HYACINTHOS 25379

[Antreas P. Hatzipolakis]:

 

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

Denote:

A", B", C"= the reflections of I in BC, CA, AB, resp.

Ma, Mb, Mc = the midpoints of AA", BB", CC", resp.

Ab, Ac = the orthogonal projections of Ma on A'C', A'B', resp.

Bc, Ba = the orthogonal projections of Mb on B'A', B'C', resp.

Ca, Cb = the orthogonal projections of Mc on C'B', C'A', resp.

Na, Nb, Nc = the NPC centers of MaAbAc, MbBcBa, McCaCb, resp.

The NPC center of NaNbNc lies on the OI line.

Point?

 

Circumcenter Version:

Let ABC be a triangle and A'B'C' the antipedal triangle of O.

Denote:

A", B", C" = the reflections of O in B'C', C'A', A'B', resp (or in A, B, C, resp.)

Ma, Mb, Mc = the midpoints of A'A", B'B", C'C", resp.

Ab, Ac = the orthogonal projections of Ma on.AC, AB, resp.

Bc, Ba = the orthogonal projections of Mb on BA, BC, resp.

Ca, Cb = the orthogonal projections of Mc on CB, CA, resp.

Na, Nb, Nc = the NPC centers of MaAbAc, MbBcBa, McCaCb, resp.

The NPC center of NaNbNc lies on the Euler line of ABC.

Point?

[César Lozada]:

1)      Z1=3*(b+c)*a^5-(3*b^2-4*b*c+3* c^2)*a^4-(b+c)*(6*b^2-5*b*c+6* c^2)*a^3+2*(3*b^4+3*c^4-b*c*( 5*b^2-b*c+5*c^2))*a^2+(b^2-c^ 2)*(b-c)*(3*b^2+b*c+3*c^2)*a- 3*(b^2-c^2)^2*(b-c)^2 : :  (trilinears)

= (2*sin(A/2)-3*sin(3*A/2))*cos( (B-C)/2)+3*(cos(A)-1)*cos(B-C) -2*cos(A)+1/2  : : (trilinears)

= (13*R+6*r)*X(1)+3*(R-2*r)*X(3)

= on lines: {1,3}, {3988,10124}, {5550,5694}

= [ 2.209850383113662, 2.11930781592039, 1.153520432525496 ]

 

2)      Z2 = 2*a^10-7*(b^2+c^2)*a^8+2*(3*b^ 4+b^2*c^2+3*c^4)*a^6+(b^2+c^2) *(4*b^4-5*b^2*c^2+4*c^4)*a^4-( b^2-c^2)^2*(8*b^4+b^2*c^2+8*c^ 4)*a^2+3*(b^4-c^4)*(b^2-c^2)^3 : : (barycentrics)

= (cos(2*A)+5/2)*cos(B-C)-3*cos( A)*cos(2*(B-C))-3*cos(A)+cos( 3*A) : : (trilinears)

= (37*R^2-10*SW)*X(3)+3*(9*R^2- 2*SW)*X(4)

= Shinagawa [3*E+40*F, -9*E+8*F]

= on lines: {2,3}, {11557,11591}

= [ 0.441655980648849, -0.42902904187740, 3.733843365984646 ]

 

Regards,

César Lozada

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