Let ABC be a triangle and P a point.
Denote:
Pa = the reflection of P in BC
A' = the orthogonal projection of Pa on AP
Similarly B', C'.
Which is the locus of P such that ABC, A'B'C' are orthologic?
O, I lie on the locus. Orthologic centers ?
[César Lozada]:
Locus = {McCay cubic } \/ {q4=circum-quartic CyclicSum[ y*z*(a^2*(c^2*y^2+b^2*z^2+2* SA*y*z)+4*S^2*x^2) ]=0 through X(110) }
q4
Z(P)=orthologic center(A->A’) and Z’(P)=orthologic center(A’->A):
Z(I)= X(79) and Z’(I)=X(6284)
Z(O) = X(6145)
Z’(O) = {20,68} /\ {30,52}
= 2*a^10-4*(b^2+c^2)*a^8+(b^2+c^ 2)^2*a^6+(b^2+c^2)*(b^4+c^4)* a^4+(b^2-c^2)^2*(b^4+c^4)*a^2- (b^4-c^4)*(b^2-c^2)^3 : : (barycentrics)
= (2*cos(2*A)+1)*cos(B-C)-cos(A) *cos(2*(B-C))-cos(3*A) : : (trilinears)
= (2*R^2-SW)*X(20)-(4*R^2-SW)*X( 68) = SW*X(6)+(2*R^2-SW)*X(382)
= On lines: {3,161}, {4,569}, {5,1495}, {6,382}, {20,68}, {22,9927}, {30,52}, {49,7574}, {113,6759}, {125,1658}, {143,11565}, {156,1568}, {265,2937}, {343,550}, {539,11412}, {1503,9967}, {1614,3153}, {1660,9833}, {2072,10282}, {2777,6293}, {3529,6515}, {3575,9730}, {5449,7488}, {5462,7576}, {5878,8538}, {5889,10116}, {5944,10224}, {6243,10112}, {6640,11202}, {6800,7547}, {7526,11550}, {7540,10110}
= reflection of X(i) in X(j) for these (i,j): (52,6146), (143,11565), (5889,10116), (6243,10112)
= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (20,11457,7689), (1614,3153,5448)
= [ 6.731259716992461, 7.62313965433352, -4.743628994319963 ]
Z(H)= O and Z’(H)=H
Z(X(110)) = X(110)
Z’(X(110)) = {110,924} /\ {476,512}
= a^2*((b^4+c^4)*a^8-2*(b^2+c^2) *(2*b^4-3*b^2*c^2+2*c^4)*a^6+ 2*(3*b^8+3*c^8-b^2*c^2*(4*b^4- 3*b^2*c^2+4*c^4))*a^4-2*(b^4- c^4)*(b^2-c^2)*(2*b^4-3*b^2*c^ 2+2*c^4)*a^2+(b^4+c^4)*(b^2-c^ 2)^4)/(b^2-c^2) : : (barycentrics)
= (2*cos(A)*cos(B-C)-cos(2*A)* cos(2*(B-C))-3/2)*csc(B-C) : : (trilinears)
= On lines: {110,924}, {146,1531}, {476,512}
= [ 0.279178092627623, -0.65261000880820, 3.963619983715770 ]
César Lozada
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