ADGEOM 1022 - ADGEOM 1044

#1022

 

 

Happy New Year and a fruitful 2014 to everybody

Here is an easy puzzle with a new triangle center:

X(__) is the only point whose reflections in the sidelines of ABC are collinear and perspective with ABC. The perspector is a triangle center with homogeneous barycentric coordinates


__________________________

and (6,9,13)-search number 0.123280682081... The line containing these reflections is _________________, and the  rectangular circum-hyperbola through this perspector has asymptotes parallel and perpendicular to _____________________.
 It intersects the circumcircle at X(___), which is __________________  (give a simple geometric description).

Best regards
Sincerely
Paul Yiu

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#1024

hints : E574, K060, K037...

 

Could we replace the reflection triangle of P by the homothetic of the pedal triangle of P, center P, ratio k ?

 

With my best wishes for 2014

 

Best regards

 

Bernard Gibert 

 

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#1025

 

Dear Paul

Happy New Year to you and to the ADGeometers

Your conditions mean that your point lies on the circumcircle (collinearity) and on the Neuberg cubic (perspectivity). Thus your point is the isogonal conjugate of the infinite point of the Euler line and the line containing the three reflections, being its Steiner line, is the perpendicular at H to the Euler line.

I fear that we need some computation for your other questions.

Kind regards

Sincerely

Jean-Pierre Ehrmann

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#1026

 

Dear Paul

in fact, if E=X(74) is the isogonal conjugate of the infinite point of the Euler line, your new point is the isogonal conjugate of the homothetic of E in (O,-1/2)

Kind regards

Sincerely

Jean-Pierre Ehrmann

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#1027

 

Dear Bernard

"Could we replace the reflection triangle of P by the homothetic of the pedal triangle of P, center P, ratio k ?"

Yes, but we have not necessarily unicity.

If L is the point of the Euler line such as HL/OL=k-1, the homothetic in (P,k) of the pedal triangle of P is degenerated and perspective with ABC if and only if the Simson line of the antipode of P on the circumcircle goes through L. (Of course, P must lie on the circumcircle)

In any case, I suspect that you knew already the result.

Kind regards

Sincerely Jean - Pierre Ehrmann

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#1029

 

Dear Bernard

I wrote : "If L is the point of the Euler line such as HL/OL=k-1, the homothetic in (P,k) of the pedal triangle of P is degenerated and perspective with ABC if and only if the Simson line of the antipode of P on the circumcircle goes through L. (Of course, P must lie on the circumcircle)"

In this case the perspector is the point Q of the line OP such as PO/OQ=k

Kind regards

Sincerely Jean - Pierre Ehrmann

 

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#1031

 

Dear Jean-Pierre,

 

 

[JPE] I wrote : "If L is the point of the Euler line such as HL/OL=k-1, the homothetic in (P,k) of the pedal triangle of P is degenerated and perspective with ABC if and only if the Simson line of the antipode of P on the circumcircle goes through L. (Of course, P must lie on the circumcircle)"

In this case the perspector is the point Q of the line OP such as PO/OQ=k

Simson lines again ! very nice and elegant.

 

a natural question now : what is locus of Q when k varies ?

 

a nice cubic curve ?

 

Best regards

 

Bernard Gibert

 

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#1032

 

A correction

I wrote "In this case the perspector is the point Q of the line OP such as PO/OQ=k"; the correct statement is

In this case the perspector is the isogonal conjugate of the point Q of the line OP such as PO/OQ=k

Kind regards

Sincerely Jean - Pierre Ehrmann

 

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#1033

 

Dear Bernard

"a natural question now : what is locus of Q when k varies ? a nice cubic curve ?"

 

I've tried unsuccessfully to delete #1029 (and #1032) : I've forgotten "isogonal conjugate of"

The isogonal conjugate Q of the perspector (OP=k.QO) moves on a circular quintic and the perspector on a bicircular septic.

 

Best regards Jean - Pierre Ehrmann

 

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#1034

 

Dear Jean-Pierre,

 

The isogonal conjugate Q of the perspector (OP=k.QO) moves on a circular quintic and the perspector on a bicircular septic.

I find that your quintic is actually Q011.

 

http://bernard.gibert.pagesperso-orange.fr/curves/q011.html

 

Best regards

 

Bernard Gibert

 

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#1036

 

Dear Bernard

Many thanks.

So, we can try to summarize :

suppose that the Simson line of a point M of the circumcircle intersects the Euler line at L.

Then, if P is the antipode of M and k=1+LH/LO, the homothetic of the pedal triangle of P in (P,k) is degenerated and perspective with ABC. The isogonal conjugate Q of the perspector is the common point of the diameter O,M,P and of the Simson line of M (thus Q lies on Q011). More over OM/OQ=k.

If k is given, this is the only way to get the points P (at least 1, at most 3) such as the homothetic of the pedal triangle of P in (P,k) is degenerated and perspective with ABC.

Best regards. Jean - Pierre Ehrmann

 

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#1037

 

Dear Jean-Pierre,

 

If k is given, this is the only way to get the points P (at least 1, at most 3) such as the homothetic of the pedal triangle of P in (P,k) is degenerated and perspective with ABC.

 

if S is such that (k-2)OS - k OH  = 0 (vectors) then these points P lie on the rectangular hyperbola through X3, X110, X2574, X2575 and S.

 

they are the intersections of (O) and pK(X6, S).

 

Best regards
Bernard Gibert

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#1038

 

Dear Bernard

 

I expressed myself poorly (as usual in English)

Of course, there are many ways to construct our points P as the common points (or the antipodes of the common points) of the circumcircle and a rectangular hyperbola.

I was meaning that my summary gives all the solutions.

More over, it's very difficult for me to visualize the points in ETC with a large number; I have to realize that X2574-2575 are the infinite points of the Jerabek hyperbola to follow your statement.

 

Best regards. Jean - Pierre Ehrmann

 

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#1039

 

Dear Bernard

if Q* (isogonal conjugate of Q) is the perspector, then PQ* is perpendicular to the Simson line of P.

