Κυριακή, 31 Μαρτίου 2013

CEVIANS, RADICAL CENTERS, EULER LINE

Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and Ma, Mb, Mc points on AA',BB',CC', resp.

Denote:

X = the radical center of the ciecles (Ma, MaB), (Mb, MbC), (Mc, McA)

Y = the radical center of the circles (Ma, MaC), (Mb, MbA), (Mc, McB)

M = the midpoint of the line segmant XY

1.

Let Ma, Mb, Mc be points such that: MaA / MaA' = MbB / MbB' = McC / McC' = t

Which is the locus of M as t varies?

For P = G, the locus is the Euler line.

For t = -1 (ie Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.)

==> M is the circumcenter O of ABC for all P's.

2. Let Ma, Mb, Mc be points such that: MaA / MaP = MbB / MbP = McC / McP = t

Which is the locus of M as t varies?

For P = O ==> M = the NPC center N

For P = G ==> The locus is the Euler Line.

3. Let Ma, Mb, Mc be points such that: MaP / MaA' = MbP / MbB' = McP / McC' = t

Which is the locus of M as t varies?

For P = G ==> the locus is the Euler line.

Antreas P. Hatzipolakis, 31 March 2013

Σάββατο, 30 Μαρτίου 2013

ORTHOPOLAR CIRCLES

Let ABCD be a quadrilateral [quadragon], A',B',C',D' the circumcenters of BCD, CDA, DAB, ABC, resp. and P a point.

Denote:

1 = the orthopole of PA' wrt BCD

2 = the orthopole of PB' wrt CDA

3 = the orthopole of PC' wrt DAB

4 = the orthopole of PD' wrt ABC

Conjecture:

The points 1,2,3,4 are concyclic. The circle passes through the Poncelet point of ABCD (=the point where the NPCs of BCD, CDA, DAB, ABC concur)

The circle (1,2,3,4) is a line when ABCD is cyclic (ie A' = B' = C'= D' = O ==> PO = PA' = PB' = PC' := L, a line psiing through the circumcenter of the cyclic ABCD)

Antreas P. Hatzipolakis, 30 March 2013

ORTHOPOLAR CIRCLES [triangle]

Let ABC be a triangle, P, Q two points and O, Q1,Q2,Q3 the circumcenters of ABC, QBC, QCA, QAB, resp.

Denote:

P0 = the orthopole of PO wrt ABC

P1 = the orthopole of PQ1 wrt QBC

P2 = the orthopole of PQ2 wrt QCA

P3 = the orthopole of PQ3 wrt QAB

We have:

1. P0, P1, P2, P3 lie on the NPCs (N),(N1),(N2),(N3) of ABC, QBC, QCA, QAB, resp. (since the respective lines pass through the circumcenters of the respective triangles)

2. The NPCs of ABC, QBC, QCA, QAB concur at the Poncelet point Q* of Q wrt ABC.

CONJECTURE:

The points P0, P1, P2, P3, Q* are concyclic.

Antreas P. Hatzipolakis, 30 March 2013.

Τετάρτη, 27 Μαρτίου 2013

SIX NPCs [N of medial, G of orthic]

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

Denote:

Nab = The NPC center of AA'B'

Nac = The NPC center of AA'C'

Nbc = The NPC center of BB'C'

Nba = The NPC center of BB'A'

Nca = The NPC center of CC'A'

Ncb = The NPC center of CC'B'

1. P = G (A'B'C' = medial triangle)

The lines NbaNca, NcbNab, NacNbc are concurrent at the NPC center of A'B'C'.

2. P = H (A'B'C' = orthic triangle)

2.1.The triangles ABC and bounded by the lines (NbcNcb, NcaNac, NabNba) are homothetic

The homothetic center is G'(centroid) of A'B'C'

2.2. The homothetic center of the medial triangle and (NbcNcb, NcaNac, NabNba) is the infinite point of the line GG'.

Antreas P. Hatzipolakis, 27 March 2013

2.3. The radical axes of ((Nab),(Nac)), ((Nbc),(Nba)),((Nca),(Ncb)) are concurrent.

Antreas P. Hatzipolakis, 1 April 2013

Δευτέρα, 25 Μαρτίου 2013

Regular Polygon

A hexakaidecagon.... !

From Black and White pictures

Naming Regular Polygons and Polyhedra:

Prof. Conway writes:

Antreas Hatzipolakis and I worked out a complete system up to the millions from which this is taken, and which has also been "vetted" by several other scholars. The most important of the reasons which make me prefer the "kai" forms is that they permit these prefixes to be unambiguously parsed even when concatenated, as they are in Kepler's names for certain polyhedra; for example, the icosidodecahedron or (20,12)-hedron, so called because it has 20 faces of one type and 12 of another. Kepler said "this particular triacontakaidihedron I call the icosidodecahedron", a remark showing that he also preferred the kai forms.

John Conway

RADICAL CENTERS, CIRCUMCIRCLE, EXCIRCLES

Let ABC be a triangle and Ia,Ib,Ic the excenters and O the circumcenter.

1.

Denote:

r1 = the rdical center of (O),(Ib),(Ic)

r2 = the rdical center of (O),(Ic),(Ia)

r3 = the rdical center of (O),(Ia),(Ib)

Perspective Triangles (?):

1.1. ABC, r1r2r3

1.2. IaIbIc, r1r2r3

2.

