HYACINTHOS 29540

[Kadir Altintas]:

 

Let ABC be a triangle, P a point and DEF the pedal triangle of P.

 

Denote:

 

Oa = the circumcenter of AFE 

Ka = the symmedian point of PEF. 

Define Ob, Oc, Kb, Kc cyclically. 

The locus of P such that the lines OaKa, ObKb, OcKc are concurrent is the McCay cubic.

[Antreas P. Hatzipolakis]:

Is this correct?

If no, which is it ?
Which are the points of concurrence for some points on the locus?

[Peter Moses]:

Hi Antreas,

> If no, which is it?

Neuberg cubic K001

X(1) -> X(1)
X(3) -> X(1216)
X(4) -> X(6750)

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X(13);

 

 MIDPOINT OF X(13) AND X(11581) =

 
= (2*a^4 - 7*a^2*b^2 + 5*b^4 - 7*a^2*c^2 - 10*b^2*c^2 + 5*c^4 - 2*Sqrt[3]*(2*a^2 + b^2 + c^2)*S)*(3*(a^2 + b^2 - c^2)*(a^2 - b^2 + c^2) + 4*S*(Sqrt[3]*a^2 + S)) : :

= lies on these lines: {13, 15}, {140, 8929}, {298, 16770}, {395, 8014}, {397, 11555}, {5478, 11624}, {11078, 11119}, {11142, 11146}, {16645, 21466}, {22797, 25163}

= midpoint of X(13) and X(11581)
= barycentric product X(13)*X(6669)
= barycentric quotient X(6669)/X(298)
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {13, 11080, 396}, {396, 11080, 11537}

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X(14);

 

MIDPOINT OF X(14) AND X(11582) =

 

 

= (2*a^4 - 7*a^2*b^2 + 5*b^4 - 7*a^2*c^2 - 10*b^2*c^2 + 5*c^4 + 2*Sqrt[3]*(2*a^2 + b^2 + c^2)*S)*(3*(a^2 + b^2 - c^2)*(a^2 - b^2 + c^2) - 4*(Sqrt[3]*a^2 - S)*S) : :

= lies on lines: {14, 16}, {140, 8930}, {299, 16771}, {396, 8015}, {398, 11556}, {5479, 11626}, {11092, 11120}, {11141, 11145}, {16644, 21467}, {22796, 25153}

= midpoint of X(14) and X(11582)
= barycentric product X(14)*X(6670)
= barycentric quotient X(6670)/X(299)
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {14, 11085, 395}, {395, 11085, 11549}

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X(15);

 

X(15)X(1154)∩X(378)X(10632) =  

 
= a^2*(Sqrt[3]*(a^2 - b^2 - c^2) - 2*S)*(4*a^4 - 7*a^2*b^2 + 3*b^4 - 7*a^2*c^2 - 6*b^2*c^2 + 3*c^4 - 2*Sqrt[3]*(b^2 + c^2)*S) : :

= lies on these lines: {15, 1154}, {378, 10632}, {396, 13350}, {11137, 11146}, {11624, 21158}
= crosspoint of X(15) and X(11146)
= crosssum of X(13) and X(11139)
= barycentric product X(i)*X(j) for these {i,j}: {15, 629}, {11146, 23302}
= barycentric quotient X(629)/X(300)

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X(16);

 

 X(16)X(1154)∩X(378)X(10633) =  

 
= a^2*(Sqrt[3]*(a^2 - b^2 - c^2) + 2*S)*(4*a^4 - 7*a^2*b^2 + 3*b^4 - 7*a^2*c^2 - 6*b^2*c^2 + 3*c^4 + 2*Sqrt[3]*(b^2 + c^2)*S) : :

 

= lies on these lines: {16, 1154}, {378, 10633}, {395, 13349}, {11134, 11145}, {11626, 21159}

= crosspoint of X(16) and X(11145)
= crosssum of X(14) and X(11138)
= barycentric product X(i)*X(j) for these {i,j}: {16, 630}, {11145, 23303}
= barycentric quotient X(630)/X(301)

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X(74);

 

X(74)X(186)∩X(125)X(32417) =  

 
= a^2*(a^2 - b^2 - c^2)*(a^4 - 2*a^2*b^2 + b^4 + a^2*c^2 + b^2*c^2 - 2*c^4)*(a^4 + a^2*b^2 - 2*b^4 - 2*a^2*c^2 + b^2*c^2 + c^4)*(2*a^10 - 2*a^8*b^2 - 3*a^6*b^4 + a^4*b^6 + 5*a^2*b^8 - 3*b^10 - 2*a^8*c^2 + 8*a^6*b^2*c^2 - a^4*b^4*c^2 - 14*a^2*b^6*c^2 + 9*b^8*c^2 - 3*a^6*c^4 - a^4*b^2*c^4 + 18*a^2*b^4*c^4 - 6*b^6*c^4 + a^4*c^6 - 14*a^2*b^2*c^6 - 6*b^4*c^6 + 5*a^2*c^8 + 9*b^2*c^8 - 3*c^10)  : :

= lies on these lines: {74, 186}, {125, 32417}, {184, 14264}, {520, 8431}, {3284, 11079}, {3470, 13367}, {5627, 13851}, {17986, 18400}

= X(110)-Ceva conjugate of X(14380)
= crosssum of X(1990) and X(3081)
= barycentric product X(7687)*X(14919)

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Best regards,
Peter Moses.