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.

Facebook

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)

César Lozada, Hyacinthos #22165

Radical axes:

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:

1st Morley Triangle

2nd Morley Triangle

3rd Morley Triangle

Roussel Triangle

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


 

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