Denote:

Ja = the excenter of the excircle respective to BC of the triangle OBC

Jb = the excenter of the excircle respective to CA of the triangle OCA

Jc = the excenter of the excircle respective to AB of the triangle OAB

R1 = the rdical center of (O),(Jb),(Jc)

R2 = the rdical center of (O),(Jc),(Ja)

R3 = the rdical center of (O),(Ja),(Jb)

Perspective Triangles (?):

2.1. ABC, R1R2R3

2.2. JaJbJc, R1R2R3

3.

Denote:

i1 = the radical center of (Ja),(Ib),(Ic)

i2 = the radical center of (Jb),(Ic),(Ia)

i3 = the radical center of (Jc),(Ia),(Ib)

j1 = the radical center of (Ia),(Jb),(Jc)

j2 = the radical center of (Ib),(Jc),(Ja)

j3 = the radical center of (Ic),(Ja),(Jb)

Perspective triangles (?):

3.1. ABC, i1i2i3

3.2. ABC, j1j2j3

3.3. i1i2i3, j1j2j3

3.4. i1i2i3, IaIbIc

3.5. i1i2i3, JaJbJc

3.6. j1j2j3, IaIbIc

3.7. j1j2j3, JaJbJc

3.8. IaIbIc, JaJbJc

Antreas P. Hatzipolakis, 25 March 2013

Κυριακή, 24 Μαρτίου 2013

ORTHOLINES

Let ABC be a triangle and 0,1,2,3 the concurrent (at S) Euler lines of ABC, BCI, CAI, ABI, resp.

Denote:

00 = the orthopole of 0 wrt ABC

01 = the orthopole of 0 wrt IBC

02 = the orthopole of 0 wrt ICA

03 = the orthopole of 0 wrt IAB

----

10 = the orthopole of 1 wrt ABC

11 = the orthopole of 1 wrt IBC

12 = the orthopole of 1 wrt ICA

13 = the orthopole of 1 wrt IAB

----

20 = the orthopole of 2 wrt ABC

21 = the orthopole of 2 wrt IBC

22 = the orthopole of 2 wrt ICA

23 = the orthopole of 2 wrt IAB

----

30 = the orthopole of 3 wrt ABC

31 = the orthopole of 3 wrt IBC

32 = the orthopole of 3 wrt ICA

33 = the orthopole of 3 wrt IAB

In short:

Denote:

Triangles ABC, IBC, ICA, IAB = (0), (1), (2), (3), resp.

Euler lines of ABC, IBC, ICA, IAB = 0,1,2,3, resp.

Orthopole of x [x in {0,1,2,3}] wrt (y) [(y) in {(0), (1), (2), (3)}] = xy

How are organized the 16 points xy ?

1. Collinearity:

The four lines: (01 02 03 04), (11 12 13 14), (21 22 23 24) (31 32 33 34)

are concurrent. Proof ?

2. Concyclicity:

The points 00, 10, 20, 30 are concyclic (?).

Is it true, for other than I points, on the Neuberg cubic (Fermat points etc)?

3. Orthology:

The triangles ABC, Triangle A'B'C' bounded by the lines [(11 12 13 14), (21 22 23 24) (31 32 33 34)] are orthologic. The one orthologic center (ABC, A'B'C') is S. The other one S' of (A'B'C', ABC) ?

Antreas P. Hatzipolakis, 24 March 2013

-----------------

When you replace point I by a random point D, then we have a random Quadrangle ABCD (system of points without restrictions). The 4 orthopoles of Component Triangles ABC, BCD, CDA, DAB are collinear.

See also Hyacinthos message #21070.

Then this is also true when D coincides with I (Incenter X(1)) and when L = some Euler Line.

In a random Quadrangle points 00, 11, 22, 33 are not concyclic.

However when D coincides with I, then 00, 11, 22, 33 are concyclic indeed (Cabri-proof).

Further 00 lies on NPC of ABC. 11, 22, 33 lie on NPC's ABD, BCD, CAD.

Chris van Tienhoven, Hyacinthos #21831

-----------------

I checked when Point D is coinciding with X(13) then points 00, 11, 22, 33 are concyclic too (Cabri-proof).

About the orthologic property between 2 triangles, it is even much more beautiful!

Denote A=P0, B=P1, C=P2, I=P3.

The points P0, P1, P2, P3 form a Complete Quadrangle.

Let L0 be the line through 00,01,02,03.

Let L1 be the line through 10,11,12,13.

Let L2 be the line through 20,21,22,23.

Let L3 be the line through 30,31,32,33.

The lines L0, L1, L2, L3 form a Complete Quadrilateral.

Now the Complete Quadrangle P0.P1.P2.P3 and the Complete Quadrilateral L0.L1.L2.L3 are orthologicically related (complete new notion to me!).

The four perpendiculars of P0,P1,P2,P3 on resp. L0,L1,L2,L3 are concurrent!

The six perpendiculars of S01, S02, S03, S12, S13, S23 on resp. P0.P1, P0.P2, P0.P3, P1.P2, P1.P3, P2.P3 form a network like the internal and external angle bisectors of a triangle producing an incenter and 3 excenters.

Last but not least, I forgot to tell that the 4 NPC's in my former message have a common point, which is the Euler-Poncelet Point (QA-P2)of the Quadrangle ABCI.

See HERE:

Chris van Tienhoven, Hyacinthos #21834

CONJECTURES:

Conjecture: Let P be a point on the Euler line of ABC, and O, Oa,Ob,Oc the circumcenters of ABC, IBC, ICA, IAB.

The orthopoles of PO, POa, POb, POc wrt ABC, IBC, ICA, IAB, resp. are concyclic.

