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


Κυριακή, 29 Ιουνίου 2014

NEUBERG CUBIC - PRIZE (REWARD)

Let ABC be a triangle and P = I the incenter.

Denote:

La,Lb,Lc = the Euler lines of PBC, PCA, PAB, resp. (concurrent at Schiffler point)

(Na), (Nb), (Nc) = the NPCs of PBC,PCA,PAB, resp. (concurrent at Poncelet Point of I)

The perpendicular to La at the NPC center Na intersects BC,CA,AB at Aa, Ab, Ac, resp.

A' = BAb /\ CAc. Similarly B', C'.

Conjecture:

The triangles ABC, A'B'C' are perspective and equivalently the points Aa, Bb, Cc are collinear.

General Conjecture: It is true for any P on the Neuberg Cubic (in this case the Euler lines of PBC,PCA, PAB are concurrent).

If the general conjecture is true: For a proof I offer the book (original edition):

Richard Heger: Elemente der analytischen Geometrie in homogenen Coordinaten. 1872

Note: It is available on-line at:

https://archive.org/details/elementederanal01hegegoog

Antreas P. Hatzipolakis, 29 June 2014

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

The conjecture is TRUE, since rhe Neuberg cubic is part of the general locus of P such that The triangles ABC, A'B'C' are perspective !

Peter Moses won the book !

Peter Moses:

Hi Antreas,

Circumcircle + Infinity + Neuberg Cubic, cyclicsum[a^2 ((S^2 - 3 SA SB) y^2 z - (S^2 - 3 SC SA) y z^2)], + maybe 3 imaginary ellipses, (-a^2 + b^2 + c^2) y z + c^2 y^2 + b^2 z^2 = 0, centered on the vertices of ABC ?

Best regards,

Peter.

Hyacinthos #22483


Πέμπτη, 5 Ιουνίου 2014

A PROBLEM

Source: Miguel Ochoa Sanchez in Facebook Group PERU GEOMETRICO

It is an old problem of unknown (to me) origin. It was published also in the Greek mathematical magazine EUCLID (June 1973)

The problem:

And the solution:


 

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