More precisely, il P' is the projection of P upon its Simson line, then Vect(PQ*) = 2k/(k+1)Vect(PP')

Kind regards Jean - Pierre Ehrmann

 

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#1040

 

Dear Paul and friends,

(6,9,13)-search number 0.123280682081 is equivalently:

cevapoint of X(13) and X(14)
isogonal conjugate of X(1511)
trilinear pole of line X(1637)X(1989)
the perspector of ABC and the reflection of the Euler triangle in the Euler line

Also, let A'B'C' be the tangential triangle of the Kiepert hyperbola.
Let A" be the intersection, other than X(3258), of the nine-point circle and line A'X(3258); define B", C" cyclically.
The lines AA", BB", CC" concur in this point.

Trilinears: sin A csc 3A/(cos A - 2 cos B cos C) : :
Barycentrics: 1/[3a^6(b^2 + c^2) - 6a^4(b^4 + c^4) + 3a^2(b^6 + c^6) - 2b^8 - 2c^8 +3b^6c^2 + 3b^2c^6 - 6b^4c^4] : :

This point lies on lines 5,1117 30,74 125,477 (at least).

Best regards,
Randy Hutson

 

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#1044

 

Dear Bernard, Peter, Jean-Pierre, Francisco Javier, and Randy,

Thank you very much for your perfect answers and generalizations to the new year puzzle. I particularly like
the description of the perspector

Y = (1/(a^2S_A-2S_{BC})(S^2-3S_{AA})) : ... : ...

as the Cevapoint of the Fermat points.

Best regards
Sincerely
Paul Yiu

 

ADGEOM 5195 * ADGEOM 5196 * ADGEOM 5197

#5195

 

Let ABC be a triangle.

 

I see that X(1), X(5), X(186), and X(1785) are concyclic.

 

Is this circle known? Which is its center?

 

Best regards

Tran Quang Hung.

 

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#5196

 

Dear Hung, 

 

the circle also goes through X11700 and X23961.

The first coordinate of the center is

a (a - b - c) (b - c) (a^5 + a^4 b - 2 a^3 b^2 - 2 a^2 b^3 + a b^4 +  b^5 + a^4 c + 2 a^2 b^2 c - 3 b^4 c - 2 a^3 c^2 + 2 a^2 b c^2 - a b^2 c^2 + 2 b^3 c^2 - 2 a^2 c^3 + 2 b^2 c^3 + a c^4 - 3 b c^4 +  c^5).

 

Best regards, 

Francisco Javier Garcia Capitan

 

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#5197

 

 Dear Tran Quang Hung the center of circle is

 

W=a (b+c-a) (b-c) (a^5+a^4 b-2 a^3 b^2-2 a^2 b^3+a b^4+b^5+a^4 c+2 a^2 b^2 c-3 b^4 c-2 a^3 c^2+2 a^2 b c^2-a b^2 c^2+2 b^3 c^2-2 a^2 c^3+2 b^2 c^3+a c^4-3 b c^4+c^5) :: (barys)

 

= on lines X(i)X(j) for these {i,j}: {522,905},{900,5901},{3738,14315}

 

=  ETC search numbers: {2.67157859398621388, 0.505268931305329665, 2.05782663993320600}

 

Best regards 

Ercole Suppa

 

ADGEOM 5185 - ADGEOM 5219

#5185

 

Dear friends, 

I introduce  four new points on Euler line: 

I found these points trying to solve a problem proposed by Kadir Altintas.

Given a triangle ABC and a point P, call DEF its circuncevian triangle. The circumcenters of six triangles PBD, PDC, PCE, PEA, PAF and PFB lie on the same conic. When this conic is a circle?

Reference:

http://garciacapitan.blogspot.com/2019/02/four-points-on-euler-line.html

Best regards,

Francisco Javier Garcia Capitan

 

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#5188

 

Hello Francisco Javier,

 

If P has trilinear coordinates tri_P = [x,y,z]

then trilinear coordinates of D,E,F are :

tri_D =  [-a*y*z, (c*y + b*z)*y, (c*y + b*z)*z]

tri_E =  [(c*x + a*z)*x, -b*x*z, (c*x + a*z)*z]

tri_F =  [(b*x + a*y)*x, (b*x + a*y)*y, -c*x*y]

 

The six points of the conic have trilinear coordinates :

tri_M1 =  [(a^3*x - a*b^2*x - a*c^2*x + a^2*c*z + b^2*c*z - c^3*z)*a,

           -(a^3*b*x - a*b^3*x + a*b*c^2*x - a^4*y + 2*a^2*b^2*y - b^4*y + 2*a^2*c^2*y + 2*b^2*c^2*y - c^4*y + 2*a^2*b*c*z),

           -(a^3*c*x + a*b^2*c*x - a*c^3*x + a^2*b^2*z - b^4*z + a^2*c^2*z + 2*b^2*c^2*z - c^4*z)]

tri_M2 =  [(a^3*x - a*b^2*x - a*c^2*x + a^2*b*y - b^3*y + b*c^2*y)*a,

           -(a^3*b*x - a*b^3*x + a*b*c^2*x + a^2*b^2*y - b^4*y + a^2*c^2*y + 2*b^2*c^2*y - c^4*y),

           -(a^3*c*x + a*b^2*c*x - a*c^3*x + 2*a^2*b*c*y - a^4*z + 2*a^2*b^2*z - b^4*z + 2*a^2*c^2*z + 2*b^2*c^2*z - c^4*z)]

tri_M3 =  [(a^4*x - a^2*b^2*x - 2*a^2*c^2*x - b^2*c^2*x + c^4*x + a^3*b*y - a*b^3*y - a*b*c^2*y),

           -(a^3*x - a*b^2*x - a*c^2*x + a^2*b*y - b^3*y + b*c^2*y)*b,

           -(2*a*b^2*c*x + a^2*b*c*y + b^3*c*y - b*c^3*y - a^4*z + 2*a^2*b^2*z - b^4*z + 2*a^2*c^2*z + 2*b^2*c^2*z - c^4*z)]

tri_M4 =  [(a^4*x - 2*a^2*b^2*x + b^4*x - 2*a^2*c^2*x - 2*b^2*c^2*x + c^4*x + a^3*b*y - a*b^3*y - a*b*c^2*y - 2*a*b^2*c*z),