Question: Is it true for any point Q (instead of I) on the Neuberg cubic?

GENERAL CONJECTURE:

Let Q be a point on the Neuberg cubic, and P a point on the Euler line of ABC.

Let O, Oa, Ob, Oc be the circumcenters of ABC, QBC, QCA, QAB.

The orthopoles of PO, POa, POb, POc wrt ABC, QBC, QCA, QAB, resp are concyclic.

APH, Hyacinthos #21835

Σάββατο, 23 Μαρτίου 2013

Radical Axes and Loci

Let ABC be a triangle, L a line and A',B',C' the orthogonal projections of A,B,C on L.

Denote:

R1 = the radical axis of (B, BB'), (C, CC')

[ie the radical axis of the circles centered at B,C with radii BB',CC', resp.]

R2 = the radical axis of (C,CC'), (A,AA')

R3 = the radical axis of (A,AA'), (B,BB')

R1, R2, R3 are concurrent at a point R(L) = the Radical Center of the Circles (A,AA'), (B,BB'), (C, CC').

Loci:

1. Let P be a point and L a line passing through P.

Which is the locus of R(L) points as L moves around P?

2. Let P be a point on the circumcircle and L the Simson line of P.

Which is the locus of R(L) as P moves on the circumcircle?

3. Let P be a point and La,Lb,Lc three lines passing trrough P ( i) parallels or (ii) perpendiculars to BC,CA,AB, resp.

Which is the locus of P such that the triangles ABC, R(La)R(Lb)R(Lc) are perspective?

Antreas P. Hatzipolakis, Hyacinthos #21815

-----------------------

1. It is an ellipse centered at the midpoint of segment OP. The ellipse degenerates when P is on NPC.

2. It is the Steiner deltoid of the medial triangle.

3 (i) A hyperbola similar to Jerabek hyperbola, but through O and and the nine point center. It intersects the Jerabek hyperbola at the isogonal conjugate of X3523.

(ii) It is the Darboux cubic of the medial triangle.

Francisco Javier, Hyacinthos #21819

The ratio squared of the homothety that carries Jerabek hyperbola into the hyperbola in 3(i) is

(p^2 - r^2 - 4 r R + 8 R^2)/(4 (p - r - 2 R) (p + r + 2 R)).

Francisco Javier, Hyacinthos #21826

Equation of hyperbola 3(i):

5 a^4 b^4 c^2 x^2 - 6 a^2 b^6 c^2 x^2 + b^8 c^2 x^2 - 5 a^4 b^2 c^4 x^2 - 3 b^6 c^4 x^2 + 6 a^2 b^2 c^6 x^2 + 3 b^4 c^6 x^2 - b^2 c^8 x^2 + a^8 c^2 x y + 2 a^6 b^2 c^2 x y - 2 a^2 b^6 c^2 x y - b^8 c^2 x y - 3 a^6 c^4 x y - 3 a^4 b^2 c^4 x y + 3 a^2 b^4 c^4 x y + 3 b^6 c^4 x y + 3 a^4 c^6 x y - 3 b^4 c^6 x y - a^2 c^8 x y + b^2 c^8 x y - a^8 c^2 y^2 + 6 a^6 b^2 c^2 y^2 - 5 a^4 b^4 c^2 y^2 + 3 a^6 c^4 y^2 + 5 a^2 b^4 c^4 y^2 - 3 a^4 c^6 y^2 - 6 a^2 b^2 c^6 y^2 + a^2 c^8 y^2 - a^8 b^2 x z + 3 a^6 b^4 x z - 3 a^4 b^6 x z + a^2 b^8 x z - 2 a^6 b^2 c^2 x z + 3 a^4 b^4 c^2 x z - b^8 c^2 x z - 3 a^2 b^4 c^4 x z + 3 b^6 c^4 x z + 2 a^2 b^2 c^6 x z - 3 b^4 c^6 x z + b^2 c^8 x z - a^8 b^2 y z + 3 a^6 b^4 y z - 3 a^4 b^6 y z + a^2 b^8 y z + a^8 c^2 y z - 3 a^4 b^4 c^2 y z + 2 a^2 b^6 c^2 y z - 3 a^6 c^4 y z + 3 a^4 b^2 c^4 y z + 3 a^4 c^6 y z - 2 a^2 b^2 c^6 y z - a^2 c^8 y z + a^8 b^2 z^2 - 3 a^6 b^4 z^2 + 3 a^4 b^6 z^2 - a^2 b^8 z^2 - 6 a^6 b^2 c^2 z^2 + 6 a^2 b^6 c^2 z^2 + 5 a^4 b^2 c^4 z^2 - 5 a^2 b^4 c^4 z^2

Francisco Javier

-----------------------------------------

Τρίτη, 19 Μαρτίου 2013

PERSPECTIVITY [Antipedal Triangles of P,P*] --2--

Let P,P* be two isogonal conjugate points and A'B'C',A"B"C" the antipedal triangles of P,P*, resp.

Denote: Abc = A'B' /\ A"C", Acb = A'C' /\ A"B"

Bca = B'C' /\ B"A", Bac = B'A' /\ B"C"

Cab = C'A' /\ C"B", Cba = C'B' /\ C"A"

Are the triangles AbcBcaCab, AcbBacCba perspective ?

Antreas P. Hatzipolakis, Hyacinthos #21788

-------------------------------------------

They are again ALWAYS perspective. The coordinates of the perspector are shown below and it is infinite when P lies on K003 [= McCay cubic].

