HYACINTHOS 27050

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of I.

Denote:

A" = AH /\ B'C'
B" = BH /\ C'A'
C" = CH /\ A'B'

The NPCs of A"HI, B"HI, C"HI are coaxial.

2nd, other than the midpoint of HI, intersection?

[César Lozada]:

 

Q2 =

= 8*p^10+4*q*p^9+2*(16*q^2-19)* p^8+(20*q^2-21)*q*p^7+(44*q^4- 115*q^2+72)*p^6+(q^2-1)*(2*(4* q^2-17)*q*p^5+(12*q^4-72*q^2+ 68)*p^4-(9*q^2-22)*q*p^3+(-11* q^4+42*q^2-32)*p^2+(3*q^2-5)* q*p+(q^2-1)*(3*q^2-6)) : : (trilinears), where p=sin(A/2), q=cos((B-C)/2)   

= 2*a^15-(b+c)*a^14-(b^2+c^2)*a^ 13+(b^3+c^3)*a^12-10*(b^2-c^2) ^2*a^11-(b^2-c^2)*(b-c)*(3*b^ 2-4*b*c+3*c^2)*a^10+(b^2-c^2)^ 2*(11*b^2+12*b*c+11*c^2)*a^9+( b^2-c^2)*(b-c)*(19*b^4+19*c^4- (b^2+16*b*c+c^2)*b*c)*a^8+2*( b^2-c^2)^2*(3*b^4+3*c^4-2*(8* b^2+b*c+8*c^2)*b*c)*a^7-(b^2- c^2)*(b-c)*(31*b^6+31*c^6+(2* b^4+2*c^4-(23*b^2+4*b*c+23*c^ 2)*b*c)*b*c)*a^6-(b^2-c^2)^2*( 11*b^6+11*c^6-(24*b^4+24*c^4-( 29*b^2-64*b*c+29*c^2)*b*c)*b* c)*a^5+(b^2-c^2)^3*(b-c)*(15* b^4+15*c^4-(b^2-56*b*c+c^2)*b* c)*a^4+2*(b^4-c^4)*(b^2-c^2)*( b-c)^2*(b^4+c^4+2*(b^2+5*b*c+ c^2)*b*c)*a^3+3*(b^4-c^4)*(b^ 2-c^2)^2*(b-c)*(b^4-10*b^2*c^ 2+c^4)*a^2+(b^4-c^4)*(b^2-c^2) ^3*(b-c)^4*a+(b^2-c^2)^5*(b-c) *(-3*b^4-3*c^4+(b^2-8*b*c+c^2) *b*c) : : (barys)

= on lines: {29, 102}, {2846, 14312}

= [ -2.9772907034177710, -3.3428234939163400, 7.3290610715808130 ]

 

César Lozada

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