           -(a^2*b*y - b^3*y + b*c^2*y - a^2*c*z - b^2*c*z + c^3*z)*b,

           -(a^2*b*c*y + b^3*c*y - b*c^3*y - a^4*z + a^2*b^2*z + 2*a^2*c^2*z + b^2*c^2*z - c^4*z)]

tri_M5 =  [(a^4*x - 2*a^2*b^2*x + b^4*x - 2*a^2*c^2*x - 2*b^2*c^2*x + c^4*x - 2*a*b*c^2*y + a^3*c*z - a*b^2*c*z - a*c^3*z),

           (a^4*y - 2*a^2*b^2*y + b^4*y - a^2*c^2*y - b^2*c^2*y - a^2*b*c*z + b^3*c*z - b*c^3*z),

           (a^2*b*y - b^3*y + b*c^2*y - a^2*c*z - b^2*c*z + c^3*z)*c]

tri_M6 =  [(a^4*x - 2*a^2*b^2*x + b^4*x - a^2*c^2*x - b^2*c^2*x + a^3*c*z - a*b^2*c*z - a*c^3*z),

           -(2*a*b*c^2*x - a^4*y + 2*a^2*b^2*y - b^4*y + 2*a^2*c^2*y + 2*b^2*c^2*y - c^4*y + a^2*b*c*z - b^3*c*z + b*c^3*z),

           -(a^3*x - a*b^2*x - a*c^2*x + a^2*c*z + b^2*c*z - c^3*z)*c]

 

Midpoints of segments M1M4, M2M5 and M3M6 are the same point :

tri_M14 =  [(2*a^4*x - 3*a^2*b^2*x + b^4*x - 3*a^2*c^2*x - 2*b^2*c^2*x + c^4*x + a^3*b*y - a*b^3*y - a*b*c^2*y + a^3*c*z - a*b^2*c*z - a*c^3*z),

            -(a^3*b*x - a*b^3*x + a*b*c^2*x - a^4*y + 3*a^2*b^2*y - 2*b^4*y + 2*a^2*c^2*y + 3*b^2*c^2*y - c^4*y + a^2*b*c*z - b^3*c*z + b*c^3*z),

            -(a^3*c*x + a*b^2*c*x - a*c^3*x + a^2*b*c*y + b^3*c*y - b*c^3*y - a^4*z + 2*a^2*b^2*z - b^4*z + 3*a^2*c^2*z + 3*b^2*c^2*z - 2*c^4*z)]

 

The circumcenter of the triangle M1M2M3 has trilinear coordinates :

tri_M =  [(a^5*x^2*y - a^3*b^2*x^2*y - a^3*c^2*x^2*y - a*b^2*c^2*x^2*y + 2*a^4*b*x*y^2 - 2*a^2*b^3*x*y^2 - 2*a^2*b*c^2*x*y^2 - b^3*c^2*x*y^2 + b*c^4*x*y^2 + a^3*b^2*y^3 - a*b^4*y^3 - a*b*c^3*x^2*z + 2*a^4*c*x*y*z - 2*a^2*b^2*c*x*y*z - 3*a^2*c^3*x*y*z - b^2*c^3*x*y*z + c^5*x*y*z + a^3*b*c*y^2*z - a*b^3*c*y^2*z - a^2*b*c^2*x*z^2),

          -y*(a^4*b*x^2 - a^2*b^3*x^2 + 2*a^3*b^2*x*y - 2*a*b^4*x*y + 2*a*b^2*c^2*x*y + a^2*b^3*y^2 - b^5*y^2 + a^2*b*c^2*y^2 + 2*b^3*c^2*y^2 - b*c^4*y^2 + a^3*b*c*x*z - a*b^3*c*x*z + a*b*c^3*x*z - a^4*c*y*z + 2*a^2*b^2*c*y*z - b^4*c*y*z + 2*a^2*c^3*y*z + 2*b^2*c^3*y*z - c^5*y*z + a^2*b*c^2*z^2),

          -(a^4*c*x^2*y + a^2*b^2*c*x^2*y - a^2*c^3*x^2*y + 2*a^3*b*c*x*y^2 + a*b^3*c*x*y^2 - a*b*c^3*x*y^2 + a^2*b^2*c*y^3 - a^2*b*c^2*x^2*z + a^3*b^2*x*y*z - a*b^4*x*y*z + a^3*c^2*x*y*z + 2*a*b^2*c^2*x*y*z - a*c^4*x*y*z + a^2*b^3*y^2*z - b^5*y^2*z + 2*a^2*b*c^2*y^2*z + 2*b^3*c^2*y^2*z - b*c^4*y^2*z - a^3*b*c*x*z^2 - a^4*c*y*z^2 + a^2*b^2*c*y*z^2 - b^4*c*y*z^2 + 2*a^2*c^3*y*z^2 + 2*b^2*c^3*y*z^2 - c^5*y*z^2)]

 

We have a circle when M14 = M ...or (by computing quadrance between these two points)  (x,y,z) satisfy the sextic form equation :

 