{a^2 (-2 a^4 b^2 c^2 x^2 y + 4 a^2 b^4 c^2 x^2 y - 2 b^6 c^2 x^2 y - 4 a^2 b^2 c^4 x^2 y - 4 b^4 c^4 x^2 y + 6 b^2 c^6 x^2 y - a^6 c^2 x y^2 + a^4 b^2 c^2 x y^2 + a^2 b^4 c^2 x y^2 - b^6 c^2 x y^2 + a^4 c^4 x y^2 - 10 a^2 b^2 c^4 x y^2 + b^4 c^4 x y^2 + a^2 c^6 x y^2 + b^2 c^6 x y^2 - c^8 x y^2 + 2 a^4 b^2 c^2 x^2 z + 4 a^2 b^4 c^2 x^2 z - 6 b^6 c^2 x^2 z - 4 a^2 b^2 c^4 x^2 z + 4 b^4 c^4 x^2 z + 2 b^2 c^6 x^2 z - 2 a^6 c^2 y^2 z - 4 a^4 b^2 c^2 y^2 z + 6 a^2 b^4 c^2 y^2 z + 4 a^4 c^4 y^2 z - 4 a^2 b^2 c^4 y^2 z - 2 a^2 c^6 y^2 z + a^6 b^2 x z^2 - a^4 b^4 x z^2 - a^2 b^6 x z^2 + b^8 x z^2 - a^4 b^2 c^2 x z^2 + 10 a^2 b^4 c^2 x z^2 - b^6 c^2 x z^2 - a^2 b^2 c^4 x z^2 - b^4 c^4 x z^2 + b^2 c^6 x z^2 + 2 a^6 b^2 y z^2 - 4 a^4 b^4 y z^2 + 2 a^2 b^6 y z^2 + 4 a^4 b^2 c^2 y z^2 + 4 a^2 b^4 c^2 y z^2 - 6 a^2 b^2 c^4 y z^2), -b^2 (-a^6 c^2 x^2 y + a^4 b^2 c^2 x^2 y + a^2 b^4 c^2 x^2 y - b^6 c^2 x^2 y + a^4 c^4 x^2 y - 10 a^2 b^2 c^4 x^2 y + b^4 c^4 x^2 y + a^2 c^6 x^2 y + b^2 c^6 x^2 y - c^8 x^2 y - 2 a^6 c^2 x y^2 + 4 a^4 b^2 c^2 x y^2 - 2 a^2 b^4 c^2 x y^2 - 4 a^4 c^4 x y^2 - 4 a^2 b^2 c^4 x y^2 + 6 a^2 c^6 x y^2 + 6 a^4 b^2 c^2 x^2 z - 4 a^2 b^4 c^2 x^2 z - 2 b^6 c^2 x^2 z - 4 a^2 b^2 c^4 x^2 z + 4 b^4 c^4 x^2 z - 2 b^2 c^6 x^2 z - 6 a^6 c^2 y^2 z + 4 a^4 b^2 c^2 y^2 z + 2 a^2 b^4 c^2 y^2 z + 4 a^4 c^4 y^2 z - 4 a^2 b^2 c^4 y^2 z + 2 a^2 c^6 y^2 z + 2 a^6 b^2 x z^2 - 4 a^4 b^4 x z^2 + 2 a^2 b^6 x z^2 + 4 a^4 b^2 c^2 x z^2 + 4 a^2 b^4 c^2 x z^2 - 6 a^2 b^2 c^4 x z^2 + a^8 y z^2 - a^6 b^2 y z^2 - a^4 b^4 y z^2 + a^2 b^6 y z^2 - a^6 c^2 y z^2 + 10 a^4 b^2 c^2 y z^2 - a^2 b^4 c^2 y z^2 - a^4 c^4 y z^2 - a^2 b^2 c^4 y z^2 + a^2 c^6 y z^2), c^2 (6 a^4 b^2 c^2 x^2 y - 4 a^2 b^4 c^2 x^2 y - 2 b^6 c^2 x^2 y - 4 a^2 b^2 c^4 x^2 y + 4 b^4 c^4 x^2 y - 2 b^2 c^6 x^2 y + 2 a^6 c^2 x y^2 + 4 a^4 b^2 c^2 x y^2 - 6 a^2 b^4 c^2 x y^2 - 4 a^4 c^4 x y^2 + 4 a^2 b^2 c^4 x y^2 + 2 a^2 c^6 x y^2 - a^6 b^2 x^2 z + a^4 b^4 x^2 z + a^2 b^6 x^2 z - b^8 x^2 z + a^4 b^2 c^2 x^2 z - 10 a^2 b^4 c^2 x^2 z + b^6 c^2 x^2 z + a^2 b^2 c^4 x^2 z + b^4 c^4 x^2 z - b^2 c^6 x^2 z + a^8 y^2 z - a^6 b^2 y^2 z - a^4 b^4 y^2 z + a^2 b^6 y^2 z - a^6 c^2 y^2 z + 10 a^4 b^2 c^2 y^2 z - a^2 b^4 c^2 y^2 z - a^4 c^4 y^2 z - a^2 b^2 c^4 y^2 z + a^2 c^6 y^2 z - 2 a^6 b^2 x z^2 - 4 a^4 b^4 x z^2 + 6 a^2 b^6 x z^2 + 4 a^4 b^2 c^2 x z^2 - 4 a^2 b^4 c^2 x z^2 - 2 a^2 b^2 c^4 x z^2 - 6 a^6 b^2 y z^2 + 4 a^4 b^4 y z^2 + 2 a^2 b^6 y z^2 + 4 a^4 b^2 c^2 y z^2 - 4 a^2 b^4 c^2 y z^2 + 2 a^2 b^2 c^4 y z^2)}

Francisco Javier, Hyacinthos #21789

-------------------------------------------

If P lies on Darboux cubic the perspector of the triangles AbcBcaCab, AcbBacCba coincides with the isotomic conjugate of the perspector of ABC and triangle bounded by (A'A", B'B ",C'C").

