HYACINTHOS 26474

 

[Tran Quang Hung]:

 

Let ABC be a triangle with orthocenter H.

Fa is the first Fermat point of triangle HBC. Similarly, we have Fb,Fc.

Let HA cut FbFc at A''. Similarly, we have the point B'',C''.

Then FaA'',FbB'',FcC'' are concurrent at a point. Which is this point ?

 

[César Lozada]:

 

Q = (-2*sqrt(3)*S-a^2+b^2+c^2)*(a^ 2-b^2+c^2)*(a^2+b^2-c^2)*((2* a^22-4*(b^2+c^2)*a^20-2*(2*b^ 4-b^2*c^2+2*c^4)*a^18+26*(b^2+ c^2)*b^2*c^2*a^16+2*(25*b^8+ 25*c^8-b^2*c^2*(27*b^4+19*b^2* c^2+27*c^4))*a^14-2*(b^2+c^2)* (40*b^8+40*c^8-b^2*c^2*(53*b^ 4-47*b^2*c^2+53*c^4))*a^12+2*( 9*b^12+9*c^12+(38*b^8+38*c^8- 11*b^2*c^2*(3*b^4-4*b^2*c^2+3* c^4))*b^2*c^2)*a^10+4*(b^4-c^ 4)*(b^2-c^2)*(11*b^8+11*c^8-b^ 2*c^2*(26*b^4-31*b^2*c^2+26*c^ 4))*a^8-2*(b^2-c^2)^2*(10*b^ 12+10*c^12-(9*b^8+9*c^8-7*b^2* c^2*(3*b^4+4*b^2*c^2+3*c^4))* b^2*c^2)*a^6-2*(b^4-c^4)*(b^2- c^2)^3*(2*b^4-3*b^2*c^2+2*c^4) *(5*b^4+3*b^2*c^2+5*c^4)*a^4+ 2*(b^2-c^2)^6*(9*b^8+9*c^8+b^ 2*c^2*(13*b^4+14*b^2*c^2+13*c^ 4))*a^2-2*(b^2-c^2)^8*(b^2+c^ 2)*(2*b^4+3*b^2*c^2+2*c^4))*S* sqrt(3)+3*a^24-15*(b^2+c^2)*a^ 22+3*(6*b^4+17*b^2*c^2+6*c^4)* a^20+6*(b^2+c^2)*(5*b^4-11*b^ 2*c^2+5*c^4)*a^18-3*(31*b^8+ 31*c^8+3*b^2*c^2*(6*b^4+b^2*c^ 2+6*c^4))*a^16+3*(b^2+c^2)*( 21*b^8+21*c^8+2*b^2*c^2*(7*b^ 4+5*b^2*c^2+7*c^4))*a^14+3*(7* b^12+7*c^12-(29*b^8+29*c^8+2* b^2*c^2*(11*b^4+16*b^2*c^2+11* c^4))*b^2*c^2)*a^12-3*(b^2+c^ 2)*(b^4+c^4)*(b^8+c^8+2*b^2*c^ 2*(b^4-12*b^2*c^2+c^4))*a^10- 3*(b^2-c^2)^2*(30*b^12+30*c^ 12-(17*b^8+17*c^8-b^2*c^2*(17* b^4+24*b^2*c^2+17*c^4))*b^2*c^ 2)*a^8+6*(b^4-c^4)*(b^2-c^2)*( 18*b^12+18*c^12-(43*b^8+43*c^ 8-2*b^2*c^2*(29*b^4-30*b^2*c^ 2+29*c^4))*b^2*c^2)*a^6-3*(b^ 2-c^2)^4*(17*b^12+17*c^12-6*b^ 2*c^2*(2*b^8+b^4*c^4+2*c^8))* a^4+9*(b^2-c^2)^6*(b^2+c^2)*( b^8+c^8-2*b^2*c^2*(b^4+c^4))* a^2+3*(b^2-c^2)^8*b^2*c^2*(b^ 2+c^2)^2) : :

= On line {4,16}

 

NOTE: Only for ABC acute.

 

César Lozada