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

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 P*a, P*b, P*c 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, P*aP*bP*c 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


Σάββατο, 19 Απριλίου 2014

RADICAL AXES

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

RADICAL AXES 1.1

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


 

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