HYACINTHOS 28980

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and P a point.

 

Denote:

 

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

 

Oa, Ob, Oc = the circumcenters of PNbNc, PNcNa, PNaNb, resp.

 

For P = I

The parallels to IOa, IOb, IOc through A, B, C, resp. are concurrent.

(the point of concurrence lies on the circumcircle and is the antipode of the reflection point of the OI line)

 

Locus?

 

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[Ercole Suppa]:

 

Hi Antreas,

 

Let Q = Q(P) the point of concurrency of the parallels to POa, POb, POc through A, B, C, resp.

 

*** For P = X(1) = I we have  Q(X(1)) = X(104)

 

*** For P=(x:y:z) the locus of concurrency points is  Γ = {sextic q6 through ETC points X(13),X(14)} U {circumcurve q11 of order 11 through ETC points X(1),X(3),X(4)}

 

q6: ∑ (a^2-b^2-c^2) (a^2-b^2-b c-c^2) (a^2-b^2+b c-c^2) x^4 y z+(a^2-b^2-b c-c^2) (a^2-b^2+b c-c^2) (a^2-b^2+c^2) x^3 y^2 z-(a^2-b^2-c^2) (a^2-b^2-a c+c^2) (a^2-b^2+a c+c^2) x^2 y^3 z+3 a^2 (a^4-a^2 b^2-b^4) x^2 y^2 z^2-a^2 (a^2-b^2-a c+c^2) (a^2-b^2+a c+c^2) y^4 z^2+a^2 (a^4+a^2 b^2-2 b^4+a^2 c^2+4 b^2 c^2-2 c^4) y^3 z^3-a^2 (a^2-a b+b^2-c^2) (a^2+a b+b^2-c^2) y^2 z^4 = 0

 

q11: the equation is very complicated to be written here

 

 

*** Some points:

 

Q(X(1)) = X(104)

 

 

Q(X(3)) = X(1)X(3520) ∩ X(4)X(145)

 

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

 

= lies on these lines: {1,3520}, {4,145}, {8,16868}, {24,8148}, {40,17506}, {108,2099}, {186,517}, {378,10247}, {403,5844}, {915,8652}, {1483,18560}, {1829,26863}, {1830,1870}, {1845,6198}, {2807,7722}, {3518,7982}, {5603,7577}, {5690,14940}, {5901,6143}, {7505,12245}, {10222,14865}, {10295,28212}, {12702,21844}, {13596,16200}, {13619,28174}

 

=(6-9-13)  search numbers:  [-4.00039015643910106, -3.57138278816751449, 7.95949417668762808]

 

 

Q(X(4)) is undefined since the points X(4),Nb,Nc are collinear.

 

*** for P = X(13) or P = X(14) the computations are very complicated.

 

 

Best regards,

Ercole Suppa