HYACINTHOS 25658

[Antreas P. Hatzipolaks]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of N.

Denote:

Ab, Ac = the orthogonal projections of A'  on NB', NC', resp.

A"B"C" = the triangle bounded by AbAc, BcBa, CaCb, resp.

 

A* = NA' intersection AbAc. Similarly B*, C*.

A*B*C* is the medial triangle of A"B"C"

Locus:

Let ABC be a triangle and A'B'C' the pedal triangle of a point P.

Denote:

Ab, Ac = the orthogonal projections of A'  on PB', PC', resp.

A"B"C" = the triangle bounded by AbAc, BcBa, CaCb, resp.

 

A* = PA' intersection AbAc. Similarly B*, C*.

Which is the locus of P such that A*B*C*and A"B"C" are perspective?

 

[César Lozada]:

 

 

Locus = {Line at infinity} \/ {cubic K044 ( Euler central cubic) through ETC’s X(3), X(4), X(5), X(52), X(68), X(155), X(5562), X(8800), X(8905), X(8906) and vertices of triangles ABC, EULER2, JOHNSON, ORTHIC}

 

ETC pairs (P,Z(P)=perspector): (3,3), (4,6748), (5,6688)

 

Z(X(52)) = {4,193} /\ {30,52}

= 2*a^10-7*(b^2+c^2)*a^8+2*(5*b^ 4+8*b^2*c^2+5*c^4)*a^6-8*(c^6+ b^6)*a^4+4*(b^2-c^2)^2*(b^4-b^ 2*c^2+c^4)*a^2-(b^4-c^4)*(b^2- c^2)^3 : : (barycentrics)

= X(20)-3*X(11245) = 3*X(3060)-X(3575) = 9*X(3060)-X(12278) = 3*X(3575)-X(12278)

= On lines: {3,11433}, {4,193}, {5,394}, {6,6823}, {20,11245}, {30,52}, {51,9825}, {68,1595}, {69,11479}, {155,1596}, {195,11799}, {235,1993}, {343,11424}, {382,6225}, {511,12241}, {524,5907}, {576,12233}, {578,6676}, {1092,6677}, {1181,1353}, {1503,10112}, {1593,6515}, {1597,11411}, {1598,6193}, {1885,5889}, {1906,11441}, {1907,11442}, {2883,3629}, {3060,3575}, {3089,3167}, {3527,7401}, {3547,11426}, {5050,7400}, {5446,6756}, {6815,9777}, {10095,10127}

= midpoint of X(i) and X(j) for these {i,j}: {1885,5889}, {10263,12370}

= reflection of X(i) in X(j) for these (i,j): (6756,5446), (12362,12241)

= {X(4), X(193)}-Harmonic conjugate of X(12164)

= [ -5.104576274673816, -4.18147310313579, 8.891488757004762 ]

 

César Lozada