a^9*b*c*x^4*y^2 - 2*a^7*b^3*c*x^4*y^2 + a^5*b^5*c*x^4*y^2 - 4*a^7*b*c^3*x^4*y^2 + 2*a^5*b^3*c^3*x^4*y^2 - 2*a^3*b^5*c^3*x^4*y^2 + 5*a^5*b*c^5*x^4*y^2 - 2*a^3*b*c^7*x^4*y^2 + a^10*c*x^3*y^3 - 4*a^6*b^4*c*x^3*y^3 + 4*a^4*b^6*c*x^3*y^3 - a^2*b^8*c*x^3*y^3 - 4*a^8*c^3*x^3*y^3 - 2*a^6*b^2*c^3*x^3*y^3 - 2*a^2*b^6*c^3*x^3*y^3 + 6*a^6*c^5*x^3*y^3 - 4*a^4*c^7*x^3*y^3 + 2*a^2*b^2*c^7*x^3*y^3 + a^2*c^9*x^3*y^3 + 2*a^9*b*c*x^2*y^4 - 4*a^7*b^3*c*x^2*y^4 + 4*a^3*b^7*c*x^2*y^4 - 2*a*b^9*c*x^2*y^4 - 4*a^7*b*c^3*x^2*y^4 + 2*a^5*b^3*c^3*x^2*y^4 - 4*a^3*b^5*c^3*x^2*y^4 + 2*a*b^7*c^3*x^2*y^4 - 4*a^3*b^3*c^5*x^2*y^4 + 4*a^3*b*c^7*x^2*y^4 + 2*a*b^3*c^7*x^2*y^4 - 2*a*b*c^9*x^2*y^4 + a^8*b^2*c*x*y^5 - 4*a^6*b^4*c*x*y^5 + 4*a^4*b^6*c*x*y^5 - b^10*c*x*y^5 - 2*a^4*b^4*c^3*x*y^5 + 2*b^8*c^3*x*y^5 - 2*a^4*b^2*c^5*x*y^5 - 2*b^4*c^7*x*y^5 + b^2*c^9*x*y^5 - a^5*b^5*c*y^6 + 2*a^3*b^7*c*y^6 - a*b^9*c*y^6 - 2*a^3*b^5*c^3*y^6 + 2*a*b^7*c^3*y^6 - a*b^5*c^5*y^6 - 2*a^7*b^2*c^2*x^4*y*z + 4*a^5*b^4*c^2*x^4*y*z - 2*a^3*b^6*c^2*x^4*y*z + 8*a^5*b^2*c^4*x^4*y*z - 6*a^3*b^2*c^6*x^4*y*z + a^10*b*x^3*y^2*z - 2*a^8*b^3*x^3*y^2*z + 2*a^4*b^7*x^3*y^2*z - a^2*b^9*x^3*y^2*z - 8*a^8*b*c^2*x^3*y^2*z + 6*a^6*b^3*c^2*x^3*y^2*z + 4*a^4*b^5*c^2*x^3*y^2*z - 2*a^2*b^7*c^2*x^3*y^2*z + 18*a^6*b*c^4*x^3*y^2*z + 2*a^4*b^3*c^4*x^3*y^2*z - 4*a^2*b^5*c^4*x^3*y^2*z - 16*a^4*b*c^6*x^3*y^2*z + 2*a^2*b^3*c^6*x^3*y^2*z + 5*a^2*b*c^8*x^3*y^2*z + a^11*x^2*y^3*z - 4*a^7*b^4*x^2*y^3*z + 2*a^5*b^6*x^2*y^3*z + 3*a^3*b^8*x^2*y^3*z - 2*a*b^10*x^2*y^3*z - 6*a^9*c^2*x^2*y^3*z - 4*a^7*b^2*c^2*x^2*y^3*z + 10*a^5*b^4*c^2*x^2*y^3*z + 14*a^7*c^4*x^2*y^3*z + 2*a^5*b^2*c^4*x^2*y^3*z - 8*a^3*b^4*c^4*x^2*y^3*z - 16*a^5*c^6*x^2*y^3*z + 4*a^3*b^2*c^6*x^2*y^3*z + 6*a*b^4*c^6*x^2*y^3*z + 9*a^3*c^8*x^2*y^3*z - 2*a*b^2*c^8*x^2*y^3*z - 2*a*c^10*x^2*y^3*z + 2*a^10*b*x*y^4*z - 3*a^8*b^3*x*y^4*z - 2*a^6*b^5*x*y^4*z + 4*a^4*b^7*x*y^4*z - b^11*x*y^4*z - 6*a^8*b*c^2*x*y^4*z + 4*a^6*b^3*c^2*x*y^4*z + 2*a^4*b^5*c^2*x*y^4*z + 4*a^6*b*c^4*x*y^4*z - 10*a^4*b^3*c^4*x*y^4*z + 2*a^2*b^5*c^4*x*y^4*z + 4*b^7*c^4*x*y^4*z + 4*a^4*b*c^6*x*y^4*z + 4*a^2*b^3*c^6*x*y^4*z - 2*b^5*c^6*x*y^4*z - 6*a^2*b*c^8*x*y^4*z - 3*b^3*c^8*x*y^4*z + 2*b*c^10*x*y^4*z + a^9*b^2*y^5*z - 2*a^7*b^4*y^5*z + 2*a^3*b^8*y^5*z - a*b^10*y^5*z - 2*a^5*b^2*c^4*y^5*z - 2*a^3*b^4*c^4*y^5*z + 4*a*b^6*c^4*y^5*z - 4*a*b^4*c^6*y^5*z + a*b^2*c^8*y^5*z - 4*a^3*b^3*c^5*x^4*z^2 - 2*a^8*b^2*c*x^3*y*z^2 + 4*a^6*b^4*c*x^3*y*z^2 - 2*a^4*b^6*c*x^3*y*z^2 + 12*a^6*b^2*c^3*x^3*y*z^2 - 4*a^2*b^6*c^3*x^3*y*z^2 - 14*a^4*b^2*c^5*x^3*y*z^2 + 4*a^2*b^2*c^7*x^3*y*z^2 - 4*a^9*b*c*x^2*y^2*z^2 + 4*a^7*b^3*c*x^2*y^2*z^2 + 2*a^5*b^5*c*x^2*y^2*z^2 - 2*a*b^9*c*x^2*y^2*z^2 + 18*a^7*b*c^3*x^2*y^2*z^2 + 4*a^5*b^3*c^3*x^2*y^2*z^2 - 6*a^3*b^5*c^3*x^2*y^2*z^2 - 28*a^5*b*c^5*x^2*y^2*z^2 + 4*a^3*b^3*c^5*x^2*y^2*z^2 + 2*a*b^5*c^5*x^2*y^2*z^2 + 18*a^3*b*c^7*x^2*y^2*z^2 + 4*a*b^3*c^7*x^2*y^2*z^2 - 4*a*b*c^9*x^2*y^2*z^2 - 2*a^10*c*x*y^3*z^2 - 2*a^8*b^2*c*x*y^3*z^2 + 6*a^6*b^4*c*x*y^3*z^2 - 2*b^10*c*x*y^3*z^2 + 9*a^8*c^3*x*y^3*z^2 + 4*a^6*b^2*c^3*x*y^3*z^2 - 8*a^4*b^4*c^3*x*y^3*z^2 + 3*b^8*c^3*x*y^3*z^2 - 16*a^6*c^5*x*y^3*z^2 + 2*a^4*b^2*c^5*x*y^3*z^2 + 10*a^2*b^4*c^5*x*y^3*z^2 + 2*b^6*c^5*x*y^3*z^2 + 14*a^4*c^7*x*y^3*z^2 - 4*a^2*b^2*c^7*x*y^3*z^2 - 4*b^4*c^7*x*y^3*z^2 - 6*a^2*c^9*x*y^3*z^2 + c^11*x*y^3*z^2 - 2*a^9*b*c*y^4*z^2 + 2*a^7*b^3*c*y^4*z^2 + 2*a^3*b^7*c*y^4*z^2 - 2*a*b^9*c*y^4*z^2 + 4*a^7*b*c^3*y^4*z^2 - 4*a^5*b^3*c^3*y^4*z^2 - 4*a^3*b^5*c^3*y^4*z^2 + 4*a*b^7*c^3*y^4*z^2 + 2*a^3*b^3*c^5*y^4*z^2 - 4*a^3*b*c^7*y^4*z^2 - 4*a*b^3*c^7*y^4*z^2 + 2*a*b*c^9*y^4*z^2 - 8*a^4*b^3*c^4*x^3*z^3 + 4*a^7*b^2*c^2*x^2*y*z^3 - 4*a^3*b^6*c^2*x^2*y*z^3 - 14*a^5*b^2*c^4*x^2*y*z^3 - 2*a*b^6*c^4*x^2*y*z^3 + 12*a^3*b^2*c^6*x^2*y*z^3 + 4*a*b^4*c^6*x^2*y*z^3 - 2*a*b^2*c^8*x^2*y*z^3 + 5*a^8*b*c^2*x*y^2*z^3 + 2*a^6*b^3*c^2*x*y^2*z^3 - 4*a^4*b^5*c^2*x*y^2*z^3 - 2*a^2*b^7*c^2*x*y^2*z^3 - b^9*c^2*x*y^2*z^3 - 16*a^6*b*c^4*x*y^2*z^3 + 2*a^4*b^3*c^4*x*y^2*z^3 + 4*a^2*b^5*c^4*x*y^2*z^3 + 2*b^7*c^4*x*y^2*z^3 + 18*a^4*b*c^6*x*y^2*z^3 + 6*a^2*b^3*c^6*x*y^2*z^3 - 8*a^2*b*c^8*x*y^2*z^3 - 2*b^3*c^8*x*y^2*z^3 + b*c^10*x*y^2*z^3 + a^9*c^2*y^3*z^3 + 2*a^7*b^2*c^2*y^3*z^3 - 2*a^3*b^6*c^2*y^3*z^3 - a*b^8*c^2*y^3*z^3 - 4*a^7*c^4*y^3*z^3 + 4*a*b^6*c^4*y^3*z^3 + 6*a^5*c^6*y^3*z^3 - 2*a^3*b^2*c^6*y^3*z^3 - 4*a*b^4*c^6*y^3*z^3 - 4*a^3*c^8*y^3*z^3 + a*c^10*y^3*z^3 - 4*a^5*b^3*c^3*x^2*z^4 - 6*a^6*b^2*c^3*x*y*z^4 - 2*a^2*b^6*c^3*x*y*z^4 + 8*a^4*b^2*c^5*x*y*z^4 + 4*a^2*b^4*c^5*x*y*z^4 - 2*a^2*b^2*c^7*x*y*z^4 - 2*a^7*b*c^3*y^2*z^4 - 2*a^3*b^5*c^3*y^2*z^4 + 5*a^5*b*c^5*y^2*z^4 + 2*a^3*b^3*c^5*y^2*z^4 + a*b^5*c^5*y^2*z^4 - 4*a^3*b*c^7*y^2*z^4 - 2*a*b^3*c^7*y^2*z^4 + a*b*c^9*y^2*z^4 = 0

