HYACINTHOS 28777

[Tran Quang Hung]:

 

Let ABC be a triangle with orthocenter H and circumcenter O.

P=H+tO. 

A"B"C" is medial triangle.

The perpendicular lines from A", B", C" to PA, PB, PC bound triangle A#B#C#.

Let H# and O# be the orthocenter and circumcenter of triangle A#B#C#.

P#=H#+tO#.

Then P# lies on Euler line of ABC. 

Which is this point in terms of P?

 

[César Lozada]

 

Put P such that OP=t*OH.

If P# is such that O#P#=t*O#H# then:

P# = 6*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(a^6-(b^2+c^2)*a^4-(b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^4)*(b^2-c^2))^2*t^6-8*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(a^6-(b^2+c^2)*a^4-(b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^4)*(b^2-c^2))^2*t^5+(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(4*a^12-7*(b^2+c^2)*a^10-2*(4*b^4-11*b^2*c^2+4*c^4)*a^8+(b^2+c^2)*(22*b^4-37*b^2*c^2+22*c^4)*a^6-(8*b^8+8*c^8+3*b^2*c^2*(5*b^4-12*b^2*c^2+5*c^4))*a^4-(b^4-c^4)*(b^2-c^2)*(7*b^4-15*b^2*c^2+7*c^4)*a^2+(4*b^4+9*b^2*c^2+4*c^4)*(b^2-c^2)^4)*t^4+4*S^2*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(a^8+(b^2+c^2)*a^6-(6*b^4-5*b^2*c^2+6*c^4)*a^4+(b^2+c^2)*(5*b^4-8*b^2*c^2+5*c^4)*a^2-(b^6-c^6)*(b^2-c^2))*t^3+((b^2+c^2)*a^14-4*(2*b^4+b^2*c^2+2*c^4)*a^12+3*(b^2+c^2)*(8*b^4-9*b^2*c^2+8*c^4)*a^10-(35*b^8+35*c^8-b^2*c^2*(17*b^4+16*b^2*c^2+17*c^4))*a^8+(b^2+c^2)*(25*b^8+25*c^8-2*b^2*c^2*(15*b^4-4*b^2*c^2+15*c^4))*a^6-2*(b^2-c^2)^2*(3*b^8+3*c^8+b^2*c^2*(15*b^4+b^2*c^2+15*c^4))*a^4-(b^4-c^4)*(b^2-c^2)^3*(2*b^4-9*b^2*c^2+2*c^4)*a^2+(b^4+3*b^2*c^2+c^4)*(b^2-c^2)^6)*t^2+((b^4+c^4)*a^12-(b^2+c^2)*(5*b^4-9*b^2*c^2+5*c^4)*a^10+(10*b^8+10*c^8-9*b^2*c^2*(b^2+c^2)^2)*a^8-2*(b^2+c^2)*(5*b^8+5*c^8-7*b^2*c^2*(b^4+c^4))*a^6+(5*b^8+5*c^8+6*b^2*c^2*(2*b^4+b^2*c^2+2*c^4))*(b^2-c^2)^2*a^4-(b^4-c^4)*(b^2-c^2)^3*(b^4+3*b^2*c^2+c^4)*a^2-(b^2-c^2)^6*b^2*c^2)*t-a^2*b^2*c^2*((b^2+c^2)*a^8-2*(b^2+c^2)^2*a^6+2*(b^2+c^2)*b^2*c^2*a^4+2*(b^6-c^6)*(b^2-c^2)*a^2-(b^4-c^4)*(b^2-c^2)^3) : : (barys)

 

For t=real number not depending of ABC or A#B#C# = a point (this occurs for P ∈ {X(1113), X(1114)}), P# lies on the Euler line of ABC.

 

ETC-pairs (P, #P): (2,381), (3,13371), (4,4), (5,5501), (1113,1312), (1114,1313)

 

Some others:

 

t=-1:

Q( X(20) ) = COMPLEMENT OF X(22049)

= SA*((-64*R^2+SA+15*SW)*S^2+4*(SB+SC)*(4*R^2-SW)*(4*SA-32*R^2+7*SW)) : : (barys)

= on the line {2, 3}

= reflection of X(i) in X(j) for these (i,j): (16273, 15948), (18017, 3)

= complement of X(22049)

= {X(15948), X(16273)}-harmonic conjugate of X(376)

= [ 13.0882396354478400, 12.1841926059181700, -10.8352717693195000 ]

 

t=1/4:

Q( X(140) ) = COMPLEMENT OF X(22050)

= 216*S^4-3*(R^2*(113*R^2+44*SW)-168*SB*SC-4*SW^2)*S^2-(7*R^2*(R^2-20*SW)-92*SW^2)*SB*SC : : (barys)

= on the line {2, 3}

= complement of X(22050)

= [ -388.5620753599945000, -388.4065587170282000, 451.8738552983090000 ]

 

Note: this configuration seems to be highly related to: Hyacinthos 28137

 

 

César Lozada