HYACINTHOS 28759

[Tran Quang Hung]:

Let ABC be a triangle with de Longchamps point L.

A'B'C' is the antipedal triangle of L.

Then de Longchamps point of A'B'C' lies on the Euler line of ABC.

Which is this point?


[Ercole Suppa]:

The de Longchamps point of A'B'C' is X(22049)

[Kadir Altintas]:

Generalization:

Let ABC be a triangle and P a point on the Euler line such that PH/OP = t (number) and A'B'C' the antipedal triangle of P.

The same to P point  P' of A'B'C' lies on the Euler line of ABC.

P = X(20) --> P ' =  X(22049)  (above)
P =  X(140) --> P' = X(22050)

P = X(547) --> P' = ?

 

[Ercole Suppa]: 

 

Dear Kadir Altintas,

 

if P = X(547), the point P' is

 

P' = 35150 a^16-224435 a^14 b^2+621425 a^12 b^4-960475 a^10 b^6+875725 a^8 b^8-439825 a^6 b^10+77435 a^4 b^12+25375 a^2 b^14-10375 b^16-224435 a^14 c^2+868714 a^12 b^2 c^2-1207083 a^10 b^4 c^2+532540 a^8 b^6 c^2+253363 a^6 b^8 c^2-202074 a^4 b^10 c^2-91925 a^2 b^12 c^2+70900 b^14 c^2+621425 a^12 c^4-1207083 a^10 b^2 c^4+516660 a^8 b^4 c^4+87335 a^6 b^6 c^4+76038 a^4 b^8 c^4+123525 a^2 b^10 c^4-217900 b^12 c^4-960475 a^10 c^6+532540 a^8 b^2 c^6+87335 a^6 b^4 c^6+97202 a^4 b^6 c^6-56975 a^2 b^8 c^6+399500 b^10 c^6+875725 a^8 c^8+253363 a^6 b^2 c^8+76038 a^4 b^4 c^8-56975 a^2 b^6 c^8-484250 b^8 c^8-439825 a^6 c^10-202074 a^4 b^2 c^10+123525 a^2 b^4 c^10+399500 b^6 c^10+77435 a^4 c^12-91925 a^2 b^2 c^12-217900 b^4 c^12+25375 a^2 c^14+70900 b^2 c^14-10375 c^16 : : (barys)

 

= 99960 S^4+S^2 (42201 R^4-158760 SB SC-38772 R^2 SW+860 SW^2)+SB SC (-2187 R^4-18468 R^2 SW+23340 SW^2) : : (barys)

 

Shinagawa coefficients:  {42201 R^4 + 99960 S^2 - 38772 R^2 SW +  860 SW^2, -3 (729 R^4 + 52920 S^2 + 6156 R^2 SW - 7780 SW^2)

 

= lies on this line: {2,3}

= ETC search numbers: [4.17424010818604528, 3.29370843448483836, -0.566167561129452966]

 

Best regards

Ercole Suppa