HYACINTHOS 29626

[Kadir Altintas]:

 

 

Let ABC be a triangle and P a point.

 

Denote:

 

Ga, Gb, Gc = the centroids of PBC, PCA, PAB, resp. 

K' = the symmedian point of GaGbGc. 

Ka, Kb, Kc = the reflections of K' in GbGc, GcGa, GaGb, resp.

 

Which is the locus of P such that ABC and KaKbKc are perspective ?

 

 

[Ercole Suppa]:

 

 

The locus of P such that ABC and KaKbKc are perspective is {Linf}∪{q2 = conic through X(6),X(3413),X(3414),X(11477),X(15534),X(32935)}

 

If P = (x:y:z) the perspector is the point Q = Q(P) = (2 b^2 x-c^2 x+2 b^2 y-c^2 y+a^2 z+3 b^2 z) (-b^2 x-3 c^2 x+a^2 y-2 c^2 y+a^2 z-2 c^2 z)  : :   (barys)

 

*** Pairs {P = X(i)∈ q1, Q(P) = X(j)} for these {i,j}: {6,76},{3413,3413},{3414,3414},{11477,262},{15534,2}

 

*** Some points:

 

Q1= Q(X(32935)) = MIDPOINT OF X(7985) AND X(9902) =

 

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

 

= X[7985]+X[9902]

 

= lies on the Kiepert circumhyperbola and these lines: {2,726}, {10,22036}, {76,4066}, {226,4135}, {516,14458}, {519,598}, {2321,11599}, {3906,4049}, {3993,21101}, {3994,30588}, {4134,14839}, {4444,30519}, {4709,13576}, {6625,17760}, {7985,9902}, {11167,17132}}

 

= isogonal conjugate of Q2

= midpoint of X(7985) and X(9902)

 

= ETC search numbers: [5.5609180703422668398, 1.0923585942098356609, 0.3178386534503697195]

 

 

Q2 = Q1* (isogonal conjugate of Q1) = X(3)X(6) ∩ X(81)X(10789) =

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

= lies on these lines: {3,6}, {81,10789}, {106,11636}, {110,727}, {385,24267}, {560,595}, {741,30554}, {1357,1412}, {2712,32694}, {4653,11364}, {4658,12194}, {6233,17222}, {7787,25526}, {21793,23095}

= isogonal conjugate of Q1

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: {58,33628,1326}, {1333,5009,58}

= ETC search numbers: [0.1694755862341742547, 0.8627568409924841824, 2.9651517829584250131]

 

Best regards 

Ercole Suppa