HYACINTHOS 28621

[Aris Pavlakis]:

 

Let ABC be a triangle, A'B'C' the pedal triangle of I and A"B"C" the pedal triangle of the Feuerbach point Fe.

Denote:

A1 = B'C' /\ A"Fe
B1 = C'A' /\ B"Fe
C1 = A'B' /\ C"Fe

A1, B1, C1 are collinear (*)

Which is this line ? (trilinear polar of which point?)

(*) Aris Pavlakis, Romantics of Geometry

 

APH

 

 

[Peter Moses]:

Hi Antreas,

The line is:  {30,1319,1387,10035,21578}.

[APH]:

> The line is:  {30,1319,1387,10035,21578}.

Is the trilinear pole of this line some interesting point?

[Peter Moses]:

Hi Antreas,

>Is the trilinear pole of this line some interesting point ?

P = (a-b) (a-c) (a+b-c) (a-b+c) (2 a^5-a^4 b-a^3 b^2-a^2 b^3-a b^4+2 b^5-a^4 c+3 a^3 b c+2 a^2 b^2 c+3 a b^3 c-b^4 c-4 a^3 c^2-a^2 b c^2-a b^2 c^2-4 b^3 c^2+2 a^2 c^3-3 a b c^3+2 b^2 c^3+2 a c^4+2 b c^4-c^5) (2 a^5-a^4 b-4 a^3 b^2+2 a^2 b^3+2 a b^4-b^5-a^4 c+3 a^3 b c-a^2 b^2 c-3 a b^3 c+2 b^4 c-a^3 c^2+2 a^2 b c^2-a b^2 c^2+2 b^3 c^2-a^2 c^3+3 a b c^3-4 b^2 c^3-a c^4-b c^4+2 c^5) : : 

= lies on these lines: {}.

isog P = 

 

= X(6)X(647)∩X(9)X(650)

 

a^2 (a-b-c) (b-c) (a^5-2 a^4 b-2 a^3 b^2+4 a^2 b^3+a b^4-2 b^5-2 a^4 c+3 a^3 b c+a^2 b^2 c-3 a b^3 c+b^4 c-2 a^3 c^2+a^2 b c^2-2 a b^2 c^2+b^3 c^2+4 a^2 c^3-3 a b c^3+b^2 c^3+a c^4+b c^4-2 c^5) : : 

 

= lies on these lines: {6,647}, {9,650}, {212,663}, {652,3217}, {654,3196}, {2423,6586}

isot P = 

 

= X(2)X(525)∩X(78)X(522)

 

(a-b-c) (b-c) (a^5-2 a^4 b-2 a^3 b^2+4 a^2 b^3+a b^4-2 b^5-2 a^4 c+3 a^3 b c+a^2 b^2 c-3 a b^3 c+b^4 c-2 a^3 c^2+a^2 b c^2-2 a b^2 c^2+b^3 c^2+4 a^2 c^3-3 a b c^3+b^2 c^3+a c^4+b c^4-2 c^5) : : 

 

= lies on these lines: {2,525}, {78,522}, {312,4391} ,{2401,25259}, {3239,21198}, {4130,25082}, {6332,6505}

Best regards,
Peter Moses.

 

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