HYACINTHOS 27070

[Antreas P. Hatzipolakis]:

Let ABC be a triangle ad A'B'C' the pedal triangle of a point P

Denote:

O' = the circumcenter of A'B'C'

MaMbMc = the midaway triangle of P
(ie Ma, Mb, Mc = the midpoints of AP, BP, CP, resp.)

Oa, Ob, Oc = the circumcenters of O'MbMc, O'McMa, O'MaMb, resp.

A'B'C', OaObOc are homothetic.
Homothetic center in terms of P?


[César Lozada]:

 

For P=u:v:w (trilinears), the homothetic center H(P) is:

H(P) = a^3*u*(u^2*b*c-2*SA*v*w)+c*( -S^2+(SB+2*SC)*SC)*u^2*v+b*(- S^2+(2*SB+SC)*SB)*u^2*w+a*b*c* u*(SB*w^2+v^2*SC)-a^2*SA*v*w*( b*v+c*w) : :

 

H(P) = inverse of P w/r to the circumcircle of MaMbMc = midpoint(P, circumcircle-inverse-of-P)

 

For P, P' mutually inverse in the circumcircle of ABC, H(P)=H(P'),

For P on the circumcircle of ABC or  in infinity, H(P)=P.

 

ETC pairs (P,H(P)): (1,1319), (2,7426), (4,403), (5,10096), (6,2030), (15,187), (16,187), (23,7426), (36,1319), (40,13528), (186,403), (187,2030), (1687,1691), (1688,1691), (2070,10096), (2077,13528), (5004,23), (5005,23), (5980,5939), (5981,5939)

 

Examples:

H( X(13) ) = circumcircle-inverse-of X(11142)

= ((4*S^2-3*(SB+SC)^2)*S+sqrt( 3)*(-12*R^2+3*SA+2*SW)*S^2- sqrt(3)*SB*SC*SW)*(sqrt(3)*SB+ S)*(sqrt(3)*SC+S) : : (barys)

= on lines: {3, 13}, {115, 11081}, {1637, 6137}, {3457, 5472}

= midpoint of X(13) and X(6104)

= circumcircle-inverse-of X(11142)

= [ -0.0047993534806196, 0.0060785278951823, 3.6386713565095400 ]

 

H( X(14) ) = circumcircle-inverse-of X(11141)

= (-(4*S^2-3*(SB+SC)^2)*S+ sqrt(3)*(-12*R^2+3*SA+2*SW)*S^ 2-sqrt(3)*SB*SC*SW)*(sqrt(3)* SB-S)*(sqrt(3)*SC-S): : (barys)

= on lines: {3, 14}, {115, 11086}, {1637, 6138}, {3458, 5471}}

= midpoint of X(13) and X(6105)

= circumcircle-inverse-of X(11141)

= [  -0.5412419993405657, 8.3407318967015090, -1.8838843699594830 ]

 

César Lozada