Does this happen only if P is in the Darboux cubic?

Some pairs of isogonal conjugate points (on Darboux cubic), and their corresponding perspector of the triangles AbcBcaCab, AcbBacCba:

(X3,X4): X512; (X20,X64): X647; (X40, X84): X663; .....

Angel Montesdeoca, Hyacinthos #21790

PERSPECTIVITY [Antipedal Triangles of P,P*] --1--

Let P,P* be two isogonal conjugate points and A'B'C',A"B"C" the antipedal triangles of P,P*, resp.

Which is the locus of P such that the triangles:

1. A"B"C", Triangle bounded by (AA',BB',CC')

2. A'B'C', Triangle bounded by (AA",BB",CC")

3. ABC, Triangle A*B*C* bounded by (A'A",B'B",C'C") are perspective ?

Antreas P. Hatzipolakis, Hyacinthos #21782

------------------------------

The loci in 1. and 2 are the same:

line at infinity + circumcircle + K004[=Darboux cubic] + three cubics, each one relative to one of the vertices.

The cubic relative to A has equation

2 b^2 c^2 x^2 y + a^2 c^2 x y^2 + b^2 c^2 x y^2 + c^4 x y^2 + 2 b^2 c^2 x^2 z + 4 b^2 c^2 x y z + 2 a^2 c^2 y^2 z + a^2 b^2 x z^2 + b^4 x z^2 + b^2 c^2 x z^2 + 2 a^2 b^2 y z^2 =0.

It intersects the line at infinity at the infinite points of the bisectors of angle A and at the infinite point of the A-altitude.

It is its isogonal conjugate, then it intersects the circumcircle at the intersections of the circumcircle and the bisectors of angle A and at the antipode of A.

With respect to 3., the triangle bounded by (A'A", B' B ",C'C") is ALWAYS perspective with ABC.

If P=(x:y:z), then the perspector S is complicated: the coordinates of its isotomic conjugate, that is infinite when P lies on K004 [= Darboux cubic], are:

{2 a^4 b^2 c^2 x^2 y - 4 a^2 b^4 c^2 x^2 y + 2 b^6 c^2 x^2 y + 4 a^2 b^2 c^4 x^2 y + 4 b^4 c^4 x^2 y - 6 b^2 c^6 x^2 y + a^6 c^2 x y^2 - a^4 b^2 c^2 x y^2 - a^2 b^4 c^2 x y^2 + b^6 c^2 x y^2 - a^4 c^4 x y^2 + 10 a^2 b^2 c^4 x y^2 - b^4 c^4 x y^2 - a^2 c^6 x y^2 - b^2 c^6 x y^2 + c^8 x y^2 - 2 a^4 b^2 c^2 x^2 z - 4 a^2 b^4 c^2 x^2 z + 6 b^6 c^2 x^2 z + 4 a^2 b^2 c^4 x^2 z - 4 b^4 c^4 x^2 z - 2 b^2 c^6 x^2 z + 2 a^6 c^2 y^2 z + 4 a^4 b^2 c^2 y^2 z - 6 a^2 b^4 c^2 y^2 z - 4 a^4 c^4 y^2 z + 4 a^2 b^2 c^4 y^2 z + 2 a^2 c^6 y^2 z - a^6 b^2 x z^2 + a^4 b^4 x z^2 + a^2 b^6 x z^2 - b^8 x z^2 + a^4 b^2 c^2 x z^2 - 10 a^2 b^4 c^2 x z^2 + b^6 c^2 x z^2 + a^2 b^2 c^4 x z^2 + b^4 c^4 x z^2 - b^2 c^6 x z^2 - 2 a^6 b^2 y z^2 + 4 a^4 b^4 y z^2 - 2 a^2 b^6 y z^2 - 4 a^4 b^2 c^2 y z^2 - 4 a^2 b^4 c^2 y z^2 + 6 a^2 b^2 c^4 y z^2, -a^6 c^2 x^2 y + a^4 b^2 c^2 x^2 y + a^2 b^4 c^2 x^2 y - b^6 c^2 x^2 y + a^4 c^4 x^2 y - 10 a^2 b^2 c^4 x^2 y + b^4 c^4 x^2 y + a^2 c^6 x^2 y + b^2 c^6 x^2 y - c^8 x^2 y - 2 a^6 c^2 x y^2 + 4 a^4 b^2 c^2 x y^2 - 2 a^2 b^4 c^2 x y^2 - 4 a^4 c^4 x y^2 - 4 a^2 b^2 c^4 x y^2 + 6 a^2 c^6 x y^2 + 6 a^4 b^2 c^2 x^2 z - 4 a^2 b^4 c^2 x^2 z - 2 b^6 c^2 x^2 z - 4 a^2 b^2 c^4 x^2 z + 4 b^4 c^4 x^2 z - 2 b^2 c^6 x^2 z - 6 a^6 c^2 y^2 z + 4 a^4 b^2 c^2 y^2 z + 2 a^2 b^4 c^2 y^2 z + 4 a^4 c^4 y^2 z - 4 a^2 b^2 c^4 y^2 z + 2 a^2 c^6 y^2 z + 2 a^6 b^2 x z^2 - 4 a^4 b^4 x z^2 + 2 a^2 b^6 x z^2 + 4 a^4 b^2 c^2 x z^2 + 4 a^2 b^4 c^2 x z^2 - 6 a^2 b^2 c^4 x z^2 + a^8 y z^2 - a^6 b^2 y z^2 - a^4 b^4 y z^2 + a^2 b^6 y z^2 - a^6 c^2 y z^2 + 10 a^4 b^2 c^2 y z^2 - a^2 b^4 c^2 y z^2 - a^4 c^4 y z^2 - a^2 b^2 c^4 y z^2 + a^2 c^6 y z^2, -6 a^4 b^2 c^2 x^2 y + 4 a^2 b^4 c^2 x^2 y + 2 b^6 c^2 x^2 y + 4 a^2 b^2 c^4 x^2 y - 4 b^4 c^4 x^2 y + 2 b^2 c^6 x^2 y - 2 a^6 c^2 x y^2 - 4 a^4 b^2 c^2 x y^2 + 6 a^2 b^4 c^2 x y^2 + 4 a^4 c^4 x y^2 - 4 a^2 b^2 c^4 x y^2 - 2 a^2 c^6 x y^2 + a^6 b^2 x^2 z - a^4 b^4 x^2 z - a^2 b^6 x^2 z + b^8 x^2 z - a^4 b^2 c^2 x^2 z + 10 a^2 b^4 c^2 x^2 z - b^6 c^2 x^2 z - a^2 b^2 c^4 x^2 z - b^4 c^4 x^2 z + b^2 c^6 x^2 z - a^8 y^2 z + a^6 b^2 y^2 z + a^4 b^4 y^2 z - a^2 b^6 y^2 z + a^6 c^2 y^2 z - 10 a^4 b^2 c^2 y^2 z + a^2 b^4 c^2 y^2 z + a^4 c^4 y^2 z + a^2 b^2 c^4 y^2 z - a^2 c^6 y^2 z + 2 a^6 b^2 x z^2 + 4 a^4 b^4 x z^2 - 6 a^2 b^6 x z^2 - 4 a^4 b^2 c^2 x z^2 + 4 a^2 b^4 c^2 x z^2 + 2 a^2 b^2 c^4 x z^2 + 6 a^6 b^2 y z^2 - 4 a^4 b^4 y z^2 - 2 a^2 b^6 y z^2 - 4 a^4 b^2 c^2 y z^2 + 4 a^2 b^4 c^2 y z^2 - 2 a^2 b^2 c^4 y z^2}

