Δευτέρα 28 Οκτωβρίου 2019

HYACINTHOS 28585

[Antreas P. Hatzipolakis]:
 

Let ABC be a triangle and P a point.

Denote:

PaPbPc = the pedal triangle of P.

P1P2P3 = the reflection of PaPbPc in P
(ie P1, P2, P3 = the reflections of Pa, Pb, Pc in P, resp.)

Ma, Mb, Mc = the midpoints of AP, BP, CP, resp.

For P = Ι, O, N:
The circumcircles of PP1Ma, PP2Mb, PP3Mc are coaxial.
(for P = I they are tangent at I)  
2nd (other than O, N) intersections?

Locus ?
 
 
--------------------------------------------------------------------------------------------
 
 
[Ercole Suppa]
 
*** If P = O the 2nd intersection point is X(12041)
 
*** If P = N the 2nd intersection point is X(546)
 
*** the locus of point P such that the circumcircles of PP1Ma, PP2Mb, PP3Mc are coaxial = { K005 = Napoleon Feuerbach cubic} 
 
-- Note: from the calculations made I also found an algebraic curve of order 4. However, since this curve is uncyclic, I believe it is not part of the locus. Can someone clarify this fact?
 
-- K005:  ∑ a^2 (a^4-2 a^2 b^2+b^4-a^2 c^2-b^2 c^2) y^2 z-a^2 (a^4-a^2 b^2-2 a^2 c^2-b^2 c^2+c^4) y z^2 = 0   (barys)
 
-- K005 passes through ETC points X(i) for these i: 1,3,4,5,17,18,54,61,62,195,627,628,2120,2121,3336,3459,3460,3461,3462,3463,3467,3468,3469,3470,3471,3489,3490,6191,6192,7344,7345,8837,8839,8918,8919,8929,8930
 
-- pairs {P=X(i) ∈ K005, Q=X(j)} for these {i,j}: {1,1},{3,12041},{5,546},{17,10611},{18,10612},{54,10610},{61,10613},{62,10614}
 
-- some points:
 
Q(X(4))= (0:0:0) undefined point
 
 
Q(X(195)) =  MIDPOINT OF X(195) AND X(1157)
 
= a^2 (2 a^20-15 a^18 b^2+49 a^16 b^4-92 a^14 b^6+112 a^12 b^8-98 a^10 b^10+70 a^8 b^12-44 a^6 b^14+22 a^4 b^16-7 a^2 b^18+b^20-15 a^18 c^2+76 a^16 b^2 c^2-153 a^14 b^4 c^2+149 a^12 b^6 c^2-55 a^10 b^8 c^2-35 a^8 b^10 c^2+69 a^6 b^12 c^2-57 a^4 b^14 c^2+26 a^2 b^16 c^2-5 b^18 c^2+49 a^16 c^4-153 a^14 b^2 c^4+164 a^12 b^4 c^4-72 a^10 b^6 c^4+35 a^8 b^8 c^4-60 a^6 b^10 c^4+71 a^4 b^12 c^4-47 a^2 b^14 c^4+13 b^16 c^4-92 a^14 c^6+149 a^12 b^2 c^6-72 a^10 b^4 c^6+22 a^8 b^6 c^6+8 a^6 b^8 c^6-42 a^4 b^10 c^6+55 a^2 b^12 c^6-28 b^14 c^6+112 a^12 c^8-55 a^10 b^2 c^8+35 a^8 b^4 c^8+8 a^6 b^6 c^8+12 a^4 b^8 c^8-27 a^2 b^10 c^8+50 b^12 c^8-98 a^10 c^10-35 a^8 b^2 c^10-60 a^6 b^4 c^10-42 a^4 b^6 c^10-27 a^2 b^8 c^10-62 b^10 c^10+70 a^8 c^12+69 a^6 b^2 c^12+71 a^4 b^4 c^12+55 a^2 b^6 c^12+50 b^8 c^12-44 a^6 c^14-57 a^4 b^2 c^14-47 a^2 b^4 c^14-28 b^6 c^14+22 a^4 c^16+26 a^2 b^2 c^16+13 b^4 c^16-7 a^2 c^18-5 b^2 c^18+c^20) : : (barys)
 
= (16 R^2-4 SB-4 SC-4 SW) S^4 + (-47 R^6+43 R^4 SB+43 R^4 SC-8 R^2 SB SC+37 R^4 SW-36 R^2 SB SW-36 R^2 SC SW+4 SB SC SW-8 R^2 SW^2+8 SB SW^2+8 SC SW^2) S^2 + (39 R^6 SB SC+35 R^4 SB SC SW-8 R^2 SB SC SW^2) : : (barys)
 
= 2*X[8254]-X[16336]
 
= lies on these lines: {3,54}, {128,24147}, {1263,18400}, {3459,6288}, {8254,16336}, {10615,16768}, {16337,24385}
 
= midpoint of X(195) and X(1157)
 
= reflection of X(i) in X(j) for these {i,j}: {16336,8254}, {21230,10615}
 
= (6-8-13) search numbers [-7.27275011534366949, 11.1608765021618583, -0.729442274045983009]
 
 
Q(X(3336)) =  X(1)X(3) ∩ X(3814)X(20292)
 
= a (a^6-3 a^4 b^2+3 a^2 b^4-b^6+2 a^4 b c+4 a^3 b^2 c-4 a^2 b^3 c-4 a b^4 c+2 b^5 c-3 a^4 c^2+4 a^3 b c^2-3 a^2 b^2 c^2+4 a b^3 c^2+b^4 c^2-4 a^2 b c^3+4 a b^2 c^3-4 b^3 c^3+3 a^2 c^4-4 a b c^4+b^2 c^4+2 b c^5-c^6) : : (barys)
 
= lies on these lines: {1,3}, {3814,20292}, {5180,10200}
 
= (6-8-13) search numbers [-3.48763695637779819, -2.58525641876342924, 7.04013598245650621]
 
 
Best regards
Ercole Suppa
 

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