Let ABC be a triangle and A'B'C' the cevian triangle of P = I
Denote:
Ab = the orthogonal projection of A' on the parallel to BB' through A
Ac = the orthogonal projection of A' on the parallel to CC' through A
Similarly (cyclically) Bc, Ba and Ca,Cb.
Conjecture: ABC and The triangle bounded by the Euler lines of AAbAc, BBcBa, CCaCb are parallelogic.
Locus ?
Antreas P. Hatzipolakis. 7 September 2014
Generalization: Hyacinthos #22570
Geometrical Loci associated with the Euler Line
Let ABC be a triangle and A'B'C' the cevian triangle of P = O
Denote:
Ab = the orthogonal projection of A' on the parallel to BB' through A
Ac = the orthogonal projection of A' on the parallel to CC' through A
Similarly (cyclically) Bc, Ba and Ca,Cb.
Conjecture: The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent.
Locus ?
Antreas P. Hatzipolakis, 7 September 2014
Generalization: Hyacinthos #22570
Geometrical Loci associated with the Euler Line
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,
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
ADDENDUM:
Denote:
Oab, Oac = the circumcenters of AA'B, AA'C, resp.
(O1) = the circumcircle of A'OabOac. Similarly (O2), (O3).
P* = the radical center of (O1),(O2), (O3)
For P = G, the G* is lyimg on the Euler line of ABC.
Locus: Which is the locus of P such that P* is lying 1. on the OP line 2. on the Euler Line of ABC?
Antreas P. Hatzipolakis, 1 September 2014
Peter Moses:
G* = X(140) = midpoint of ON
GENERALIZATION
Let P is a point on Euler line of triangle ABC and DEF is pedal triangle of P. Let Oab,Oac be the circumcenters of triangle DAB,DAC and (Oa) is circumcircle of triangle DOabOac. Similarly, we have circle (Ob),(Oc). Then radical center of (Oa),(Ob),(Oc) lies on Euler line, too.
