What Kimberling point is this? Any chance it is the new one?

Anton Zakharov shared this question 2 years ago

Answered

Hello!

I am fairly new to this and have some amateurish understanding of the basic concepts (Brocard points, conjugations, trilinear coordinates, etc) and a bit of geometric imagination. Frankly I don't quite get it is there some way to identify coordinates of the given point with the help of computer, or it should always be done manually (that task is obviously far beyond my current competence)

Specifically, I found the following construction:

The point D is the circumcenter of the triangle ABC. Then E,F,G are the circumcenters of the triangles ABD, BDC, ADC accordingly. Continuations of the sides of the triangles ABC and EFG always meet in 3 points J,H,I that apparently always lie on a common line. In such case (according to the converse to Desargues' theorem) lines BG, AF, CE always intersect at point K.

K must be a well known triangle center. I suspect it is probably already included in the Kimberling database. Could someone please give a hint on how exactly to look for whether this point is already well-known or the new one...

Probably not going to annoy you with the further requests!

Files: Desargues Point...

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Comments (3)

rami ● 2 years ago

Your construction has 2 Kimberling points.

x(1)
x(1127)

This is determined with the four (red) list objects l1,l2,l3,l4.

l1 contains the list of the points to be examined (for the triangle A,C,B).

The points are examined up to Kimberling number 3053.

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Principle:

All Kimberling points up to Kimberling-number 3053 are calculated for the triangle ACB and for those calculated points occurring in l1 the Kimberling number is stored in l2.

l3 calculates the Kimberling point for all numbers in l2

l4 creates a text for each point in l3

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Note:

I don't know if the direction of rotation is relevant for the command TriangleCenter(). I used counterclockwise (i.e. triangle ACB)

Files: Desargues Point...

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Anton Zakharov ● 2 years ago

Thank you. nice to see!

I also found that it is possible to construct a point related to the circumcenter D in the similar manner. If E,F,G are the circumcenters of the triangles ABD, BDC, ADC, continuations of the sides of the triangles ABC and EFG meet in 3 points J,H,I that apparently always lie on a common line. In such case (according to the converse to Desargues' theorem) lines BG, AF, CE always intersect at point K.

And apparently the same construction can be applied to the 9-point center of ABC as well.

Sadly my notebook battery discharged and I lost the .ggb files for these two constructions.

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Anton Zakharov ● 2 years ago

and that brings me to the hypothesis that perhaps the points X(1127) and also X(??), X(???) (both defined in my previous comment) + the original vertices of the triangle ABC can be connected by lines that meet at some other known point X(?)

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