HYACINTHOS 25317

 

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, P a point, PaPbPc the pedal triangle of P and A'B'C' the cevian triangle of G.

Denote:

Oa, Ob, Oc = the circumcenters of AA'Pa, BB'Pb, CC'Pc, resp.

Which is the locus of P such that the centroid of OaObOc lies on the Euler line?

The Euler line?

 

[César Lozada]:

 

Yes. The Euler line except O.

 

If OP=t*OH and Go(P) is the given centroid then OGo(P)=((1+3*t)/12)*OH. The coincident point is X(3524) (the point for which Go(P)=P), but P and Go(P) are not symmetrical with respect to this point.

 

Go(t) = -3*(S^2-3*SB*SC)*t+11*S^2-9* SB*SC) : : (barycentrics)

 

ETC pairs (P,Go(P)): (2,549), (4,2), (20,8703), (376,3), (381,140), (382,5066), (469,1346), (546,11540), (549,3530), (3146,3845), (3524,3524), (3529,3534), (3543,5), (3545,5054), (3627,10109), (3830,547), (3839,11539), (3845,10124), (5071,631), (6953,470), (11001,376)

 

Go(N) = complement of X(5066)

= 14*a^4-(19*(b^2+c^2))*a^2+5*( b^2-c^2)^2 : : (barycentrics)

= On lines: {2,3}, {551,11278}, {590,6481}, {615,6480}, {952,4745}, {1327,6412}, {1328,6411}, {1385,4669}, {3054,11648}, {3653,5690}, {5008,9300}, {5097,10168}, {5462,11592}, {6432,8981}, {11055,11171}

= midpoint of X(i) and X(j) for these {i,j}: {3,547}, {140,549}, {376,546}, {381,548}, {3530,10124}, {5066,8703}

= reflection of X(i) in X(j) for these (i,j): (2,11540), (3530,549), (3628,10124), (3850,547), (3860,10109), (3861,11737), (10109,2), (10124,140), (11737,3628)

= [ 4.186741580629078, 3.30617692772791, -0.580569505271440 ]

 

César Lozada