ADGEOM 4000 * ADGEOM 4001

#4000

Dear geometers,

 

Let ABC be a triangle with centroid G.

 

A'B'C' is cevian triangle of G.

 

A''B''C'' is antipedal triangle of G wrt A'B'C',

 

Let (L) and (L'') be the van Lamoen circles of triangles ABC and A''B''C''.

 

Then midpoint of LL'' lies on Euler line of ABC. Which is this point ?

 

Best regards,

Tran Quang Hung.

 

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#4001

 

 

L/\L”  [read (LL")/\Euler-line-of-ABC] =

= 2*a^8-12*(b^2+c^2)*a^6+(29*b^4+20*b^2*c^2+29*c^4)*a^4-(b^2+c^2)*(27*b^4-53*b^2*c^2+27*c^4)*a^2+(8*b^4-17*b^2*c^2+8*c^4)*(b^2-c^2)^2 : : (barys)

= (39*S^2-SW^2)*X(3)+(33*S^2+SW^2)*X(4)

= On line {2, 3}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (24, 5066, 10011), (421, 1113, 6973), (547, 5066, 10011), (4196, 8364, 1559), (4246, 8356, 5079), (6873, 7388, 7461)

= [ -0.507702217114054, -1.37588280874362, 4.827522834321064 ]

 

Notes:

L = X(1153)

L” = 2*a^8-3*(b^2+c^2)*a^6-6*(b^4+b^2*c^2+c^4)*a^4+(b^2+c^2)*(11*b^4-20*b^2*c^2+11*c^4)*a^2-4*(b^2-c^2)^4 :: (barys)

= 7*X(3851)-X(5171)

= on lines: {4, 7769}, {5, 7830}, {30, 7619}, {115, 5097}, {381, 511}, {575, 5475}, {3091, 7898}, {3830, 9734}, {3843, 9737}, {3851, 5171}, {7694, 11645}

= midpoint of X(3830) and X(9734)

= [ -5.927961358617093, -5.74068363709368, 10.350966088487890 ]

 

César Lozada