 

Dominique

dominique laurain

 

-----------------------------------------------------------------

#5210

Hello,

 

I recall the original subject : given a reference triangle ABC and a point P inside it, the pedal triangle of P is DEF (vertices are intersections of AP,BP,CP lines with circumcircle).

Then six points M1,M2,M3,M4,M5,M6 are circumcenters of triangles PBD, PDC,PCE,PEA,PAF,PFB.

 

The original post is : the six points are lying on an ellipse.

I have found more about that ellipse : be x:y:z trilinear coordinates of P then center of the ellipse is the point with barycentric coordinates c*(a*y + b*x) , a*(b*z + c*y) and b*(a*z + c*x) with respect to the points M1,M3,M5

 

The ellipse is the inellipse of a triangle vertices H1,H2,H3 with barycentric coordinates

z*(a*y + b*x),x*(b*z + c*y),-y*(a*z + c*x),

-z*(a*y + b*x),x*(b*z + c*y),y*(a*z + c*x)

and z*(a*y + b*x),-x*(b*z + c*y),y*(a*z + c*x)

with respect to M1,M3,M5.

 

The foci of the ellipse are the points which can be deduced from Linfield theorem (see https://en.wikipedia.org/wiki/Marden%27s_theorem) using the cutting ratios (a/b)*(y/x) , (a/c)*(z/x) and 1 given complex coordinates z1,z2,z3 of the M1,M3,M5 points.

 

Cuting ratios for edges of H1H2H3 triangle were used to deduce Hi points from Mi points :

H1M3 / H2M3 = (c*y)/(b*z) =>  (c*y + b*z) M3 =  (b*z) H1 + (c*y) H2

H2M5 / H3M5 = (a*z)/(c*x) =>  (a*z + c*x) M5 =  (c*x) H2 + (a*z) H3

H3M1 / H1M1 = (b*x)/(a*y) =>  (b*x + a*y) M1 =  (a*y) H3 + (b*x) H1

 

Dominique

dominique laurain

 

-----------------------------------------------------------------

#5211

Dear all,

 

I would like to add a comment: 

 

Four circles (M1M2M3), (M1M5M6), (M2M6M4), (M3M4M5) are concurrent.

