## Κυριακή, 10 Φεβρουαρίου 2013

### A PONCELET POINT ON THE EULER LINE

Let ABC be a triangle, (N),(N1),(N2),(N3) the NPCs of ABC,NBC,NCA,NAB, resp. [concurrent at pN]. The NPCs of the triangles N1N2N3, NN1N2, NN2N3, NN3N1 concur at point ppN on the Euler Line of ABC.

Generalization:

Let ABC be a triangle, P a point, (N),(N1),(N2),(N3) the NPCs of ABC, PBC, PCA, PAB resp. [concurrent at pP]. If P is on the Euler line of ABC, then the NPCs of N1N2N3, PN1N2, PN2N3, PN3N1 concur at point ppP on the Euler Line of ABC.(??)

Antreas P. Hatzipolakis, 10 Febr. 2013

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pP is the center of the rectangular circumhyperbola through P.

ppP does not, in general, lie on the Euler line.

Some results:

P=X(1), the NPCs are concurrent, with center = non-ETC 1.121590125545969 (on lines 1,5 3,962).

P=X(2), ppP=non-ETC 1.690358502447462

P=X(3), ppP=X(140)

P=X(4), ppP=undefined

P=X(5), ppP=X(3628)

P=X(6), ppP=non ETC 0.780037257060191

P=X(7), ppP=non ETC 0.750876768572663

P=X(8), ppP=non ETC 2.966801160450799

P=X(9), ppP=non-ETC 0.972023454564163

P=X(10), ppP=non-ETC 2.238481946743318

P=X(20), ppP=non-ETC 6.363850996796102

P=X(21), ppP=non-ETC -1.717011738240629

P=X(22), ppP=non-ETC -4.036288926987237

Of these, only X(140) and X(3628) lie on the Euler line.Β The ppP for points P on the Euler line do not even lie on the same conic.

Locus?

Randy Hutson, Hyacinthos #21519