ADGEOM 5139 * ADGEOM 5140 * ADGEOM 5141 * ADGEOM 5142

#5139

Let ABC be a triangle.

 

OI line of ABC meets BC, CA, AB at A', B', C'.

 

Perpendicular lines from A', B', C' to BC, CA, AB bound triangle A"B"C".

 

Then X(56) of A"B"C" lies on OI line of ABC.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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#5140

X(56) of A"B"C" lies on OI line of ABC, is X(23981)

 

 

Let P1 and P2 be the two points on line X(1)X(3) whose trilinear polars are parallel to line X(1)X(3). X(23981) is the barycentric product P1*P2.

 

Angel Montesdeoca

 

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#5141

 

Thank you very much. I see the same problem, I hope you will help me again.

 

Let ABC be a triangle.

 

X(1)X(4) line of ABC meets BC, CA, AB at A', B', C'.

 

Perpendicular lines from A', B', C' to BC, CA, AB bound triangle A"B"C".

 

Then X(84) of A"B"C" lies on OI line of ABC.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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#5142

[Tran Quang Hung]:
    
    Let ABC be a triangle.

    X(1)X(4) line of ABC meets BC, CA, AB at A', B', C'.

    Perpendicular lines from A', B', C' to BC, CA, AB bound triangle A"B"C".

    Then X(84) of A"B"C" lies on OI line of ABC.

    Which is this point?

*** W =  X(1)X(3) /\ X(109)X(13138)

 X(1) = incenter
 X(3) = circumcenter
 
 X(109) = Point where the line that passes through incenter and the antipode of anticomplement of Feuerbach point intersects the circumcircle again.
 X(13138) =  Trilinear pole of the line that passes through circumcenter and  Mittenpunkt.
 
W = a (a-b) (a^2-(b-c)^2) (a-c) (2 a^7-a^6 (b+c)+a^4 (b-c)^2 (b+c)-(b-c)^2 (b+c)^5-6 a^3 (b^2-c^2)^2+4 a (b^2-c^2)^2 (b^2+c^2)+a^2 (b-c)^2 (b^3+7 b^2 c+7 b c^2+c^3)) : ... : ...

Angel Montesdeoca