Four circles (M4M5M6), (M4M2M3), (M5M3M1), (M6M1M2)  are  also concurrent.

 

This holds whenever  M1M4, M2M5 and M3M6 share the same midpoint.  I discovered this fact when solving another problem by Kadir Altintas.

 

Best regards,

 

Vu Thanh Tung

 

-----------------------------------------------------------------

#5212

 

Dear Vu Thanh Tung and Francisco,

 

The opposite sides of the hexagon are parallel and hence the

circumcenters M1,M2,M3,M4,M5,M6  are lying on a conic

and we can prove with center Q that is the mid point of the diagonals

M1M4, M2M5, M3M6 and also mid point of PO (O=circumenter of ABC).

If we try to find when the conic is a circle by finding the locus of P

when QM1 = QM2 = .  . .

we get three coaxial  circles and the points are X(5000), X(5001).

The contruction of these circles is as follows:

Let A1B1C1 be the orthic triangle

Let A2B2C2 be the cevian triangle of K the lemoine point.

Let A3, B3, C3 be the intersections of the line sides of A2B2C2

with the sides of ABC. A3B3C3 is a line the trilinear polar of K.

The circumcircles of the triangles AA1A3, BB1B3, CC1C3

are the three circles that give the two points iff ABC is acute angled

for which the conic of M1,M2,M3,M4,M5,M6 is a circle.

 

Nikolaos Dergiades

 

 

-----------------------------------------------------------------

#5213

 

Dear all,

About the concurrency,  we  can have generalization. 

 

Let A , B , C , D ,E , F  be 6 points on a circle (O).

Let P be another point on the plane.

Let  M1 , M2 , M3 , M4 , M5 , M6  be circumcenter of PAB, PBC, PCD , PDE , PEF, PFA.

Then: 

 

(1) M1M4 ,  M2M5, M3M6 are concurrent at a point X

 

(2) if additionally AD , BE , CF are concurrent at point K then:

(2.1)  O , K , X are collinear.  

(2.2)   if <OPK = 90° (or P on the circle of diameter OK) then X is midpoint of OK

(2.3)(original problem) if M=K then X is also the midpoint of M1M4 ,  M2M5, M3M6.

 

Best regards,

 

Vu Thanh Tung

 

-----------------------------------------------------------------

#5214

 

Dear all,

 

The concurrency problem was already posted in AoPS:

 

https://artofproblemsolving.com/community/c6h303504

 

https://artofproblemsolving.com/community/u239653h626275p3755959

Best regards,

 

Vu Thanh Tung

 

-----------------------------------------------------------------

#5219

 

The ellipse is degenerated to a circle when foci of the ellipse collapse into the same point.

P point has trilinear coordinates x:y:z

My equation in complex numbers from the Linfield theorem is :

 

f'(z) = alpha*z^2 + beta*z + gamma

with f(z) = (z-z1)^i (z-z2)^j (z-z3)^k

whose roots are

z1 = ( z*(a*y + b*x)*m1 + x*(b*z + c*y)*m3 - y*(a*z + c*x)*m5 ) /(2*b*x*z)

z2 = (-z*(a*y + b*x)*m1 + x*(b*z + c*y)*m3 + y*(a*z + c*x)*m5 ) /(2*c*x*y)

z3 = ( z*(a*y + b*x)*m1 - x*(b*z + c*y)*m3 + y*(a*z + c*x)*m5 ) /(2*a*y*z)

equivalent complex numbers in the complex plane for H1,H2,H3

 

m1,m3,m5 are complex numbers for M1,M3,M5

 

i = (a/b)*(y/x) ; j = (a/c)*(z/x) ; k = 1

alpha = i + j + k ; beta = -((z2 + z3)*i + (z3 + z1)*j + (z1 + z2)*k) ; gamma = z2*z3*i + z3*z1*j + z1*z2*k

 

Square discriminant of the equation f'(z) = 0 (whose roots are foci) is : 

disc2 =  (c*x*y*(m3 - m5)^2 + b*x*z*(m1 - m3)^2 + a*y*z*(m1 - m5)^2)*(b*x + a*y)*(c*x + a*z)*(c*y + b*z)/(b^2*c^2*x^3*y*z)

 

So unique root when disc2 = 0 or

 

c*x*y*(m3 - m5)^2 + b*x*z*(m1 - m3)^2 + a*y*z*(m1 - m5)^2 = 0

 

Dominique

 

 

ADGEOM 5177 * ADGEOM 5182

#5177

Let ABC be a triangle with centroid G. A'B'C' is pedal triangle of G.

 

Then reflections of Brocad axis of triangles GBC, GCA, GAB in GA', GB', GB' respectively are concurrent.

 

Which is this concurrent point?

 

Best regards,

Tran Quang Hung.

---------------------------------------------------

#5182

[Tran Quang Hung]

 

Let ABC be a triangle with centroid G. A'B'C' is pedal triangle of G.

 

Then reflections of Brocad axis of triangles GBC, GCA, GAB in GA', GB', GB' respectively are concurrent.

 

Which is this concurrent point?

 

Best regards,

Tran Quang Hung.

 

 

--------------------------------------------------------------------------------------------

 

 

[Ercole Suppa]

 

 

Then reflections of Brocad axis of triangles GBC, GCA, GAB in GA', GB', GB' respectively concur at point:

 

 

X = 77 a^10-205 a^8 b^2+101 a^6 b^4+116 a^4 b^6-124 a^2 b^8+35 b^10-205 a^8 c^2+107 a^6 b^2 c^2+90 a^4 b^4 c^2+47 a^2 b^6 c^2-91 b^8 c^2+101 a^6 c^4+90 a^4 b^2 c^4+90 a^2 b^4 c^4+56 b^6 c^4+116 a^4 c^6+47 a^2 b^2 c^6+56 b^4 c^6-124 a^2 c^8-91 b^2 c^8+35 c^10 :: (barys)

 

= (1188 R^2-27 SB-27 SC-414 SW)S^4 + (-324 R^2 SB SC+162 SB SC SW-9 SB SW^2-9 SC SW^2+16 SW^3)S^2 +12 SB SC SW^3 :: (barys)

 

= on lines X(i)X(j) for these {i,j}: {15694,24206}

 

= (6-9-13) search numbers [0.887705658890037360, -0.587071572915610678, 3.63738834366901541]

 

 

Best regards

Ercole Suppa

 

 

ADGEOM 5178 * ADGEOM 5181

#5178

Let ABC be a triangle P=X(35).

