ADGEOM 2538 * ADGEOM 2539 * ADGEOM 2540 * ADGEOM 2541 * ADGEOM 2542

 

#2538

Dear friends:

May someone please explain me what are McCay circles?

According to http://mathworld.wolfram.com/McCayCircles.html, the A-McCay circle is the circle through the centroid G of a triangle ABC and the vertices B’ and C’ of the 2^nd Brocard triangle A’B’C’ of ABC. In this way, indicated center function and radius  of the A-McCay circle are both right.

But I don´t get this affirmation in the same web page:

>> The circumcircle of their centers (i.e., of the second Brocard triangle) is therefore the Brocard circle. The circle through the McCay-circles centers is  not the the Brocard circle (circle with diameter [OK], O=circumcenter and K=symmedian point )

More:

There is another web page:

http://mathworld.wolfram.com/McCayCirclesRadicalCircle.html

 

In this last page, the radical circle of McCay circles is described and its center<>G and radius<>0 are given.

How is it possible if McCay circles all pass through G?

Thanks in advance

César Lozada

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#2539

Hi César,

This is one of several errors in Wolfram MathWorld.  I have attempted to send them corrections in the past, but never get a response.  I don't know if the site is being actively managed.  The circle through the centers of the McCay circles is a new circle (the 'McCay circumcircle'?), with center on lines 2,99 3,1153 5,524 381,2080 (at least).

Best regards,
Randy Hutson

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#2540

Hello Randy,

Thank you very much.

.. and what about the circle described in http://mathworld.wolfram.com/McCayCirclesRadicalCircle.html ? What circle is it?

Thanks again

César Lozada

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#2541

Hi César,

This, too, seems to be in error.  Since the McCay circles all include G, this would be the center of their radical circle, not the function given in MathWorld.  This function, in fact, turns out to be X(182) of the 'McCay triangle' (triangle formed by the centers of the McCay circles), and has barycentrics 5a^6 + 2b^6 + 2c^6 - 6a^4(b^2 + c^2) - 18a^2b^2c^2 - 3b^4c^2 - 3b^2c^4 : :, on line {2,6}, ETC search: 3.465193324606316.  MathWorld also states 'The McCay circles radical circle is real only in the case that the McCay circles do not intersect.'  But again, since they all pass through G, they always intersect.

Some interesting results for the McCay triangle, MaMbMc and McCay circles:

MaMbMc is perspective to ABC at: isogonal conjugate of X(576) = trilinear product of centers of McCay circles, on the Kiepert hyperbola and lines 2,575 3,671 4,187 5,598 76,140 (at least).  ETC search: 1.888683841922814.

Let Ma' be the reflection of Ma in line BC, and define Mb', Mc' cyclically.
Ma'Mb'Mc' is perspective to ABC at: isogonal conjugate of X(575) = trilinear product Ma'*Mb'*Mc', on the Kiepert hyperbola and lines 2,576 3,598 4,574 5,671 (at least).  ETC search: 0.809176513111329

MaMbMc is perspective to the excentral triangle at: {X(6191),X(6192)}-harmonic conjugate of X(3496), on lines 1,576 4,9 (at least). ETC search: 6.147351386644087

MaMbMc is similar to the 4th Brocard triangle with similitude center X(111).

X(2) of MaMbMc = X(2)
X(6) of MaMbMc = X(182)
X(23) of MaMbMc = X(111)
X(69) of MaMbMc = X(1352)
X(98) of MaMbMc = antipode of X(381) in McCay circumcircle, on lines 2,6 3,543 98,6233 (at least)
X(99) of MaMbMc = X(381)
X(325) of MaMbMc = X(114)
X(385) of MaMbMc = X(98)
X(542) of  MaMbMc = X(524)

Let Sa be the insimilicenter of the B- and C-McCay circles, and define Sb, Sc cyclically.
SaSbSc is perspective to MaMbMc at combo: Ma/Ra + Mb/Rb + Mc/Rc, where Ra, Rb, Rc are the respective radii of the McCay circles.  On lines 1,576 2,2783 37,517 (at least).  ETC search: 3.065607549874190.

The exsimilicenters of pairs of McCay circles lie on the respective sidelines of the excentral triangle.

Let Sa' be the insimilicenter of the A-McCay circle and A-Neuberg circle, and define Sb', Sc' cyclically.
Sa'Sb'Sc' is perspective to ABC at: isogonal conjugate of X(1351), on lines 2,3167 3,2996 4,230 5,5395 69,6036 76,631 94,7493 98,5033 (at least).  ETC search: 3.673650171303900.

The exsimilicenter of the A-McCay circle and A-Neuberg circle is the midpoint of BC, and cyclically...

The triangle bounded by the radical axes of the McCay circles and the corresponding Neuberg circle, is the reflection of ABC in X(6).

Let Ta be the radical trace of the A-McCay circle and A-Neuberg circle, and define Tb, Tc cyclically.
TaTbTc is perspective to ABC at X(2996).

Also, the McCay circumcircle is orthogonal to the Parry circle.

inverse-in-McCay-circumcircle of X(2) = X(111)


Best regards,
Randy Hutson

 

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#2542

Hello, Randy

 

Thanks again for your exhaustive analysis.

 

I’d like to add:

 

1)  The tangent circle to McCay circles has:

Radius Rm=r*sqrt(|3*S^2-SW^2|)/|-3*s^2+4*SW|

and center 

Om = a^3-(b+c)*a^2 - (b^2+c^2-6*b*c)*a - 3*(b^2-c^2)*(b-c) :  :  (barycentrics)

     = On lines: (2,846), (7,1738), (10,4862), (277,5805) , (346,3836) and others

     = ( 4.753992682847906, -3.62712034320988, 3.957597327430649 )

No ETC-centers lie on this circle.

The triangle limited by the tangents to this circle at touchpoints with McCay circles is perspective to the excentral triangle of ABC at:

Z = a*(a^4-3*(b+c)*a^3+6*a^2*b*c+3*(b^2-c^2)*(b-c)*a-(b^2+c^2)*(b-c)^2) : : (barycentrics)

  = On lines: (9,4454), (373,4414), (1445,2347)

  = ( 6.757543407894820, 8.30760797988599, -5.229622615503528 )

 

2)  The bisector circle of McCay circles has:

Radius R’m=sqrt(|(3-cot(o)^2)*( 27*R^2-S*(6*cot(o)+cot(o)^3))|)/9

where o is the Brocard angle of ABC. It has center:

O’m = a^4 + 2*(b^2+c^2)*a^2 - 5*b^4 + 14*b^2*c^2 - 5*c^4 : : (barycentrics)

    = Reflection of: (376/5569)

    = On lines: (2,99), (4,3849), (6,3363), (262,538), (376,5569) and others

    = ( 3.456133116674531, -1.58714230336286, 3.144316945770420 )

No ETC-centers lie on this circle.

The triangle bounded by the radical axis of this circle and McCay circles is perspective to 1^st and 2^nd Brocard triangle of ABC at X(2)

 

Best regards

César Lozada