## Τετάρτη, 3 Σεπτεμβρίου 2014

### MOBY LOCI PROBLEMS and REWARD (PRIZE)

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.

REWARD:

For a complete solution I offer the rare book of J. Neuberg, Sur les projections et contre-projections d' un triangle fixe, Bruxelles 1890.

Antreas P. Hatzipolakis, 4 September 2014

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1: Euler line and this conic:

2 a^2 (a^2 - b^2 - c^2) (a^6 b^2 - 3 a^4 b^4 + 3 a^2 b^6 - b^8 + a^6 c^2 + 2 a^2 b^4 c^2 - 3 b^6 c^2 - 3 a^4 c^4 + 2 a^2 b^2 c^4 + 8 b^4 c^4 + 3 a^2 c^6 - 3 b^2 c^6 - c^8) x^2 + (3 a^12 - 11 a^10 b^2 + 15 a^8 b^4 - 10 a^6 b^6 + 5 a^4 b^8 - 3 a^2 b^10 + b^12 - 11 a^10 c^2 + 21 a^8 b^2 c^2 + 6 a^6 b^4 c^2 - 33 a^4 b^6 c^2 + 19 a^2 b^8 c^2 - 2 b^10 c^2 + 15 a^8 c^4 + 6 a^6 b^2 c^4 + 56 a^4 b^4 c^4 - 16 a^2 b^6 c^4 - b^8 c^4 - 10 a^6 c^6 - 33 a^4 b^2 c^6 - 16 a^2 b^4 c^6 + 4 b^6 c^6 + 5 a^4 c^8 + 19 a^2 b^2 c^8 - b^4 c^8 - 3 a^2 c^10 - 2 b^2 c^10 + c^12) y z + cyclic

2.1: A nasty cubic.

2.2: circumconic thru X(3519)

(b^2-c^2) (a^2-b^2-c^2) (a^8-3 a^6 b^2+4 a^4 b^4-3 a^2 b^6+b^8-3 a^6 c^2-11 a^4 b^2 c^2+3 a^2 b^4 c^2-4 b^6 c^2+4 a^4 c^4+3 a^2 b^2 c^4+6 b^4 c^4-3 a^2 c^6-4 b^2 c^6+c^8) y z + cyclic

Peter Moses 4 September 2014

Suppose we parameterize a point on the Euler lines as a^2 SA + k SB SC::, then the concurrence is

1 / (a^2 SA (S^2 + 5 SA^2) + k (3 S^2 - SA^2) SB SC)::

Thus:

1): L, k = -1

concurrence = 1/(a^2 SA (S^2+5 SA^2)-(3 S^2-SA^2) SB SC)::

2): O, k = 0

concurrence = 1/(a^2 SA (S^2+5 SA^2)):: on lines {{4,3521},{93,403},...}

3): G, k = 1

concurrence = 1/(a^2 SA (S^2+5 SA^2)+(3 S^2-SA^2) SB SC)::

4): N, k = 2

concurrence = 1/(a^2 SA (S^2+5 SA^2)+2 (3 S^2-SA^2) SB SC)::

5): H, k = Infinity

concurrence = 1/((3 S^2-SA^2) SB SC):: = X(3519).

6): Schiffler, k = R/(r+R)

concurrence = 1/(a^2 (r+R) SA (S^2+5 SA^2)+R (3 S^2-SA^2) SB SC)::

Peter Moses 5 September 2014