HYACINTHOS 25381

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A',B',C' three points on BC,CA,AB, resp.

Denote:

Ma, Mb, Mc = the midpoints of BC, CA, AB, resp.

M1, M2, M3 = the midpoints of B'C', C'A' A'B', resp.

Mi, Mii, Miii = the midpoints of AA' BB', CC', resp.

The triangles MaM1Mi, MbM2Mii, McM3Miii share the same centroid Gp.

If A'B'C' is the cevian or pedal triangle of a point P, which is Gp in terms of P?

 

And which is its locus as P moves on the Euler line?

 

[César Lozada]:

Assume that A’, B’, C’ divide BC, CA, AB, resp., such that BA’/BC=p, CB’/CA=q and AC’/AB= r. Then the barycentric coordinates of the common centroid G* are:

 

G*(p,q,r) = p-q-2 : q-r-2 : r-p-2

 

When A’, B’, C’ are the traces of a point P=x:y:z (barycentrics) we have:

p=1/(1+y/z)

q=1/(1+z/x)

r=1/(1+x/y)

 

and the centroid G* is:

 

G*(P) =  (y+z)*(3*x^2+y*z+2*(y+z)*x) : :

or

G*(P) = (1-X)*(X^2 +Y*Z+2) : :, if P = X:Y:Z are the normalized barycentric coordinates.

 

ETC pairs ( P, G*(P) ): (1,10180), (2,2), (4,5943), (6,10191), (7,3742), (8,3740), (20,10192), (69,3819), (99,10190), (144,10164), (148,10189), (190,10196), (264,10184), (523,10278), (598,10183), (10405,3817)

 

Some others:

 

G*( O ) = complement of X(11197)

= (S^2+SB*SC)*(4*SA^2+(8*R^2-6* SW)*SA-S^2+SW^2-4*R^2*SW) : :  (barycentrics)

= cos(B-C)*(4*(cos(A)-cos(3*A))* cos(B-C)+cos(2*(B-C))-cos(2*A) -3*cos(4*A)+3) : : (trilinears)

= on lines: {2,10184}, {3,275}, {418,10003}, {549,1154}

= reflection of X(10184) in X(2)

= [ -7.458117798896138, 8.38132180452797, 1.280419139801694 ]

 

G*(N) = (10*SA^2-8*(R^2+SW)*SA+21*S^2- SW^2+4*R^2*SW)*(3*S^2-SB*SC) : :  (barycentrics)

= on lines: {547,11197}, {1656,3462}

= [ 11.001827159907200, -4.70427030808035, 1.819700621390529 ]

 

>>And which is its locus as P moves on the Euler line?

The cubic with parametric barycentrics equation:

x(t,a,b,c) = ((5*a^8-5*(b^2+c^2)*a^6-(6*b^ 4-17*b^2*c^2+6*c^4)*a^4+7*(b^ 4-c^4)*(b^2-c^2)*a^2-(b^4+7*b^ 2*c^2+c^4)*(b^2-c^2)^2)*t^2+(- 8*a^8+15*(b^2+c^2)*a^6-(3*b^4+ 16*b^2*c^2+3*c^4)*a^4-7*(b^4- c^4)*(b^2-c^2)*a^2+3*(b^4+c^4) *(b^2-c^2)^2)*t+3*a^8-8*(b^2+ c^2)*a^6+7*(b^4+b^2*c^2+c^4)* a^4-2*(b^4-c^4)*(b^2-c^2)*a^2- (b^2-c^2)^2*b^2*c^2)*((2*a^4-( b^2+c^2)*a^2-(b^2-c^2)^2)*t-( b^2+c^2)*a^2+(b^2-c^2)^2)

 

and cyclically y(t,b,c,a) and z(t,c,a,b).

 

Regards,

César Lozada