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

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I get the cubic

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

Francisco Javier, Hyacinthos #21747

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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

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

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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

ADDENDUM (9/9/19)

Q = X(34224) = X(3)X(70)∩X(4)X(6)
S = X(34225) = ISOGONAL CONJUGATE OF X(34224)

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

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

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

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