A'B'C' is pedal triangle of P.

Then reflections of Euler lines of triangles PBC, PCA, PAB in PA', PB', PC' respectively are concurrent.

Which is this concurrent point?

Best regards,
Tran Quang Hung.

---------------------------------------------------

#5181

 

[Tran Quang Hung]

 

Let ABC be a triangle P=X(35).

 

A'B'C' is pedal triangle of P.

 

Then reflections of Euler lines of triangles PBC, PCA, PAB in PA', PB', PC' respectively are concurrent.

 

Which is this concurrent point?

 

Best regards,

Tran Quang Hung.

 

--------------------------------------------------------------------------------------------

 

 

[Ercole Suppa]

 

 

Then reflections of Euler lines of triangles PBC, PCA, PAB in PA', PB', PC' respectively concur at point:

 

 

X = midpoint of X(21) and X(3871)

 

= a^2 (a^5-b^5-b^4 c+b^3 c^2+b^2 c^3-b c^4-c^5-a^4 (b+c)-2 a^3 (b^2+b c+c^2)+a (b^2+b c+c^2)^2+2 a^2 (b^3+b^2 c+b c^2+c^3)) :: (barys)

 

= 2 R S^2 + (5 a R^2-b R^2-c R^2-a SB+b SB+c SB-a SC+b SC+c SC)S -2 R SB SC :: (barys)

 

= on lines X(i)X(j) for these {i,j}: {1,27086},{8,21},{30,11491},{35,758},{100,442},{191,3811},{404,11281},{1392,6767},{1621,6675},{2475,3085},{2646,4996},{2802,3746},{3241,10267},{3647,3935},{3649,14882},{3651,5758},{3874,14799},{3957,14798},{4421,6175},{4428,15671},{5010,16126},{5428,5844},{5703,11507},{7676,17768},{9780,11517},{10528,15680},{11510,20057},{12867,13615},{13607,17009},{15674,19843},{25440,26725}

 

= midpoint of X(i) and X(j) for these {i,j}: {21,3871}

 

= reflection of X(i) in X(j) for these {i,j}: {24390,6675}

 

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: 

 

= (6-9-13) search numbers [2.26338369616380577, 2.58854507743396244, 0.803956183916033209]

 

 

Best regards

Ercole Suppa

 

 

ADGEOM 5172 * ADGEOM 5175 * ADGEOM 5176

#5172

Let ABC be a triangle with orthocenter H.

 

Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively are concurrent.

 

Which is this concurrent point?

 

Best regards,

Tran Quang Hung.

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#5175

 

[Tran Quang Hung]

 

Let ABC be a triangle with orthocenter H.

 

Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively are concurrent.

 

Which is this concurrent point?

 

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[Ercole Suppa]

 

Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively concur at point

 

X = midpoint of X(382) and X(13512)

 

= a^16-3 a^14 b^2+a^12 b^4+6 a^10 b^6-10 a^8 b^8+9 a^6 b^10-7 a^4 b^12+4 a^2 b^14-b^16-3 a^14 c^2+4 a^12 b^2 c^2+4 a^10 b^4 c^2-8 a^8 b^6 c^2+11 a^4 b^10 c^2-13 a^2 b^12 c^2+5 b^14 c^2+a^12 c^4+4 a^10 b^2 c^4-3 a^8 b^4 c^4-7 a^4 b^8 c^4+15 a^2 b^10 c^4-10 b^12 c^4+6 a^10 c^6-8 a^8 b^2 c^6+6 a^4 b^6 c^6-6 a^2 b^8 c^6+11 b^10 c^6-10 a^8 c^8-7 a^4 b^4 c^8-6 a^2 b^6 c^8-10 b^8 c^8+9 a^6 c^10+11 a^4 b^2 c^10+15 a^2 b^4 c^10+11 b^6 c^10-7 a^4 c^12-13 a^2 b^2 c^12-10 b^4 c^12+4 a^2 c^14+5 b^2 c^14-c^16 :: (barys)

 

= S^4 + (3 R^4-5 R^2 SB-5 R^2 SC+SB SC+R^2 SW+2 SB SW+2 SC SW-SW^2)S^2 -27 R^4 SB SC+23 R^2 SB SC SW-5 SB SC SW^2 :: (barys)

 

= on circumcircle of JohnsonTriangle

 

= on lines X(i)X(j) for these {i,j}: {3,128},{4,11671},{5,49},{30,930},{137,381},{140,23237},{382,13512},{523,14980},{539,16337},{546,1263},{550,6592},{1209,11016},{1478,3327},{1479,7159},{2070,23181},{2072,23319},{3153,20957},{3627,14073},{3652,10747},{3850,25147},{3851,23516},{3853,23238},{11459,13504},{12111,13505},{12254,14071},{13556,22823},{14050,18439},{14142,21230},{16336,18400}

 

= combo: X[3]-2*X[128], 3*X[4]-X[11671], 2*X[137]-3*X[381], 2*X[140]-3*X[23237], X[382]+X[13512], 2*X[546]-X[1263], X[550]-2*X[6592], X[3627]+X[14073], 4*X[3850]-3*X[25147], 7*X[3851]-6*X[23516], 2*X[3853]+X[23238], 3*X[11459]-X[13504], X[12111]+X[13505], X[12254]-2*X[14071]

 

= midpoint of X(i) and X(j) for these {i,j}: {382,13512},{12111,13505}

 

= reflection of X(i) in X(j) for these {i,j}: {3,128},{550,6592},{930,14072},{1141,5},{1263,546},{14674,14769},{19553,16337}

 

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: 

 

= (6-9-13) search numbers [1.68062122615268063, -2.17060108179101817, 4.36771697261538524]

 

 

Best regards

Ercole Suppa

 

 

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#5176

 

Let ABC be a triangle with orthocenter H.

