HYACINTHOS 29498

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, P a point and A'B'C', A"B"C" the cevian, circumcevian triangles of P, resp.

Denote:

(Oa), (O1) = the circles with diameters BC, A'A", resp.
(Ob), (O2) = the circles with diameters CA, B'B", resp.
(Oc), (O3) = the circles with diameters AB, C'C", resp.

Ra = the radical axis of (Oa), (O1)
Rb = the radical axis of (Ob), (O2)
Rc = the radical axis of (Oc), (O3)

A*B*C* = the triangle bounded by Ra, Rb, Rc.

1. For P = H:
ABC, A*B*C* are perspective.
Perspector = X(34285)
See also HG140919

I think that for:

2. P = O
3. P = I

The triangles ABC, A*B*C* are also perspective.  

 

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[Ercole Suppa] 

 

Hi Antreas, 

 

2. If P=X(3)=O The triangles ABC, A*B*C* are perspective at

 

Q = Q(X(3)) = X(3)X(1075) ∩ X(97)X(3164) 

 

= (a^10 b^2-4 a^8 b^4+6 a^6 b^6-4 a^4 b^8+a^2 b^10-a^10 c^2-a^8 b^2 c^2+2 a^6 b^4 c^2+2 a^4 b^6 c^2-a^2 b^8 c^2-b^10 c^2+4 a^8 c^4-2 a^6 b^2 c^4-4 a^4 b^4 c^4-2 a^2 b^6 c^4+4 b^8 c^4-6 a^6 c^6+2 a^4 b^2 c^6+2 a^2 b^4 c^6-6 b^6 c^6+4 a^4 c^8+a^2 b^2 c^8+4 b^4 c^8-a^2 c^10-b^2 c^10) (a^10 b^2-4 a^8 b^4+6 a^6 b^6-4 a^4 b^8+a^2 b^10-a^10 c^2+a^8 b^2 c^2+2 a^6 b^4 c^2-2 a^4 b^6 c^2-a^2 b^8 c^2+b^10 c^2+4 a^8 c^4-2 a^6 b^2 c^4+4 a^4 b^4 c^4-2 a^2 b^6 c^4-4 b^8 c^4-6 a^6 c^6-2 a^4 b^2 c^6+2 a^2 b^4 c^6+6 b^6 c^6+4 a^4 c^8+a^2 b^2 c^8-4 b^4 c^8-a^2 c^10+b^2 c^10) : : (barys)

 

= S^4 + (64 R^4-8 R^2 SB-8 R^2 SC-2 SB SC-16 R^2 SW+2 SB SW+2 SC SW)S^2 -256 R^8 -128 R^6 SB-128 R^6 SC-96 R^4 SB SC+256 R^6 SW+96 R^4 SB SW+96 R^4 SC SW+32 R^2 SB SC SW-96 R^4 SW^2-24 R^2 SB SW^2-24 R^2 SC SW^2-2 SB SC SW^2+16 R^2 SW^3+2 SB SW^3+2 SC SW^3-SW^4 : : (barys)

 

= on lines X(i)X(j) for these {i,j}: {3,1075},{97,3164},{394,8613}

 

= lies on circumconics: {A,B,C,X(i),X(j)} for these {i,j}: {2, 3}, {4, 8613}, {64, 8794}, {92, 6360}  (computed only for i=1,2,...,100)

 

= lies on Q124 (Wallace quartic)

 

= ETC search numbers: [-9.1222335988569843884,-7.2805104551222545880,12.8912795349260866125]

 

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3. If P = I The triangles ABC, A*B*C* are perspective at

 

Q = Q(X(1)) = X(15446) 

 

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4. Let P = (x:y:z) (barys). The locus of points P such that ABC and A*B*C* are perspective is: {sidelines} ∪ {Darboux cubic K004} ∪ {q9 = circular-circum-curve of order 9}

 

K004:   ∑ c^2 (a^4+2 a^2 b^2-3 b^4-2 a^2 c^2+2 b^2 c^2+c^4) x^2 y+c^2 (3 a^4-2 a^2 b^2-b^4-2 a^2 c^2+2 b^2 c^2-c^4) x y^2 = 0 (barys)

 

through ETC centers X(i) for these i: {1,3,4,20,40,64,84,1490,1498,2130,2131,3182,3183,3345,3346,3347,3348,3353,3354,3355,3472,3473,3637}

 

 

q9:     ∑ b^2 c^4 (a^2-b^2+c^2) x^6 y^3+(a-c) c^4 (a+c) (a^2-b^2+c^2) x^5 y^4-(b-c) c^4 (b+c) (a^2-b^2-c^2) x^4 y^5-a^2 c^4 (a^2-b^2-c^2) x^3 y^6-b^2 c^2 (a^4-2 a^2 b^2+b^4-a^2 c^2-b^2 c^2) x^6 y^2 z-c^2 (a^6-3 a^4 b^2+3 a^2 b^4-b^6-2 a^4 c^2+a^2 c^4+b^2 c^4) x^5 y^3 z+(a^2+b^2) c^2 (a^2-b^2-c^2) (a^2-b^2+c^2) x^4 y^4 z+c^2 (a^6-3 a^4 b^2+3 a^2 b^4-b^6+2 b^4 c^2-a^2 c^4-b^2 c^4) x^3 y^5 z-a^2 c^2 (a^4-2 a^2 b^2+b^4-a^2 c^2-b^2 c^2) x^2 y^6 z-(a^2-b^2-b c-c^2) (a^2-b^2+b c-c^2) (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) x^5 y^2 z^2-(a^8-3 a^6 b^2+3 a^4 b^4-a^2 b^6-2 a^6 c^2+a^4 b^2 c^2+2 a^2 b^4 c^2-b^6 c^2+3 a^2 b^2 c^4+b^4 c^4+2 a^2 c^6+b^2 c^6-c^8) x^4 y^3 z^2+(a^6 b^2-3 a^4 b^4+3 a^2 b^6-b^8+a^6 c^2-2 a^4 b^2 c^2-a^2 b^4 c^2+2 b^6 c^2-a^4 c^4-3 a^2 b^2 c^4-a^2 c^6-2 b^2 c^6+c^8) x^3 y^4 z^2+2 a^2 b^2 (a^4-2 a^2 b^2+b^4-2 a^2 c^2) x^3 y^3 z^3 = 0 (barys)

 

 

 

Best regards

Ercole Suppa