HYACINTHOS 24566

[Antreas P. Hatzipolakis]:

 

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

Denote:

O', Oa, Ob, Oc = the circumcenters of A'B'C', PBC, PCA, PAB, resp.

Which is the locus of P such that the reflections of O'Oa, O'Ob, O'Oc in BC, CA, AB, resp. are concurrent?

H lies on the locus. Point of concurrence = Ο

O lies on the locus. Point of concurrence?

[Peter Moses]:

Hi Antreas,

>O lies on the locus. Point of concurrence?

The coords are:

a^4 (a^12-4 a^10 b^2+5 a^8 b^4-5 a^4 b^8+4 a^2 b^10-b^12-4 a^10 c^2+9 a^8 b^2 c^2-5 a^6 b^4 c^2+a^4 b^6 c^2-3 a^2 b^8 c^2+2 b^10 c^2+5 a^8 c^4-5 a^6 b^2 c^4+2 a^4 b^4 c^4-a^2 b^6 c^4-b^8 c^4+a^4 b^2 c^6-a^2 b^4 c^6-5 a^4 c^8-3 a^2 b^2 c^8-b^4 c^8+4 a^2 c^10+2 b^2 c^10-c^12)::
on lines {{3,8157},{4,54},{49,52},{110, 2888},{154,9704},{156,9927},{ 182,6689},{206,576},{539, 10201},{569,6145},{1092,7691}, {1147,1154},{1209,6639},{1971, 9697},{2904,9707},{3518,7730}, {6288,10254},{9813,9827},{ 10182,10203}}.

 

Midpoint of X(195) and X(2917).

2 X[54] + X[6759], 3 X[154] - X[9920].

X[4] + (J^2 - 2) X[54], (J = OH/R).

X(324)-Ceva conjugate of X(571).

{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (54,3574,578).

 

X(74) is on the locus, point of concurrence = X(110).

Best regards,

Peter Moses.