HYACINTHOS 25064

[Tran Quang Hung]:

 

 

Let ABC be a triangle inscribed in circle (O).

 

Incircle (I) touches BC,CA,AB at A',B',C'.

 

(AIA'),(BIB'),(CIC') cuts (O) again at A'',B'',C''.

 

A*,B*,C* are midpoints of B'C',C'A',A'B'.

 

Then circles (AA*A''),(BB*B''),(CC*C'') are coaxal with radical axis OI.

 

Which are their intersection points on OI ?

 

 
[César Lozada]:

Very complicated trilinear coordinates:

 

Z1= (-b*a^2-a^2*c+2*a*b*c+b^3-b^2* c-b*c^2+c^3)*((4*R*r+2*c^2-2* c*s+4*r^2)*K+48*R^2*r+16*R*c^ 2-16*R*c*s-2*c^2*r-4*r^3+S*c- 4*R*r^2)/R/S/(R-2*r)+16*(a-c)* (a-b+c) : (a^3-a^2*c-a*b^2+2*a*b*c-a*c^ 2-b^2*c+c^3)*((4*R*r+2*c^2-2* c*s+4*r^2)*K+48*R^2*r+16*R*c^ 2-16*R*c*s-2*c^2*r-4*r^3+S*c- 4*R*r^2)/R/S/(R-2*r)+16*(b-c)* (-a+b+c) : ((4*R*r+2*c^2-2*c*s+4*r^2)*K+ 48*R^2*r+16*R*c^2-16*R*c*s-2* c^2*r-4*r^3+S*c-4*R*r^2)/R/S/( R-2*r)*(a^3-b*a^2-(b-c)^2*a+( b^2-c^2)*b)

where K= sqrt((4*R+r)*(12*R+r))

 

for the other, Z2, just change K->-K in Z1.

 

ETC-6-9-13 searc numbers:

Z1 = [ -6.287558967332868, -4.89722521619106, 9.933001462654897 ]

Z2 = [ 1.380039257040124, 1.43411107696340, 2.010877156145043 ]

Both on line IO (checked numerically) and none of them is a triangle center.

 

[APH]:

So, they are (Z1 and Z2)  bicentric points.

The midpoint of Z1Z2 should be a triangle center.

[César Lozada]:

Yes. It is:

M=midpoint(Z1,Z2)=radical- trace-of(BB*B”,CC*C”) =
= (a^3-(b+c)*a^2-(b^2-3*b*c+c^2) *a+(b^2-c^2)*(b-c))*((b+c)*a^ 4-2*(b^2-5*b*c+c^2)*a^3-4*b*c* (b+c)*a^2+2*(b^2-3*b*c+c^2)*( b-c)^2*a-(b^2-c^2)*(b-c)^3) : : (trilinears)

= (4*R^2-r^2)*X(1)-r*(8*R-r)*X( 3)

= On lines: {1,3}, {142,5087}, {226,10156}, {513,2473}, {910,5053}, {971,3911}, {5057,9776}, {5281,10569}, {5435,10167}, {5439,11111}, {5745,10855}, {5777,6970}, {7288,9856}, {9588,9850}, {10178,11019}

= incircle-inverse-of-X(10980)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (1155,1319,5537), (2446,2447,10980), (5122,5126,3)

= [ -2.453759855146372, -1.73155706961383, 5.971939309399970 ]