 

Then reflections of line X(3)X(54) of triangles HBC, HCA, HAB in HA, HB, HC respectively are concurrent.

 

Which is this concurrent point?

 

Best regards,

Tran Quang Hung.

 

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Q = X(3)X(128) ∩ X(5)X(49)

= S^4-(R^2*(3*R^2+5*SA-4*SW)-2*SA^2+3*SB*SC+SW^2)*S^2+(R^2*(27*R^2-23*SW)+5*SW^2)*SB*SC : : (barys)

= (6*cos(2*A)+4*cos(4*A)+5)*cos(B-C)-2*(cos(A)+cos(3*A))*cos(2*(B-C))+(2*cos(2*A)+1)*cos(3*(B-C))-cos(5*A)-3*cos(A)-cos(3*A) :: (trilinears)

= 3*X(3)-4*X(13372), 3*X(4)-X(11671), 3*X(5)-2*X(12026), 9*X(5)-8*X(25339), 3*X(128)-2*X(13372), 2*X(137)-3*X(381), 2*X(140)-3*X(23237), 4*X(3850)-3*X(25147), 7*X(3851)-6*X(23516), 2*X(3853)+X(23238), 3*X(11459)-X(13504), 3*X(12026)-4*X(25339)

= lies on these lines: {3, 128}, {4, 11671}, {5, 49}, {30, 930}, {137, 381}, {140, 23237}, {382, 13512}, {523, 14980}, {539, 16337}, {546, 1263}, {550, 6592}, {1209, 11016}, {1478, 3327}, {1479, 7159}, {2070, 23181}, {2072, 23319}, {3153, 20957}, {3627, 14073}, {3652, 10747}, {3850, 25147}, {3851, 23516}, {3853, 23238}, {11459, 13504}, {12111, 13505}, {12254, 14071}, {13556, 22823}, {14050, 18439}, {14142, 21230}, {16336, 18400}

= midpoint of X(i) and X(j) for these {i,j}: {382, 13512}, {3627, 14073}, {12111, 13505}

= reflection of X(i) in X(j) for these (i,j): (3, 128), (550, 6592), (930, 14072), (1141, 5), (1263, 546), (12254, 14071), (14674, 14769), (19553, 16337)

= reflection of X(265) in the line X(4)X(14050)

= (Ehrmann-side)-isogonal conjugate of-X(1510)

= (Johnson)-isogonal conjugate of-X(1154)

= [ 1.6806212261526800, -2.1706010817910170, 4.3677169726153860 ]

 

César Lozada

ADGEOM 5137 * ADGEOM 5144

#5137

Dear geometers,

 

Let ABC be a triangle.

 

d is a line passing through circumcenter of ABC.

 

A' is reflection of A in line d.

 

A" is reflection of A' in BC.

 

Define similarly the points B', B", C', C".

 

Then triangles ABC and A"B"C" are symmetric in point P on NPC of ABC. 

 

Circle (A"B"C") passes through orthocenter H of ABC.

 

Then Simson line d1 of H wrt A"B"C" is perpendicular to line d.

 

The line passing through H and is parallel to d which meets circle (A"B"C") at H'.

 

Steiner line d2 of H' wrt A"B"C" passes through circumcenter O of ABC. 

 

Orthopole Q of d2 wrt ABC lies on d1.

 

If d=Euler line, d=OI line, which is the point P, H'and Q?

 

Best regards,

Tran Quang Hung.

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#5144

 

 [Tran Quang Hung]:

Let ABC be a triangle.

d is a line passing through circumcenter of ABC.

A' is reflection of A in line d.
A" is reflection of A' in BC.

Define similarly the points B', B", C', C".

Then triangles ABC and A"B"C" are symmetric in point P on NPC of ABC.

Circle (A"B"C") passes through orthocenter H of ABC.

Then Simson line d1 of H wrt A"B"C" is perpendicular to line d.

The line passing through H and is parallel to d which meets circle (A"B"C") at H'.

Steiner line d2 of H' wrt A"B"C" passes through circumcenter O of ABC.

Orthopole Q of d2 wrt ABC lies on d1.

If d=Euler line, d=OI line, which is the point P, H'and Q?

*** If d=Euler line:
P=X(125). 
H' =X(17511).
Q=X(3258) = reflection of X(125) in Euler line.

*** If   d=OI line: 
 P=X(11).
  
 H'= a^6+6 a^4 b c-2 a^5 (b+c)-2 a^3 b c (b+c)-(b-c)^4 (b+c)^2+a^2 b c (3 b^2-5 b c+3 c^2)+a (b-c)^2 (2 b^3-3 b^2 c-3 b c^2+2 c^3) : ... : ...
  lies on lines X(i)X(j) for these {i, j}: {4,8}, {11,901}, {100,3259}, {149,513}, {497,14115}, {528,14513}, {953,5840}, {2829,14511}, {3025,13274}, {3583,10774}, {4380,17036}, {5854,17101}, {6075,10707}, {6789,11813}, {8047,23813}, {13273,13756},
and is the reflection of X(i) in X(j), for these {i, j}: {100,3259},  {901,11} .

 Q=X(3259) = complement of X(901).
 
 *** If   d=Brocard axis:
 P=X(115).
 
 H' = a^2 (a^8 b^2 c^2-2 a^2 b^4 c^4 (b^2+c^2)-a^6 (b^6+b^4 c^2+b^2 c^4+c^6)+a^4 (b^8+4 b^4 c^4+c^8)-b^2 c^2 (b^8-3 b^6 c^2+3 b^4 c^4-3 b^2 c^6+c^8)) : ... : ...,
 lies on lines X(i)X(j) for these {i, j}: {4,69}, {99,2679}, {115,805}, {148,512}, {543,14509}, {671,6071}, {2549,14113}, {2698,23698}, {2794,14510}, {14061,22103},
and is the reflection of X(i) in X(j), for these {i, j}: {99,2679}, {805,115}.
 
 Q=X(2679).
 
 Angel Montesdeoca