## Πέμπτη, 4 Σεπτεμβρίου 2014

### MOBY LOCI PROBLEMS and REWARD (PRIZE) ---- ADDENDUM ---

[APH]

Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.

Denote:

Pab, Pac = the orthogonal projections of Pa on OB,OC, resp.

(N1) = the NPC of PaPabPac. Similarly (N2),(N3)

Ra = the radical axis of (N2),(N3. Similarly Rb, Rc.

Sa = the parallel to Ra through A. Similarly Sb, Sc

1. Which is the locus of P such that Sa,Sb,Sc are concurrent? The Euler Line + + ??

2. Let P be a point on the Euler Line.

2.1. Which is the locus of the radical center P' of (N1),(N2),(N3) [point of concurrence of Ra,Rb,Rc] as P moves on the Euler line?

2.2. Which is the locus of the point of concurrence P" of Sa,Sb,Sc (if concur) as P moves on the Euler line?

If the loci are new I name them 1st MOBY Locus and 2nd MOBY Locus and the points P',P" as P'-Moby Point and P"-Moby point.

[Tran Quang Hung]

I have some idea as following

Problem 1. Let ABC be a triangle, P a point on Euler line and PaPbPc the pedal triangle of P. O is circumcenter of ABC.

Denote:

Pab, Pac = the orthogonal projections of Pa on OB,OC, resp.

Na = NPC center of PaPabPac. Similarly Nb,Nc.

Actually, in the original problem then triangle ABC and NaNbNc are orthogonal and more if the lines passing through Na,Nb,Nc and perpendicular to BC,CA,AB concur at point Q, then Q lies on a fixed line when P move.

If Oa = circumcenter of PaPabPac. Similarly Ob,Oc, then triangle ABC and OaObOc are orthogonal. If the lines passing through Oa,Ob,Oc and perpendicular to BC,CA,AB concur at point R, then R lies on Euler line, too.

Further more, if Ka divide OaNa in ratio t. Similarly Kb,Kc, then triangle ABC and KaKbKc are orthogonal. If the lines passing through Ka,Kb,Kc and perpendicular to BC,CA,AB concur at point Q(t), then Q(t) lies on a fixed line d(t). Note that d(t) passes through O.

[APH]

Problem 2. Let ABC be a triangle, P a point on Euler line and PaPbPc the pedal triangle of P. H is orthocenter of ABC.

Denote:

Pab, Pac = the orthogonal projections of Pa on HB,HC, resp.

[Tran Quang Hung]:

Oa = circumcenter of PaPabPac

Ha = orthocenter of PaPabPac

Ka divide OaHa in ratio t. Similarly Kb,Kc.

Then triangle ABC and KaKbKc are orthogonal. If the lines passing through Ka,Kb,Kc and perpendicular to BC,CA,AB concur at point Q(t), then Q(t) lies on a fixed line d(t). Note that d(0) = Euler line.

Best regards,