ,

    so that { Triangle R-S-T } is an { equilateral triangle } , as per this diagram below .

    

    ( hint : use the quadrilaterals as marked in different-colors in the diagram above , with the 3 angles at { Point P } being [ 120 degrees ] . )

Let us now take a look at the 3 lines { Line H-P } , { Line I-P } & { Line G-P } :

• for the purpose of determining whether { Line H-P-E } , { Line I-P-F } & { Line G-P-D } are all straight-lines .

• For the { diagram on-the-left } , above :

  the { 4 angles } of the quadrilateral { Quadrilateral P-F-R-H } always add-up to [ 360 degrees ] :

  • but we already know that { Angle F-R-H } is [ 240 degrees ] ,
    • because the outside-angle of { Angle F-R-H } is [ 120 degrees ] , { Point R } being the centroid of an equilateral triangle ,

    so that :

    • the other { 3 angles } must add-up to [ 120 degrees ] .

  • but because { Triangle F-R-P } & { Triangle P-R-H } are { isosceles triangles } , ( equal-radius as sides ) :
    • { Angle F-P-H } is always equal to one-half of the [ 120 degrees ] , or simply [ 60 degrees ] .

• For the { diagram in-the-middle } :

  We can apply the same logical sequence to quadrilateral { Quadrilateral P-E-S-I } to arrive at the conclusion that :

  • { Angle E-P-I } is always equal to [ 60 degrees ] .

• For the { diagram on-the-right } :

  We can also apply the same logical sequence to quadrilateral { Quadrilateral P-E-T-G } to arrive at the conclusion that :

  • { Angle E-P-G } is always equal to [ 60 degrees ] .

• We can then conclude that :
  • { Point G } , { Point P } & { Point D } are colinear points so that { Line G-P-D } is a straight-line ,
  • { Point H } , { Point P } & { Point E } are colinear points so that { Line H-P-E } is a straight-line ,
  • { Point I } , { Point P } & { Point F } are colinear points so that { Line I-P-F } is a straight-line ,

    because : [ 60 degrees ] + [ 120 degrees ] = [ 180 degrees ] .

  as per the diagram on-the-left , below , with the [ 60 degrees ] angles as marked in different-colors .

  And because we have established that { Line G-P-D } , { Line H-P-E } & { Line I-P-F } are straight-lines ,

  • we may now use the traditional { Fermat Point Construction Method } of joining the opposite-vertices , instead , to find { Point P } .

Let us now construct 3 new lines :

• the 1st line being perpendicular to { Line P-H } and bisecting { Line P-H } at the same time ,
• the 2nd line being perpendicular to { Line P-I } and bisecting { Line P-I } at the same time ,
• the 3rd line being perpendicular to { Line P-G } and bisecting { Line P-G } at the same time ,

  so that these 3 lines intersect one-another to form the triangle { Triangle X-Y-Z } , as per the diagram in-the-middle , above .

We note here that :

• the 1st line passes thru { Point R } because { Point R } is equi-distant from { Point P } & { Point H } :
  • and all equi-distant points from { Point P } & { Point H } are on this 1st line .

• the 2nd line passes thru { Point S } because { Point S } is equi-distant from { Point P } & { Point I } :
  • and all equi-distant points from { Point P } & { Point I } are on this 2nd line .

• the 3nd line passes thru { Point T } because { Point T } is equi-distant from { Point P } & { Point G } :

  • and all equi-distant points from { Point P } & { Point G } are on this 3rd line .

And because { Triangle R-S-T } is an { equilateral triangle } , the diagram on-the-right , above , tell us that :

• { Triangle X-Y-Z } is also an { equilateral triangle } , because :
  • both { Line R-S } & { Side Y-Z } are perpendicular to { Line G-P-D } and are therefore parallel ,
  • both { Line S-T } & { Side Z-X } are perpendicular to { Line I-P-F } and are therefore parallel ,
  • both { Line T-R } & { Side X-Y } are perpendicular to { Line H-P-E } and are therefore parallel ,

    so that the 3 sides of { Triangle X-Y-Z } also intersects at [ 60 degrees ] .

• { Point R } , { Point S } & { Point T } are the mid-points of { Side Z-X } , { Side X-Y } & { Side Y-Z } , respectively :
  • This is because the { equilateral triangles X-S-R / Y-T-S / Z-R-T } are congruent to { Equilateral Triangle R-S-T } ,
    • because each { equilateral triangle } shares a common-side with { Equilateral Triangle R-S-T } ,

    so that all sides of the 4 { equilateral triangles } are of-the-same-length .

  so that :

  • the { lengths of the sides } for { Equilateral Triangle X-Y-Z } is twice the { length of the sides } for { Equilateral Triangle R-S-T } .

We then construct the { Reference Circle of radius } for { Triangle D-E-F } and { Triangle G-H-I } :

• using our derivation above for { Triangle R-S-T } & { Triangle X-Y-Z } , and knowing that { Point P } is the { Fermat Point } for both triangles .

  

  We see here that :

  • { radius } for the { Reference Circle } for { Triangle G-H-I } is twice that for { Triangle D-E-F } ,
    • because { Triangle X-Y-Z } is twice the size of { Triangle R-S-T } , length-of-side-wise ,

  • the { Polar North Direction } flips [ 180 degrees ] ,
    • because of the orientation of { Triangle X-Y-Z } vs. { Triangle R-S-T } flipped [ 180 degrees ] .

This then confirms our findings in { Topic #3 } above that :

• { angle } flip-flops between a [ Fixed Value ] & [ the Fixed Value + 180 degrees ] ,

• the ratio of ( [ length of { Line O-P } ] : [ radius for the { Referencing Circle } ] ) converges towards [ zero ] ,

  because { Radius } keeps-on doubling while the { length of Line O-P } remains constant .

The { infinite series of triangles } , therefore , converges toward the { equilateral triangle } , at { Point O } .

Let us now compare our { infinite series of triangles } here with this well established { infinite series of numbers } :

To certain extent [ flipping the { Polar-North Direction } ] in our { infinite series of triangles } is liken to the [ changing of signs ( +/- ) ] for our { infinite series of numbers } .

• More thoughts here , for the author & readers alike ?!

Any one wishing to try an [ infinite series of { C-T Tetrahedra } ] may now do so , but solely at his/her own risks !!!

• while others may attempt to [ 'spiral-in' on { Point O } ] .