Francisco Javier, Hyacinthos #21782

POINT ON THE CIRCUMCIRCLE

Let ABC be a triangle, A'B'C', A"B"C" the antipedal tiangles of H, O [antimedial, tangential, resp.].

The triangles:

3. ABC, Triangle A*B*C* bounded by the lines (A'A",B'B",C'C") are perspective.

[the perspector is on the circumcircle(ABC) = NPC(A'B'C') = Incircle(A"B"C")]

Coordinates?

Antreas P. Hatzipolakis, Hyacinthos #21780

Generalization

Σάββατο, 16 Μαρτίου 2013

REFLECTIONS OF PARALLELS TO OP LINE

1.

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

Let La, Lb. Lc be the parallels to OI line through A',B',C', resp. and Ma,Mb,Mc the reflections of La,Lb,Lc in AA', BB',CC', resp.

The lines Ma,Mb,Mc are concurrent.

Locus:

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

Let La, Lb, Lc be the parallels to OP line through A',B',C', resp. and Ma,Mb,Mc the reflections of La,Lb,Lc in AA', BB',CC', resp.

Which is the locus of P such that the lines Ma,Mb,Mc are concurrent ?

2. Orthic Triangle version:

Let ABC be a triangle, A'B'C' the pedal triangle of H (orthic triangle) and A"B"C" the circumcevian triangle of H wrt A'B'C'.

Let La,Lb,Lc be the parallels to HN line (Euler line) through A",B",C", resp. and Ma,Mb,Mc the reflections of La,Lb,Lc in HA',HB',HC', resp.

The lines Ma,Mb,Mc are concurrent.

Locus:

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

Let La,Lb,Lc be the parallels to PQ,, where Q is the center of the pedal circle of P, through A",B",C", resp. and Ma,Mb,Mc the reflections of La,Lb,Lc in the lines PA',PB',PC', resp.

Which is the locus of P such that Ma,Mb,Mc are concurrent?

Antreas P. Hatzipolakis, 16 March 2013

Παρασκευή, 15 Μαρτίου 2013

ORTHOLOGIC TRIANGLES - LOCUS

1.

Let ABC be a triangle and A'B'C' the pedal triangle of O (= medial triangle).

Denote:

A" = (O, OA')/\((A'B'C') - A') = the other than A' intersection of the circle centered at O with radius OA' and the pedal circle of O(=NPC)

B" = (O, OB')/\((A'B'C') - B')

C" = (O, OC')/\((A'B'C') - C')

The triangles ABC, A"B"C" are orthologic (with orthologic centers O1,O2 on NPC and circumcircle)

Locus:

Let ABC be a triangle, A'B'C' the pedal triangle of point P and A",B",C" the second intersections of the circles (P,PA'), (P,PB'), (P,PC') with the pedal circle of P, resp.

Which is the locus of P such that the triangles ABC, A"B"C" are orthologic?

2.

Let ABC be a triangle, A'B'C' the pedal triangle of H (=orthic triangle) and A*B*C* the circumcevian triangle of H wrt A'B'C'.

