Topic #1 | Introduction & Summary for the Section |
Topic #2 | Setting up the { 9 Circles } |
Topic #3 | The { Fermat Point } & { Equi-distant Point } |
Topic #4 | The { Centroids } |
Topic #5 | { Perpendicular Lines to the Opposite Sides } |
Topic #6 | Let's Try Something New |
This Section summarizes the relations between our findings in [ Sections II thru VI ] , for review & consolidation purposes ,
The { 9 Circles } of the { Triangles } is then made up of :
as per our discussion below .
We first set up , in general , a triangle { Triangle D-E-F } , that is neither an { equilateral triangle } nor an { isosceles triangle } :
We then construct three (3) sets of { 3 concentric circles } , of radii { L1 } , { L2 } & { L3 } respectively ,
as per this set of 3 diagrams below :
We then have this set of 3 diagrams below , marking { 9 intersection points } on each side of { Line D-E } , { Line E-F } & { Line F-D } respectively :
and there are 6 intersection-points on this line , each point being equi-distant from { Point D } & { Point E } ,
and there are 6 intersection-points on this line , each point being equi-distant from { Point E } & { Point F } ,
and there are 6 intersection-points on this line , each point being equi-distant from { Point F } & { Point D } .
And each pair of [ yellow-color lines ] in the 3 diagrams above are always parallel to the [ white-color line ] in the same diagram ,
We shall be making use of these 9 [ yellow/white-color lines ] for our discussion below .
Let us now construct the { Fermat Point } , using the traditional { Fermat Point Construction Method } of attaching 3 { equilateral triangles } to the 3 { Sides } , as per the diagram on-the-left , below :
Let us now bring back our { Shape Classification System } based on the { Fermat Point } from [ Section V ] and :
as per the diagram in-the-middle , above .
We then also mark-off the { Equi-distant Point } for { Triangle D-E-F } , being the intersection of the 3 { white-color lines } , as per the diagram on-the-right , above .
We simply note here that all 3 diagrams above are based on the 3 [ white-color lines ] .
Let us now find the { centroid } using a procedure similar to the traditional { Fermat Point Construction Method } :
We also note here that { Point O } is the { centroid } of { Triangle A-B-C } , as per the diagram in-the-middle , above .
And the diagram on-the-right then shows us why { Point O } splits each of the 3 line { Line A-D } , { Line B-E } & { Line C-F } at the ratio :
( hint : the 4 smaller { congruent triangles } are { similar triangles } of { Triangle A-B-C } ) .
We also make a special note here that :
( A brief outline of the proof here is provided in [ Appendix B --- Centroid for { Triangle D-E-F } & { Triangle R-S-T } ] ) .
Let us now find the { intersection of the 3 Perpendicular Lines to the Opposite Sides } using a procedure similar to the traditional { Fermat Point Construction Method } :
Let us also look at the { interesection of the Perpendicular Lines to the Opposite Sides } for { Triangle A-B-C } , as per the diagram in-the-middle , above .
We then have the diagram on-the-right , showing the 3 pairs of { parallel yellow-color lines } responsible for identifying both these { intersection points } , as mark-off in [ light-pink-color ] .
Let us now bring back the { Equi-distant Point } for { Triangle D-E-F } , as marked-off in { blue-color } in the diagrams on-the-left & in-the-middle , below .
And the diagram on-the-left , above , then shows us why the { blue-color point } is the mid-point of the [ Line joining the 2 { light-pink-color } points ] ,
And the diagram on-the-right , above , shows us how the 3 pairs of { parallel yellow-color lines } took responsibility for identifying the { Centroid } .
In essence , the { Nine Circles } are quite powerful in explaining quite a few things ,
Let us now construct a pair of { equilateral triangles } on both side of { Line F-D } , { Line D-E } & { Line E-F } respectively ,
We then also identify the { Centroids } of the { equilateral triangles } & the mid-points of { Line F-D } , { Line D-E } & { Line E-F } respectively ,
Let us now look at the 3 triangles , { Triangle G-H-I } , { Triangle R-S-T } and the { Triangle formed by the mid-points } , as per the diagram on-the-left , below :
Let us also look at the 2 triangles , { Triangle R'-S'-T' } & { Triangle G'-H'-I' } , as per the diagram on-the-right , above .
Our empirical / graphic evidence here then shows that { Point O } is the { Centroid } of all 5 triangles .
Let us now take { Triangle D-E-F } and each of the { 5 triangles } , and join the opposite vertices to form 3 lines , as marked-off in { red-color } in the next 5 diagrams below :
It happens that each set of { 3 lines } met at a single point and this point is marked-off in { blue-color } in each of the 5 diagrams , above .
What happened here ??? ( reader's call )
But let us move on now to the next Section --- Mapping { Triangles } onto { Triangles } .