ADGEOM 1550 * ADGEOM 1555 * ADGEOM 1556

#1550

 

Dear geometers,

Let ABC be a triangle with three Feuerbach points Fa,Fb,Fc. Let Fa' be isogonal conjugate of Fa with respect to triangle AFbFc. Smilarly we have point Fb',Fc'. Then FaFa',FbFb',FcFc' are concurrent.

Did we know about this point ?

Best regards,
Tran Quang Hung.

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#1555

Dear Tran Quang Hung and Francisco Javier

 

 

[TQH]: 

 

Let ABC be a triangle with three Feuerbach points Fa,Fb,Fc. Let Fa' be isogonal conjugate of Fa with respect to triangle AFbFc. Similarly we have point Fb',Fc'. Then FaFa',FbFb',FcFc' are concurrent.

 

 

****  My calculations give that the point of intersection of the lines FaFa',FbFb',FcFc' has barycentric coordinates:

 

 

a^11(b - c)^2+ a^10 (b + c) (b^2 - 6 b c + c^2)+ a^9( -5 b^4 - 5 b^3 c -12 b^2 c^2 - 5 b c^3 - 5 c^4)- a^8(b + c) (5 b^4 - 6 b^3 c + 6 b^2 c^2 - 6 b c^3 + 5 c^4) a^7 (10 b^6 + 22 b^5 c + 25 b^4 c^2 + 14 b^3 c^3 + 25 b^2 c^4 + 22 b c^5 + 10 c^6)+ a^6 (b + c) (10 b^6 + 18 b^5 c + 27 b^4 c^2 + 6 b^3 c^3 + 27 b^2 c^4 + 18 b c^5 + 10 c^6)+ a^5( -10 b^8 - 16 b^7 c + 22 b^6 c^2 + 57 b^5 c^3 + 54 b^4 c^4 + 57 b^3 c^5 + 22 b^2 c^6 - 16 b c^7 - 10 c^8)- 2a^4 (b + c) (5 b^8 + 15 b^7 c + 5 b^6 c^2 - 19 b^5 c^3 - 22 b^4 c^4 - 19 b^3 c^5 + 5 b^2 c^6 + 15 b c^7 + 5 c^8)+ a^3(b - c)^2 (b + c)^2 (5 b^6 - 4 b^5 c - 42 b^4 c^2 - 59 b^3 c^3 - 42 b^2 c^4 - 4 b c^5 + 5 c^6)+ a^2 (b - c)^2 (b + c)^3 (5 b^6 + 12 b^5 c - 8 b^4 c^2 - 26 b^3 c^3 - 8 b^2 c^4 + 12 b c^5 + 5 c^6)- a(b - c)^4 (b + c)^6 (b^2 - 7 b c + c^2) -(b - c)^6 (b + c)^7 : :

 

with (6-9-13)-search number -0.0381537327593557454260317

 

 

This point is not on the line X(1)X(21).

 

 

Best regards, 

 

Angel Montesdeoca

 

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#1556

 Dear Tran Quang Hung, Francisco Javier, and Angel,

I get the same search value as Angel, on lines 5,191 30,5948 442,5947 (at least).

Best regards,
Randy Hutson