Δευτέρα 28 Οκτωβρίου 2019

HYACINTHOS 28578

[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 = I:
 
MaMbMc, P1P2P3 are perspective.
Perspector = the Feuerbach point X(11)
 
For P = N:
 
1. MaMbMc, P1P2P3 are perspective.
Perspector ?
 
2. MaP1 = MbP2 = McP3 =: k = ?
 
 
Locus of P such that MaMbMc, P1P2P3 are perspective?
 
--------------------------------------------------------------------------------------------
 
 
[Ercole Suppa]
 
*** If P = N 
 
1. the perspector is X(11801) and 
 
2. MaP1 = MbP2 = McP3 = (1/4)sqrt(9R^2-2*SW)
 
 
*** the locus of points P such that MaMbMc, P1P2P3 are perspective is K005 = Napoleon - Feuerbach cubic
 
-- 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}: {1,11},{3,30},{4,4},{5,11801},{195,22051}
 
-- some points:
 
 
Q(X(54)) =  MIDPOINT OF  X(54) AND X(1141)
 
= (a^4-2 a^2 b^2+b^4-a^2 c^2-b^2 c^2) (a^4-a^2 b^2-2 a^2 c^2-b^2 c^2+c^4) (2 a^14-6 a^12 b^2+7 a^10 b^4-7 a^8 b^6+8 a^6 b^8-4 a^4 b^10-a^2 b^12+b^14-6 a^12 c^2+10 a^10 b^2 c^2-3 a^8 b^4 c^2-9 a^6 b^6 c^2+12 a^4 b^8 c^2+a^2 b^10 c^2-5 b^12 c^2+7 a^10 c^4-3 a^8 b^2 c^4+8 a^6 b^4 c^4-8 a^4 b^6 c^4+5 a^2 b^8 c^4+9 b^10 c^4-7 a^8 c^6-9 a^6 b^2 c^6-8 a^4 b^4 c^6-10 a^2 b^6 c^6-5 b^8 c^6+8 a^6 c^8+12 a^4 b^2 c^8+5 a^2 b^4 c^8-5 b^6 c^8-4 a^4 c^10+a^2 b^2 c^10+9 b^4 c^10-a^2 c^12-5 b^2 c^12+c^14) : : (barys)
 
= X[128]-2*X[6689]
 
= lies on these lines: {5,49},{128,6689},{137,18400},{1154,24147},{6592,25042},{10610,25150},{18370,24144},{20424,25044}
 
= midpoint of X(54) and X(1141)
 
= reflection of X(128) in X(6689)
 
= (6-8-13) search numbers [-3.38131839767579538, 6.14437931967307705, 0.947471905676462089]
 
 
Q(X(3336)) =  MIDPOINT OF X(79) AND X(3336)
 
= a^5 b^2-a^4 b^3-2 a^3 b^4+2 a^2 b^5+a b^6-b^7+6 a^5 b c+5 a^4 b^2 c-a^3 b^3 c-6 a^2 b^4 c-5 a b^5 c+b^6 c+a^5 c^2+5 a^4 b c^2+6 a^3 b^2 c^2+4 a^2 b^3 c^2-a b^4 c^2+3 b^5 c^2-a^4 c^3-a^3 b c^3+4 a^2 b^2 c^3+10 a b^3 c^3-3 b^4 c^3-2 a^3 c^4-6 a^2 b c^4-a b^2 c^4-3 b^3 c^4+2 a^2 c^5-5 a b c^5+3 b^2 c^5+a c^6+b c^6-c^7 : : (barys)
 
= X[10543]-2*X[20323]
 
= lies on these lines: {5,79},{11,1354},{12,11544},{30,4325},{392,11263},{442,3828},{517,3649},{2475,9657},{3647,17575},{3654,5499},{3813,15679},{4309,16117},{4317,10525},{4338,16159},{5221,16116},{6175,9710},{6701,17529},{9711,11684},{10543,20323}
 
= midpoint of X(79) and X(3336)
 
= reflection of X(10543) in X(20323)
 
= (6-8-13) search numbers [-0.0437387620004458985, 0.117427680465678781, 3.57955551635449959]
 
 
Best regards
Ercole Suppa
 

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου