## Σάββατο, 27 Δεκεμβρίου 2014

### ANOPOLIS PRIMES

Pn = n + Qk, n is a natural number (including 0) ie n belongs to No = {0, 1, 2, 3, ....} with n > 0

where Qk is the smallest number in the set No - {Q1, Q2,.... Qk-1} such that Pn is prime

n = 1

Q1 = the smallest number in the set No - {Q0} = No - {0} such that 1 + Q1 is prime = 1

P1 = 1 + 1 = 2

n = 2

Q2 = the smallest number in the set No - {Q0, Q1} = No - {0, 1}such that 2 + Q2 is prime = 3

P2 = 2 + 3 = 5

n = 3

Q3 = the smallest number in the set No - {Q0, Q1, Q2} = No - {0, 1, 3} such that 3 + Q3 is prime = 2

P3 = 3 + 2 = 5

n = 4

Q4 = the smallest number in the set No - {Q0, Q1, Q2, Q3} = No - {0, 1, 3, 2} such that 4 + Q4 is prime = 7

P4 = 4 + 7 = 11

and so on.

Pn : 2, 5, 5, 11, 11, 11, 11, 17, 17, 23, 23, 23, 23, 29, 29, 37, 37, 37, 37, 37, 37, 47, 47, 47, 47, 53, 53, 59, 59, 59, 59, 59, 67, 67, 67, 67, 73, 73, 79, 79, 83, 83, 83, 83, 89, 89, 97, 97, 97, 97, 97, 97,.....

Anopolis Prime Numbers Sequence:

An = the missing primes from Pn: 3, 7, 13, 19, 31, 41, 43, 61, 71,.............................

Antreas P Hatzipolakis, 27 Dec. 2014

## 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

### 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

## 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,

## 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

1. = X(34223)
2. = X(15424)

## 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.

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:

## Τετάρτη, 4 Ιουνίου 2014

### CONCYCLIC CIRCUMCENTERS - 2

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

Ab,Ac = the reflections of A' in CC',BB', resp. (lying on AC, AB, resp.)

Similarly Bc,Ba = the reflections of B' in AA',CC', resp. and Ca,Cb = the reflections of C' in BB',AA', resp.

A1B1C1 = the antipedal triangle of I wrt triangle A'AbAc

Similarly A2B2C2 = the antipedal triangle of I wrt triangle B'BcBa and A3B3C3 = the antipedal triangle of I wrt triangle C'CaCb,

Oa,Ob,Oc = the circumcenters of the triangles A1B1C1, A2B2C2, A3B3C3, resp.

The circumcenter O of ABC and the circumcenters Oa,Ob,Oc of A1B1C1,A2B2C2,A3B3C3, resp. are concyclic.

Which point is the center X of the circle?

Antreas P. Hatzipolakis, 4 June 2014

X is X(5495)

Peter Moses 5 June 2014

## Πέμπτη, 29 Μαΐου 2014

### APPLES ( A problem in Facebook)

Μια φορά περνούσανε τρεις από έναν τόπο που ήταν μια μηλιά φορτωμένη με μήλα. Κόβει ο ένας το ένα τρίτο των μήλων. Απ' όσα μείνανε κόβει ο δεύτερος το ένα τρίτο. Και απ' όσα μείνανε κόβει ο τρίτος το ένα τρίτο. Όσα μείνανε τελικά τα μοιράσανε με τέτοιο τρόπο που και οι τρεις πήρανε τον ίδιο αριθμό μήλων συνολικά.

Πόσα μήλα είχε η μηλιά και πόσα πήρε ο καθένας; Ζητείται η μικρότερη λύση.

Ας υποθέσουμε ότι είχε x μήλα. Ο πρώτος κόβει x/3 και μένουν στην μηλιά x - (x/3) = 2x/3. Ο δεύτερος κόβει το 1/3 από αυτά δηλαδή (2x/3)/3 = 2x/9. Μένουν τώρα στη μηλιά (2x/3) - (2x/9) = 4x/9. Από αυτά ο τρίτος κόβει το 1/3 δηλαδή (4x/9)/3 = 4x/27. και μένουν στην μηλιά πάνω 4x/9 - 4x/27 = 8x/27.

Συνοψίζω: Ο πρώτος έκοψε x/3, ο δεύτερος 2x/3, ο τρίτος 4x/27 και μείνανε στην μηλιά 8x/27.

Αυτά που μείνανε θα τα μοιράσουν σε τρία μέρη a,b,c έτσι ώστε ο πρώτος να πάρει a, o δεύτερος b και ο τρίτος c και να έχουν τον ίδιο αριθμό μήλων τελικά ο καθένας.

Έτσι έχουμε:

(x/3)+ a = (2x/3) + b = (4x/27) + c. Επειδή τώρα ζητούμε τον μικρότερο αριθμό μήλων, θέτουμε στον a την μικρότερη ακέραιη θετική τιμή δηλαδή 0. Και οι εξισώσεις μας γίνονται: x/3 = (2x/3) + b = (4x/27) + c. Λύνοντας ως προς b και c βρίσκουμε ότι:

b = x / 9 και c = 5x / 27. Επειδή τώρα ζητούμε τον μικρότερο ακέραιο x έτσι ώστε και οι b = x / 9 ,c = 5x / 27 να είναι ακέραιοι, αυτός είναι ο 27. Έτσι ο πρώτος πήρε 9 + 0 ο δεύτερος 8 + 1 και ο τρίτος 4 + 5.

## Σάββατο, 3 Μαΐου 2014

### ORTHOLOGIC - PERSPECTIVE TRIANGLES

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

Denote:

Ab, Ac = The circumcenters of APB', APC', resp.

Bc, Ba = The circumcenters of BPC', BPA', resp.

Ca, Cb = The circumcenters of CPA', CPB', resp.

1. M1a,M1b,M1c = The midpoints of AbAc,BcBa,CaCb, resp.

Which is the locus of P such that:

1.1. ABC, M1aM1bM1c are perspective?

1.2. ABC, M1aM1bM1c are orthologic?

1.3. The perpendicular bisectors of AbAc,BcBa,CaCb are concurrent?

For P = G:

1.2. ABC, M1aM1bM1c are orthologic.

Orthologic center (M1aM1bM1c, ABC) = N

Orthologic center (ABC, M1aM1bM1c) : Anopolis #1284, #1295

1.3. The perpendicular bisectors concur at van Lamoen Circle Center X(1153)

2. M2a,M2b,M2c = The midpoints of BcCb, CaAc, AbBa, resp.

Which is the locus of P such that:

2.1. ABC, M2aM2bM2c are perspective?

2.2. ABC, M2aM2bM2c are orthologic?

2.3. The perpendicular bisectors of BcCb, CaAc, AbBa are concurrent?

For P = G:

2.2. ABC, M2aM2bM2c are orthologic.

Orthologic center (M2aM2bM2c, ABC) = ?

Orthologic center (ABC, M2aM2bM2c) = G

2.3. The perpendicular bisectors concur at van Lamoen Circle Center X(1153)

3. M3a,M3b,M3c = The midpoints of BaCa, CbAb, AcBc, resp.

Which is the locus of P such that:

3.1. ABC, M3aM3bM3c are perspective?

3.2. ABC, M3aM3bM3c are orthologic?

3.3. The perpendicular bisectors of BaCa, CbAb, AcBc are concurrent?

For P = G

3.2. ABC, M3aM3bM3c are orthologic.

Orthologic center (M3aM3bM3c, ABC) = O

Orthologic center (ABC, M3aM3bM3c) = ?

3.3. The perpendicular bisectors concur at van Lamoen Circle Center X(1153)

4. Which is the locus of P such that:

4.1. M1aM1bM1c, M2aM2bM2c

4.2. M1aM1bM1c, M3aM3bM3c

4.3. M2aM2bM2c, M3aM3bM3c

are perspective/orthologic ?

4.4. The Euler lines of M1aM1bM1c, M2aM2bM2c, M3aM3bM3c are concurrent?

For P = G ??

5. Which is the locus of P such that:

4.1. M1aM2aM3a, M1bM2bM3b

4.2. M1aM2aM3a, M1cM2cM3c

4.3. M1bM2bM3b, M1cM2cM3c

are perspective/orthologic ?

4.4. The Euler lines of M1aM2aM3a, M1bM2bM3b, M1cM2cM3c are concurrent?

For P = G ??

Antreas P. Hatzipolakis, 4 May 2014

## Δευτέρα, 21 Απριλίου 2014

### EULER LINES OF TRIAGLES BOUNDED BY REFLECTED PARALLEL LINES

Dao Thanh Oai:

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C respectively. The reflections of L1,L2,L3 in BC,CA,AB, resp. bound a triangle A1B1C1. The Euler line of A1B1C1 passes through a fixed point (as the three lines L1,L2,L3 move around A,B,C, being parallel)

Francisco Javier García Capitán:

The point is (f(a,b,c):f(b,c,a):f(c,a,b)) where f(a,b,c) is

a^22 - 8 a^20 b^2 + 28 a^18 b^4 - 56 a^16 b^6 + 70 a^14 b^8 - 56 a^12 b^10 + 28 a^10 b^12 - 8 a^8 b^14 + a^6 b^16 - 8 a^20 c^2 + 42 a^18 b^2 c^2 - 92 a^16 b^4 c^2 + 106 a^14 b^6 c^2 - 62 a^12 b^8 c^2 + 7 a^10 b^10 c^2 + 13 a^8 b^12 c^2 - 8 a^6 b^14 c^2 + 4 a^4 b^16 c^2 - 3 a^2 b^18 c^2 + b^20 c^2 + 28 a^18 c^4 - 92 a^16 b^2 c^4 + 113 a^14 b^4 c^4 - 62 a^12 b^6 c^4 + 17 a^10 b^8 c^4 - 9 a^8 b^10 c^4 + 5 a^6 b^12 c^4 - 6 a^4 b^14 c^4 + 13 a^2 b^16 c^4 - 7 b^18 c^4 - 56 a^16 c^6 + 106 a^14 b^2 c^6 - 62 a^12 b^4 c^6 + 4 a^10 b^6 c^6 + 4 a^8 b^8 c^6 + 8 a^6 b^10 c^6 - 6 a^4 b^12 c^6 - 18 a^2 b^14 c^6 + 20 b^16 c^6 + 70 a^14 c^8 - 62 a^12 b^2 c^8 + 17 a^10 b^4 c^8 + 4 a^8 b^6 c^8 - 12 a^6 b^8 c^8 + 8 a^4 b^10 c^8 + 3 a^2 b^12 c^8 - 28 b^14 c^8 - 56 a^12 c^10 + 7 a^10 b^2 c^10 - 9 a^8 b^4 c^10 + 8 a^6 b^6 c^10 + 8 a^4 b^8 c^10 + 10 a^2 b^10 c^10 + 14 b^12 c^10 + 28 a^10 c^12 + 13 a^8 b^2 c^12 + 5 a^6 b^4 c^12 - 6 a^4 b^6 c^12 + 3 a^2 b^8 c^12 + 14 b^10 c^12 - 8 a^8 c^14 - 8 a^6 b^2 c^14 - 6 a^4 b^4 c^14 - 18 a^2 b^6 c^14 - 28 b^8 c^14 + a^6 c^16 + 4 a^4 b^2 c^16 + 13 a^2 b^4 c^16 + 20 b^6 c^16 - 3 a^2 b^2 c^18 - 7 b^4 c^18 + b^2 c^20

Reference: Facebook Group "Short Mathematical Idea"

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and let A'B'C' be the triangle bounded by the reflections of L1,L2,L3, in BC,CA,AB, resp.

Where the lines L1,L2,L3 move around A,B,C, being parallel, the Euler line of A'B'C' passing through a point Q which we shall call the Parry-Pohoata point. Barycentric coordinates for Q, of degree 22 in a,b,c, were found by J. F. Garcia Captitán (Hyacinthos #15827, Nov. 19, 2007) and are included in Pohoata's article Cosmin Pohoata, "On the Parry reflection point," Forum Geometricorum 8 (2008), 43-48

Angel Montesdeoca, Hyacinthos #22171

Conjecture:

Let A'B'C' be the antimedial triangle of ABC. The reflections of the parallel lines L1,L2,L3 trough A,B,C, resp. in the sidelines of A'B'C' (instead of ABC) bound a triangle whose the Euler line passes through a fixed point.

In general:

Let ABC, A'B'C' be two homothetic triangles. Let L1,L2,L3 be three parallel lines through A,B,C, resp. The reflections of L1,L2,L3 in the sidelines B'C',C'A',A'B', resp. of A'B'C' bound a triangle whose the Euler line passes through a fixed point.

Antreas P. Hatzipolakis, 21 April 2014

Special case:

If A’B’C’ is the antimedial triangle of ABC, the fixed point on Euler lines has trilinear coordinates:

(2*f(6)+6*f(4)+14*f(2)+7)*g(1) -2*(2*f(5)+2*f(3)+5*f(1))*g(2) +(2*f(2)+2*f(4)+3)*g(3) -2*(f(3)+f(1))*g(4) -(f(7)+f(5)+6*f(3)+6*f(1)) : :

Where f(n)=cos(n*A) and g(n)=cos(n*(B-C))

This point is on line (1141,3484) and has ETC-6-9-13 numbers:

(4.923838044604385591130, 4.248841971282335390013, -1.573382134182338747412)

Let (Oa), (Na) be the circumcenter, NPC center, resp. of the triangle A1B1C1 (=the triangle bounded by the reflections in BC,CA,AB of the parallels through A,B,C, resp.).

Conjecture:

The radical axis of (Oa), (Na) passes through a fixed point.

Generalization:

Let ABC, A'B'C' be two homothetic triangles. Let L1,L2,L3 be three parallel lines through A,B,C, resp. The reflections of L1,L2,L3 in the sidelines B'C',C'A',A'B', resp. of A'B'C' bound a triangle A1B1C1

Let (Oa), (Na) be the circumcenter, NPC center of A1B1C1. The radical axis of (Oa),(Na) passes through a fixed point.

Antreas P. Hatzipolakis, 22 April 2014

The point Y is (f(a,b,c):f(b,c,a):f(c,a,b)) where f(a,b,c) is

a^10(b^2+c^2) -a^8(3b^4+4b^2c^2+3c^4) +a^6(2b^6+5b^4c^2+5b^2c^4+2c^6) +2a^4(b^8-3b^6c^2+b^4c^4-3b^2c^6+c^8) -a^2(b^2-c^2)^2(3b^6-4b^4c^2-4b^2c^4+3c^6) +(b^2-c^2)^6+2a^4(b^8-3b^6c^2+b^4c^4-3b^2c^6+c^8)

Angel Montesdeoca, Hyacinthos #22171 and in HG

CONJECTURES:

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and A'B'C' the triangle bounded by the reflections of L1,L2,L3 in the sidelines BC,CA,AB of ABC, resp.

Conjecture 1.

Let P be a fixed point. As the three lines L1,L2,L3 move around A,B,C, being parallel, the point P wrt triangle A'B'C' moves on a fixed circle. Note: We have seen the cases of the Euler lines and the radical axes of (O) and (N). Conjecture 2.

Let L be a fixed line. As the lines L1,L2,L3 move around A,B,C, being parallel, the line line L wrt A'B'C' passes through a fixed point (the envelope of the lines is a degenerated circle).

Problem: Which are the envelopes of fixed circles of A'B'C' (circumcircle, NPC, incircle ...) ?

Antreas P. Hatzipolakis, 25 April 2014

## Κυριακή, 20 Απριλίου 2014

### HOMOTHETIC TRIANGLES and EULER LINES

Theorem 1.

Let ABC be an equilateral triangle and P a point. The Euler lines of the triangles PBC,PCA,PAB are concurent.Denote the point of concurrence with P*.

Reference:

APH, Hyacinthos #21592

Locus Problems:

As P moves on a line or on a circle [special case the incircle of ABC] which is the locus of P*?

Application:

Let ABC be a triangle and A',B',C' the apices of the equilateral triangles erected out/inwardly ABC, P a point and Pa, Pb, Pc the respective points of concurrences of the Euler lines wrt equil. triangles A'BC,B'CA,C'AB.

Which is the locus of P such that ABC, PaPbPc are perspective or orthologic?

Theorem 2 (generalization of Th. 1).

Let ABC, A'B'C' be two homothetic equilateral triangles. The Euler lines of the triangles AB'C', BC'A', CA'B' (and of the triangles A'BC, B'CA, C'AB) are concurrent.

Application to Morley Configuration with homothetic equilateral triangles:

Theorem 3.

Let ABC, A'B'C' be two dilated triangles with scale factor 1. The Euler lines of AB'C', BC'A', CA'B' (and of the triangles A'BC, B'CA, C'AB) are concurrent.

Application:

Let ABC be a triangle and (O1),(O2),(O3) the reflections of the circumcircle (O) in BC,CA,AB, resp. Denote:

Ab = the second intersection of (O2) and the reflection of AN in the bisector of the angle HAC.

Ac = the second intersection of (O3) and the reflection of AN in the bisector of the angle HAB.

Similarly (cyclically) Bc, Ba and Ca, Cb

Oa, Ob, Oc = the circumcenters of the triangles ABcCb, BCaAc, CAbBa

The triangles ABC, OaObOc are dilated triangles with scalar facror 1. The Euler lines of AObOc, BOcOa, COaOb are concurrent and also the Euler lines of OaBC, ObCA, OcAB.

Note: The circumradii of the triangles ABcCb, BCaAc, CAbBa are equal.

Antreas P. Hatzipolakis, 20 April 2014

## Σάββατο, 19 Απριλίου 2014

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

Denote:

(Oa) = the circumcircle of (PBC)

(O1) = the reflection of (Oa) in BC

(O'1) = the reflection of (O1) in PA'

Ra = the radical axis of the pedal circle of P and (Oa)

R1 = the radical axis of the pedal circle of P and (O1)

R'1 = the radical axis of the pedal circle of P and (O'1)

Similarly Rb,Rc, R2,R3, R'2,R'3

Sa = the radical axis of the antipedal circle of P and (Oa)

S1 = the radical axis of the antipedal circle of P and (O1)

S'1 = the radical axis of the antipedal circle of P and (O'1)

Similarly Sb,Sc, S2,S3, S'2,S'3

Ab, Ac =

1. the orthogonal projections of Oa on Rb,Rc, resp.

APH, Hyacinthos #22150

2. the orthogonal projections of Oa on R2,R3, resp.

APH, Hyacinthos #22153

3. the orthogonal projections of Oa on R'2,R'3, resp.

APH, Hyacinthos #22153

4. the orthogonal projections of Oa on Sb,Sc, resp.

5. the orthogonal projections of Oa on S2,S3, resp.

6. the orthogonal projections of Oa on S'2,S'3, resp.

Similarly Bc,Ba and Ca, Cb

Which is locus of P such that the perpendicular bisectors of the line segments BaCa, CbAb,AcBc are concurrent?

1.

The perpendicular bisectors of the line segments BaCa, CbAb, AcBc are concurrent for all P.

If P=(x:y:z), the perpendicular bisectors concurrent in Q with first barycentric coordinate:

a^6y^2z^2(4x^2+4y*z+3x(y+z)) + a^4x*y*z(c^2y(5x^2+2x*y+2(y-z)z) + b^2z(5x^2+2x z+2y(-y+z))) - a^2x(c^4y^2(z^2(y+z)+2x*z(3y+2z)+2x^2(y+3z)) + b^4z^2(y^2(y+z)+2x^2(3y+z)+2x*y(2y+3z)) + 2b^2c^2y z(4x^2(y+z)-y*z(y+z) + x(y^2+4y*z+z^2))) + (b^2-c^2)x^3(c^4y^2(2y-z)+3b^2c^2y(y-z)z + b^4(y-2z)z^2)

The unique pairs of points {P, Q}, both being in ETC are, {X(1),X(3)} and {X(4),X(546)}.

If P=(x:y:z) lies on the circumcircle or line at infinity, the construction does not make sense.

However, as the coordinates sum of Q is 4(a-b-c)(a+b-c)(a-b+c)(a+b+c)xyz(x+y+z)(c^2xy+b^2xz+a^2yz), then Q lies on the line at infinity.

In fact, if P is in the circumcircle then Q is the isogonal conjugate of P.

Angel Montesdeoca, Hyacinthos #22152

## Τρίτη, 15 Απριλίου 2014

### CIRCUMCENTERS ON THE LINES OH (Euler Line), OI

Let ABC be a triangle and

1. A'B'C' the pedal triangle of H (orthic triangle)

Denote:

(Oa) = the circumcircle of OBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in HA'.

(Ob) = the circumcircle of OCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in HB'.

(Oc) = the circumcircle of OAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in HC'.

The circumcenter of the triangle O'1O'2O'3 lies on the OH line (Euler line)

-----

2. A'B'C' the pedal triangle of I.

Denote:

(Oa) = the circumcircle of IBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in IA'.

(Ob) = the circumcircle of ICA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in IB'.

(Oc) = the circumcircle of IAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in IC'.

The circumcenter of the triangle O'1O'2O'3 lies on the OI line

Generalizations (Loci):

Let ABC be a triangle, P,P* two isogonal conjugate points and A'B'C',A"B"C" the pedal triangles of P,P*.

Denote:

(Oa) = the circumcircle of PBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in PA'.

(O"1) = the reflection of (O1) in P*A"

(Ob) = the circumcircle of PCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in PB'.

(O"2) = the reflection of (O2) in P*B"

(Oc) = the circumcircle of PAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in PC'.

(O"3) = the reflection of (O3) in P*C"

R' = the circumcenter of O'O'2O'3

R" = the circumcenter of O"1O"2O"3

Which is the locus of P such that:

1. O, P, R'

2. O, P, R"

3. O, R', R"

4. P, P*, R'

5. P, R', R"

are collinear ?

The McCay cubic?

Antreas P. Hatzipolakis, 16 April 2014.

### RADICAL CENTERS ON THE LINES OH (Euler line),OI

Let ABC be a triangle and

1. A'B'C' the pedal triangle of H (orthic triangle)

Denote:

(Oa) = the circumcircle of OBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in HA'.

(Ob) = the circumcircle of OCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in HB'.

(Oc) = the circumcircle of OAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in HC'.

The radical center of (O'1),(O'2),(O'3) lies on the OH line (Euler line)

-----

2. A'B'C' the pedal triangle of I.

Denote:

(Oa) = the circumcircle of IBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in IA'.

(Ob) = the circumcircle of ICA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in IB'.

(Oc) = the circumcircle of IAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in IC'.

The radical center of (O'1),(O'2),(O'3) lies on the OI line.

Generalizations (Loci):

Let ABC be a triangle, P,P* two isogonal conjugate points and A'B'C',A"B"C" the pedal triangles of P,P*.

Denote:

(Oa) = the circumcircle of PBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in PA'.

(O"1) = the reflection of (O1) in P*A"

(Ob) = the circumcircle of PCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in PB'.

(O"2) = the reflection of (O2) in P*B"

(Oc) = the circumcircle of PAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in PC'.

(O"3) = the reflection of (O3) in P*C"

R' = the radical center of (O'1),(O'2),(O'3)

R" = the radical center of (O"1),(O"2),(O"3)

Which is the locus of P such that:

1. O, P, R'

2. O, P, R"

3. O, R', R"

4. P, P*, R'

5. P, R', R"

are collinear ?

The McCay cubic?

Antreas P. Hatzipolakis, 15 April 2014.

## Παρασκευή, 11 Απριλίου 2014

### CONCURRENT CIRCLES -- EULER LINE

Theorem:

Let ABC be a triangle and A',B',C' three points. If the circumcircles of A'BC, B'CA, C'AB are concurrent, then also the circumcircles of AB'C', BC'A', CA'B' are concurrent.

Corollary:

Let ABC be a triangle and P a point. If A',B',C' are arbitrary points on the circumcircles of PBC,PCA,PAB, resp. then the circumcircles of AB'C', BC'A', CA'B' are concurrent.

Applications:

Let P,Q be two points and PaPbPc, Q1Q2Q3 the antipedal, pedal triangles of P,Q, resp. The orthogonal projections A',B',C' of Pa,Pb,Pc on PQ1,PQ2,PQ3, resp. lie on the circumcircles of PBC,PCA,PAB, resp.

The circumcircles of AB'C', BC'A', CA'B' concur at a point D.

1. For P = O, Q = H:

The point D lies on the Euler line of ABC

2. For P = Q = I:

The point D lies on the Euler line of ABC

3. For P = Q = H:

The Euler lines L,L1,L2,L3 of ABC, DBC,DCA,DAB are concurrent at a point D' on the Neuberg cubic.

The parallels to L1,L2,L3 through A,B,C, resp. are concurrent at a point D"

Coordinates of the points D's, D' and D"?

More pairs (P,Q) such that the circumcircles concur on the Euler line?

Antreas P. Hatzipolakis, 11 April 2014

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

It seems that the general case for D has truly appalling barycentrics.

But, for the cases you mention we have:

1) PQ = OH, D= X(186)

2) PQ = II, D = X(1325)

3) PQ = HH, D = X(1157)

Also PQ = OI, D = X(36)

Peter Moses, 14 April 2014

## Πέμπτη, 10 Απριλίου 2014

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

Ab, Ac = the reflections of A' in AB, AC. resp.

A2, A3 = the reflections of A' in BB', CC', resp.

Oab, Oac = the circumcenters of BAbA2, CAcA3, resp.

Similarly (cyclically):

Obc, Oba and Oca, Ocb.

1. P = O.

The radical axes R1 =:((Oab),(Oac)), R2 =:((Obc),(Oba)), R3 =:((Oca),(Ocb))are concurrent at O.

The reflections of R1,R2,R3 in BC,CA,AB are the lines AN,BN,CN, resp.

2. P = H.

The radical axes R1 =:((Oab),(Oac)), R2 =:((Obc),(Oba)), R3 =:((Oca),(Ocb)) are concurrent on the Euler line of ABC.

The reflections of the radical axes S1 =:((Obc), (Ocb)), S2 =:((Oca),(Oac)), S3 =: ((Oab), (Oba)) in BC,CA,AB, resp. are concurrent.

The reflections of the radical axes T1 =: ((Oba), (Oca)), T2 =:((Ocb),(Oab)), T3 =:((Oac), (Obc)) in Bc,CA,AB are concurrent.

The triangles: bounded by the lines (T1,T2,T3) and the orthic A'B'C' are paralle;ogic.

Antreas P. Hatzipolakis, 10 April 2014