How to check what triangle centers are lying on a circle?

Anton Zakharov shared this question 5 months ago
Answered

For instance, I found that points X(110) and X(112) are lying on a circle with center at X(1576):

https://www.geogebra.org/ge...

I suspect that some other triangle centers are possibly lying on this circle as well. Is there a simple way to check whether this conjecture is true or not? Perhaps there is a special GeoGebra function or some online script that can be used for this purpose. Any help would be much appreciated.

Best Answer
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I mistakenly marked X (110) as X (140) in my sketch. Can't edit that comment as of now.

Hopefully this final version of the construction clarifies the issue:


Let 3 cevians AA1, BB1, CC1 intersect at X (110). Then draw 6 circles:

1) Through 3 points: A1, X (110), C with center at point Ac.

2) Through 3 points: A1, X (110), B with center at point Ab.


3) Through 3 points: B1, X (110), C with center at Bc. And so on ...

Then 3 points (Ab, Ca, Bc) are collinear. 3 points (Ac, Cb, Ba) are collinear. (conjectured)

Their intersection gives us X (1576). Points Y, Z are the intersection points of 3 circles (red, green, purple).

Finally, the ' yellow dotted circle' is defined by 4 points X (110), X (112), Y, Z and X (1576) is its center.

Comments (19)

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1

This "d-circle" has a following property:

A', B', C' are lying on the circumcircle of the triangle ABC. Where A' = inverse-in-d-circle of A, B' = inverse-in-d-circle of B, C' = inverse-in-d-circle of C. Also triangles ABC and A'B'C' are perspective and X(1576) is the perspector.

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or in other words, this "d-circle" is orthogonal to the circumcircle of the triangle ABC.

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Bonjour ,

en utilisant la commande Séquence(TriangleCentre(A, B, C, n), n, 1, 5000) on peut faire des conjectures qu'il faut ensuite confirmer . Aucun de ces points ne se trouvent sur le cercle indiqué .

Cordialement

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Hopefully some point in the interval (5000-40000) belongs to this circle...

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or do you actually mean that segments X(1576)X(110) and X(1576)X(112) are not equal ?

The exact coordinate of X(112) is 0.34335865611170964779. [First coordinate (of normalized trilinears): triangle centers 1-39460 evaluated at (a,b,c) = (9,13,6)] Point X was constructed by me under assumption that it belongs to a circle with center at X(1576) and radius X(1576)X(110). Its coordinate = 0.34335865611171.

So this point X must be X(112) I guess... This can't be just a coincidence !

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Pour les points X110 et X112 , je pense qu'ils sont bien sur le cercle mais je n'en vois pas d'autres .

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wow, so you just found that X(113) is the center of a circle that goes through X(107), X(110),X(4).

and this purple circle apparently is not even included into the Encyclopedia? impressive...

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Ce sont GeoGebra et ETC qui sont impressionnants

X(113) = center of rectangular circumhyperbola that passes through X(110)X(113) = center of rectangular hyperbola {{X(3),X(4),X(110),X(155),X(1351),X(1352),X(2574,X(2575)}}

Mais je n'ai pas trouvé que X107 était sur le cercle de diamètre X4-X110 . Donc j'ai un doute pour ce point .

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2

según los valores de l6 del adjunto solo hay tres pares de puntos que equidisten de x1576

una pareja es X7 X506 (por el bug https://help.geogebra.org/t... )


otra pareja es X506 X507 y la otra pareja puede ser la encontrada por vosotros y no hay más que puedan ser comprobados con GG

Files: foro.ggb
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bug now fixed in v604

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Les calculs de distances par GeoGebra n'ont pas une précision infinie .

En ce qui concerne ETC , je pense que ne sont répertoriés que les colinéarités de points et pas les "cocyclicités"

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ya pensé en la precisión y el cuarto valor no es lo bastante pequeño como para ser un cero mal calculado (creo), así que solo hay tres posibles valores 0 o redondeados a cero lo que da un máximo de tres puntos con otro posterior que equidiste.

puedes buscar qué puntos son los que dan una diferencia de distancia de 0.00000063 y ver si en realidad

podrían estar en un circulo pero yo lo dudo mucho

no afirmo que haya tres pares de puntos sobre un circulo de centro X1576, lo que afirmo es que como mucho hay tres y que dos son errores por el bug de creacion de x7 x506 y x507, solo quedaría uno por analizar y hay que tener en cuenta que GG calcula menos de 3200 puntos

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The final construction: X (1576) is the center of the circumcircle of Y, Z, X (140).

Yellow dotted circle is defined by 4 points now...

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oops. I mistakenly marked X(110) as X(140) in my sketch... Can't edit the comment as of now.

Hopefully this final version of the construction clarifies the issue:

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2

Le centre de l'hyperbole équilatère passant par X1576 est sur le cercle d'Euler mais n'est pas dans ETC<3000

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I should have shared the original construction that brought my attention to X(1576)

Let 3 cevians AA1, BB1,CC1 intersect at X(140). Then draw 6 circles:

1) Through 3 points A1, X(140), C with center at point Ac.

2) Through 3 points. A1, X(140), B with center at point Ab.

through B1, X(140), C with center at Bc, and so on... Then we get X(1576) and 1 unknown point.

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Updated construction: 3 points (Ab, Ca, Bc) are collinear. 3 points (Ac, Cb, Ba) are collinear. (conjectured)

Their intersection gives us X (1576). Points Y, Z are the intersection points of 3 circles (red, green, purple).

Finally, X (1576) is the circumcenter of the triangle X (140) YZ.

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1

I mistakenly marked X (110) as X (140) in my sketch. Can't edit that comment as of now.

Hopefully this final version of the construction clarifies the issue:


Let 3 cevians AA1, BB1, CC1 intersect at X (110). Then draw 6 circles:

1) Through 3 points: A1, X (110), C with center at point Ac.

2) Through 3 points: A1, X (110), B with center at point Ab.


3) Through 3 points: B1, X (110), C with center at Bc. And so on ...

Then 3 points (Ab, Ca, Bc) are collinear. 3 points (Ac, Cb, Ba) are collinear. (conjectured)

Their intersection gives us X (1576). Points Y, Z are the intersection points of 3 circles (red, green, purple).

Finally, the ' yellow dotted circle' is defined by 4 points X (110), X (112), Y, Z and X (1576) is its center.

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Remarquable comme construction . Bravo .

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