Παρασκευή, 3 Οκτωβρίου 2014

I - CONCURRENT EULER LINES

Let ABC be a triangle and A'B'C' the cevian triangle of P = I (incenter).

Denote:

Oab, Oac = the circumcenters of ABA', ACA', resp.

Oa = the circumcenter of AOabOac. Similarly Ob, Oc

1. The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent.

2. The Orthocenter of OaObOc is lying on the Euler Line of ABC.

Note:

For A'B'C' = cevian triangle of a point P, the circumcircles of AOabOac, BObcOba, COcaOcb are concurrent at O of ABC.

By Miquel theorem in the quadrilateral (AB,BC,CA,AA'), O is lying on the circumcircle of AOabOac. Similarly O is lying on the circumcircles of BObcOba, COcaOcb

Antreas P. Hatzipolakis, 3 October 2014

*** The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent in X(186) = Inverse in circumcircle of orthocenter.

*** The Orthocenter of OaObOc is

X=(a^2 (a^2 - b^2 - b c - c^2) (a^5 (b + c) - 2 a^3 (b^3 + c^3) - a^2 b c (b^2 + c^2) + a (b^5 - b^4 c - b c^4 + c^5)+ b c (b^2 - c^2)^2) : ... :...)

lying on the Euler Line of ABC, and with (6-9-13)-search number 1.269044824987441168348286504.

X = (r^2 + 2 r R - R^2 + s^2)X(3) + R^2 X(4)

Angel Montesdeoca. Hyacinthos #22604

1) X(186)

2) a^2 (a^2-b^2-b c-c^2) (a^5 b-2 a^3 b^3+a b^5+a^5 c-a^2 b^3 c-a b^4 c+b^5 c-2 a^3 c^3-a^2 b c^3-2 b^3 c^3-a b c^4+a c^5+b c^5)::

on lines {{2,3},{35,500},{55,5453},{511,5495},{3724,5492}},

Search = 3.7222371386671506172

Agree with barys, however I think the search number is suspect. Reckon it should be 3.7222371386671506172.

Peter Moses


Τετάρτη, 1 Οκτωβρίου 2014

PARALLEL NN-LINES

10. Let ABC be a triangle and P a point.

Denote:

Ab, Ac = the orthogonal projections of A on PB,PC, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc.

Similarly Nb1, Nb2 and Nc1, Nc2.

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel.

11. If P = I,

the lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

21. Let ABC be a triangle.

Denote:

Na1 = the NPC center of IBC

Na2 = the NPC center of Na1BC.

Similarly Nb1,Nb2, Nc1,Nc2

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

31. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Ab, Ac = the orthogonal projections of A on IaIc, IaIb, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

41. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Na1 = the NPC center of IaBC

Oa = the circumcenter of IaBC

Nao1 = The NPC center of OaBC.

Similarly Nb1, Nbo1, Nc1, Nco1.

The lines Na1Nao1, Nb1Nbo1, Nc1Nco1 are parallel to OI line of ABC.

Antreas P. Hatzipolakis, 1 October 2014


Σάββατο, 6 Σεπτεμβρίου 2014

EULER LINES - PARALLELOGIC TRIANGLES - I

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


CONCURRENT EULER LINES - O

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


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,

Tran Quang Hung.


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

************************************

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


Δευτέρα, 1 Σεπτεμβρίου 2014

RADICAL CENTER - EULER LINE

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

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.

Tran Quang Hung.

Response


 

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