Denote:

A" = (H, HA*) /\ (A*B*C* - A*) = the other intersection of the circle centered at H with radius HA* and the circle (A*B*C*) = (A'B'C') = pedal circle of H.

B" = (H, HB*) /\ (A*B*C* - B*)

C" = (H, HC*) /\ (A*B*C* - C*)

The triangles ABC, A"B"C" are orthologic (with orthologic centers H1,H2 on NPC and circumcircle)

Locus:

Let ABC be a triangle, A'B'C' the pedal triangle of point P, A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with the pedal circle of P, resp.

Which is the locus of P such that the triangles ABC, A"B"C" are orthologic?

Antreas P. Hatzipolakis, 15 March 2013

Pedal and Antipedal Circles tangent

Are there real points P such that the pedal and antipedal circles of P are tangent?

Antreas P. Hatzipolakis, Hyacinthos #21746

------------------------------------

I get the cubic

S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0.

Francisco Javier, Hyacinthos #21747

------------------------------------

Yes, if P is on an octic or on the cubic K191="circumcircle pedal cubic, nK(X6, X6,?)".

If P is on the cubic K191, then the point of the contact of the two circles are on the circumcircle.

K191: S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0,

or equivalently

K191: S^2 xyz + CyclicSum[ a^2 x (c^2 y^2 + b^2 z^2)] = 0.

S= 2*area(ABC) (In CTC of Bernad Gibert, S=area(ABC))

Angel Montesdeoca, Hyacinthos #21750

The cubic S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0 is not K191

It will be K634 in Bernard Gibert's list

Πέμπτη, 14 Μαρτίου 2013

MCCAY CUBIC (Circumcevian Triangle) +

Let ABC be a triangle, P a point, A'B'C' the pedal triangle of P, A"B"C" its antipodal triangle (in the pedal circle) and A*B*C* the circumcevian triangle of P.

Which is the locus of P such that A*B*C*, A"B"C" are perspective?

Antreas P. Hatzipolakis, Hyacinthos #21738

--------------------

A*B*C*, A"B"C" are perspective if P is on the circumcircle (the circumcevian triangle degenerates) or on the K003="McCay cubic" or on K191="circumcircle pedal cubic".

If P is on K191="circumcircle pedal cubic", the perspector of the triangles A*B*C* and A"B"C" is a point in the circumcircle.

Angel Montesdeoca, Hyacinthos #21743

The cubic is S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0 (not K191) and will be K634 in Bernard Gibert's list.

REFLENTING PERPENDICULARS

Let ABC be a triangle and A'B'C' the pedal triangle of H (orthic triangle).

Denote:

La = the reflection of HA' in B'C'

Lb = the reflection of HB' in C'A'

Lc = the reflection of HC' in A'B'

Ma = the reflection of La in OA.

Mb = the reflection of Lb in OB.

Mc = the reflection of Lb in OC.

The lines Ma,Mb,Mc concur at a point Q. Denote:

A" = BC /\ Ma

B" = CA /\ Mb

C" = AB /\ Mc

The triangles ABC, A"B"C" are orthologic, with orthologic centers Q,S.

Locus:

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

Denote:

La = the reflection of PA' in B'C'

Lb = the reflection of PB' in C'A'

Lc = the reflection of PC' in A'B'

Ma = the reflection of La in AP*.

Mb = the reflection of Lb in BP*.

Mc = the reflection of Lb in CP*.

Denote:

A" = BC /\ Ma

B" = CA /\ Mb

C" = AB /\ Mc

(Ma,Mb,Mc are perpendiculars to BC,CA,AB, resp.)

Which is the locus of P such that:

1. ABC, triangle bounded by (Ma,Mb,Mc) are perspective? Special case: Ma,Mb,Mc be concurrent.

2. ABC, A"B"C" are orthologic?

Antreas P. Hatzipolakis, 14 March 2013

Τετάρτη, 13 Μαρτίου 2013

REFLECTIONS - COLLINEARITY

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

Denote:

A* = BC /\ (Reflection of B'C' in AA')

B* = CA /\ (Reflection of C'A' in BB')

C* = AB /\ (Reflection of A'B' in CC')

Which is the locus of P such that the A*,B*,C* are collinear?

The incenter I is on the locus

Antreas P; Hatzipolakis, 13 March 2013

Τρίτη, 12 Μαρτίου 2013

CIRCUMCIRCLE

Let ABC be a triangle and P, Q two isogonal conjugate points and P1P2P3, Q1Q2Q3 the pedal triangles of P,Q, resp.

Denote:

R1 = the radical axis of the circles ((P1, P1Q)[=circle centered at P1 with radius P1Q],(Q1, Q1P))

R2 = the radical axis of the circles ((P2, P2Q),(Q2, Q2P))

R3 = the radical axis of the circles ((P3, P3Q),(Q3, Q3P))

The triangles ABC, Triangle A'B'C' bounded by (R1,R2,R3) are perspective at a point D on the circumcircle.

Antreas P. Hatzipolakis, Hyacinthos #21733

Παρασκευή, 8 Μαρτίου 2013

MCCAY CUBIC. Locus Problems (Parallels to cevians)

Let ABC be a triangle, P a point and A1B1C1, A2B2C2 the pedal triangles of P and its isogonal conjugate P*.

The parallels through A1,B1,C1 to AP,BP,CP bound a triangle A3B3C3.

The above parallels intersect the pedal circle of A1B1C1 (and A2B2C2)at A4,B4,C4.

