ADGEOM 3999 * ADGEOM 4003

#3999

Dear geometers,

 

Let ABC be a triangle with centroid G.

 

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

 

Perpendicular bisector of GA',GB',GC' bound triangle A''B''C''.

 

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

 

Euler line of ABC meets LL'' at P then PL=2PL''.

 

Thus point divives LL'' in ratio 2 lies on Euler line of ABC. Which is this point ?

 

Best regards,

Tran Quang Hung.

 

--------------------------------------------------------------

 

#4003

 

(LL”) /\ Euler-line of ABC =

= 5*a^8-24*(b^2+c^2)*a^6+(47*b^4+32*b^2*c^2+47*c^4)*a^4-(b^2+c^2)*(39*b^4-77*b^2*c^2+39*c^4)*a^2+(11*b^4-23*b^2*c^2+11*c^4)*(b^2-c^2)^2 :: (barys)

= (63*S^2-SW^2)*X(3)+(45*S^2+SW^2)*X(4)

= on line {2,3}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (471, 1889, 5054), (632, 3540, 2566), (1344, 6934, 11290), (4245, 13737, 14038), (7405, 7522, 7453), (7444, 13861, 11818), (7488, 7892, 14119), (11323, 11346, 11585)

= [ 0.537988119420203, -0.33295102972048, 3.622866831981538 ]

 

Notes:

 

L = X(1153)

L” = 5*(b^2+c^2)*a^6-2*(9*b^4+7*b^2*c^2+9*c^4)*a^4+(b^2+c^2)*(19*b^4-36*b^2*c^2+19*c^4)*a^2-6*(b^2-c^2)^4 : : (barys)

= 13*X(5079)-X(5171)

= On lines: {5, 141}, {381, 9734}, {5072, 9737}, {5079, 5171}, {7603, 11171}

= [ -1.649296283064187, -1.99388555438393, 5.782260457895189 ]

 

César Lozada