Which is the locus of P such that:

1. A2B2C2, A3B3C3

2. A2B2C2, A4B4C4

3. A3B3C3, A4B4C4

4. A1B1C1, A3B3C3

5. ABC, A3B3C3

6. ABC, A4B4C4

are perspective?

Antreas P. Hatzipolakis, Hyacinthos #20745

1. A2B2C2, A3B3C3: line at infinity + circumcircle + McCay cubic.

2. A2B2C2, A4B4C4: The whole plane. The perspector R (1) is collinear with P and P* when P is on the McCay cubic.

3. A3B3C3, A4B4C4: line at infinity + McCay cubic + sextic

4. A1B1C1, A3B3C3: Same as previous.

5. ABC, A3B3C3: sides of anticomplementary triangle + circumcircle + McCay cubic

6. ABC, A4B4C4: line at infinity + circumcircle + McCay cubic.

(1) For P=(x:y:z) the perspector R is the point

{y z (a^2 b^2 x^2 - b^4 x^2 + a^2 c^2 x^2 + 2 b^2 c^2 x^2 - c^4 x^2 + a^4 x y - a^2 b^2 x y + a^2 c^2 x y + a^4 x z + a^2 b^2 x z - a^2 c^2 x z + 2 a^4 y z), x z (-a^2 b^2 x y + b^4 x y + b^2 c^2 x y - a^4 y^2 + a^2 b^2 y^2 + 2 a^2 c^2 y^2 + b^2 c^2 y^2 - c^4 y^2 + 2 b^4 x z + a^2 b^2 y z + b^4 y z - b^2 c^2 y z), x y (2 c^4 x y - a^2 c^2 x z + b^2 c^2 x z + c^4 x z + a^2 c^2 y z - b^2 c^2 y z + c^4 y z - a^4 z^2 + 2 a^2 b^2 z^2 - b^4 z^2 + a^2 c^2 z^2 + b^2 c^2 z^2)}

Francisco Javier, Hyacinthos #20745

Τετάρτη, 6 Μαρτίου 2013

CONCURRENT EULER LINES

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

Denote:

Ab, Ac = the orthogonal projections of A on BB',CC', resp.

Bc, Ba = the orthogonal projections of B on CC',AA', resp.

Ca, Cb = the orthogonal projections of C on AA',BB', resp.

Abc = the orthogonal projection of Ab on CC'

Acb = the orthogonal projection of Ac on BB'

Bca = the orthogonal projection of Bc on AA'

Bac = the orthogonal projection of Ba on CC'

Cab = the orthogonal projection of Ca on BB'

Cba = the orthogonal projection of Cb on AA'

For P = I, we have Bca = Cba =: A*, Cab = Acb =: B* and Abc = Bac =: C*

The Euler lines of AB*C*, BC*A*, CA*B* are concurrent.

Which is the locus of P such that the Euler lines of AAbcAcb, BBcaBac, CCabCba are concurrent?

Note: The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at Feuerbach point.

(APH, Hyacinthos)

Orthic Triangle Version:

Let ABC be a triangle and A'B'C' the cevian/pedal triangle of P = H (orthic triangle)

Denote:

A'b, A'c = the orthogonal projections of A' on BB',CC', resp.

B'c, B'a = the orthogonal projections of B' on CC',AA', resp.

C'a, C'b = the orthogonal projections of C' on AA',BB', resp.

A'bc = the orthogonal projection of A'b on CC'

A'cb = the orthogonal projection of A'c on BB'

B'ca = the orthogonal projection of B'c on AA'

B'ac = the orthogonal projection of B'a on CC'

C'ab = the orthogonal projection of C'a on BB'

C'ba = the orthogonal projection of C'b on AA'

For P = H, we have B'ca = C'ba =: A*, C'ab = A'cb =: B* and A'bc = B'ac =: C*

The Euler lines of A'B*C*, B'C*A*, C'A*B* are concurrent.

Let A'B'C' be the cevian (or pedal) triangle of P. Which is the locus of P such that the Euler lines of A'A'bcA'cb, B'B'caB'ac, C'C'abC'ba are concurrent?

Summary:

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

1. Denote:

Ab, Ac = the orthogonal projections of A on BB',CC', resp.

Bc, Ba = the orthogonal projections of B on CC',AA', resp.

Ca, Cb = the orthogonal projections of C on AA',BB', resp.

Abc = the orthogonal projection of Ab on CC'

Acb = the orthogonal projection of Ac on BB'

Bca = the orthogonal projection of Bc on AA'

Bac = the orthogonal projection of Ba on CC'

Cab = the orthogonal projection of Ca on BB'

Cba = the orthogonal projection of Cb on AA'

Which is the locus of P such that the Euler lines of AAbcAcb, BBcaBac, CCabCba are concurrent?

2. Denote:

A'b, A'c = the orthogonal projections of A' on BB',CC', resp.

B'c, B'a = the orthogonal projections of B' on CC',AA', resp.

C'a, C'b = the orthogonal projections of C' on AA',BB', resp.

A'bc = the orthogonal projection of A'b on CC'

A'cb = the orthogonal projection of A'c on BB'

B'ca = the orthogonal projection of B'c on AA'

B'ac = the orthogonal projection of B'a on CC'

C'ab = the orthogonal projection of C'a on BB'

C'ba = the orthogonal projection of C'b on AA'

Which is the locus of P such that the Euler lines of A'A'bcA'cb, B'B'caB'ac, C'C'abC'ba are concurrent?

Antreas P. Hatzipolakis, 6 March 2013

 

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