Perspective

Let ABC be a triangle, HaHbHc the cevian triangle of H (orthic triangle), GaGbGc the cevian triangle of G (medial triangle), H1H2H3, G1G2G3 the circumcevian triangles of H and G, resp.

Denote:

A1 = the 2nd intersection of H1Ga and the circumcircle (other than H1)
A2 = the 2nd intersection of G1Ha and the circumcircle (other than G1)

B1 = the 2nd intersection of H2Gb and the circumcircle (other than H2)
B2 = the 2nd intersection of G2Hb and the circumcircle (other than G2)

C1 = the 2nd intersection of H3Gc and the circumcircle (other than H3)
C2 = the 2nd intersection of G3Hc and the circumcircle (other than G3)

The lines A1A2, B1B2, C1C2 bound a triangle A*B*C*
(A* = B1B2 /\ C1C2, B* = C1C2 /\ A1A2, C* = A1A2 /\ B1B2)

The triangles ABC, A*B*C* are perspective.

Perspector?

Generalization:
Locus of P instead of 1.H, 2.G

APH 29 January 2012

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

The coordinates are:

{a^2 (a^2 + b^2 - c^2) (a^2 - b^2 + c^2) (3 a^4 - 6 a^2 b^2 + 3 b^4 +
c^4) (3 a^4 + b^4 - 6 a^2 c^2 + 3 c^4),
b^2 (a^2 + b^2 - c^2) (-a^2 + b^2 + c^2) (3 a^4 - 6 a^2 b^2 + 3 b^4 +
c^4) (a^4 + 3 b^4 - 6 b^2 c^2 + 3 c^4),
c^2 (a^2 - b^2 + c^2) (-a^2 + b^2 + c^2) (3 a^4 + b^4 - 6 a^2 c^2 +
3 c^4) (a^4 + 3 b^4 - 6 b^2 c^2 + 3 c^4)}

Not in ETC

********************
ADDENDUM (11/09/19)

Now X(34233)
********************

The result is true for any P, Q.
The Isogonal Conjugate of the perspector for P=(u:v:w) and Q(x:y:z) is:.

{u x (w^2 y^2 - v w y z + v^2 z^2), v y (w^2 x^2 - u w x z + u^2 z^2), wz (v^2 x^2 - u v x y + u^2 y^2) }

Francisco Javier García Capitán
30 January 2012

LOCUS

Let ABC be triangle, P a point, A1B1C1 the cevian triangle of P and A2B2C2 the circumcevian triangle of P.

Denote:

Ab = the other than A2 intersection of the circumcircle and A2B1
Ac = the other than A2 intersection of the circumcircle and A2C1

Bc = the other than B2 intersection of the circumcircle and B2C1
Ba = the other than B2 intersection of the circumcircle and B2A1

Ca = the other than C2 intersection of the circumcircle and C2A1
Cb = the other than C2 intersection of the circumcircle and C2B1

The lines AbAc, BcBa, CaCb bound a triangle A3B3C3.

Which is the locus of P such that:

1. ABC, A3B3C3

2. A1B1C1, A3B3C3

3. A2B2C2, A3B3C3

are perspective?


APH, 26 January 2012

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

For 1. and 3. the locus is the whole plane.
For 2, the locus is (symmedians) + circumcircle + (12th degree curve whose isogonal conjugate is 6th degree curve).

For 1. the perspector is

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

For 2. the perspector is

{a^2 (-c^16 x^8 y^8 - 6 c^14 x^7 y^7 (b^2 x + a^2 y) z -
c^12 x^6 y^6 (17 b^4 x^2 + 30 a^2 b^2 x y + 14 a^4 y^2) z^2 -
c^10 x^5 y^5 (30 b^6 x^3 + 69 a^2 b^4 x^2 y + 54 a^4 b^2 x y^2 +
14 a^6 y^3) z^3 -
3 b^2 c^8 x^5 y^4 (12 b^6 x^3 + 33 a^2 b^4 x^2 y +
31 a^4 b^2 x y^2 + 10 a^6 y^3) z^4 -
c^6 x^3 y^3 (30 b^10 x^5 + 99 a^2 b^8 x^4 y +
107 a^4 b^6 x^3 y^2 + 21 a^6 b^4 x^2 y^3 - 30 a^8 b^2 x y^4 -
14 a^10 y^5) z^5 -
c^4 x^2 y^2 (b^2 x + a^2 y)^3 (17 b^6 x^3 + 18 a^2 b^4 x^2 y -
12 a^4 b^2 x y^2 - 14 a^6 y^3) z^6 -
6 c^2 x y (b^2 x - a^2 y) (b^2 x + a^2 y)^6 z^7 - (b^2 x -
a^2 y) (b^2 x + a^2 y)^7 z^8),
b^2 (-c^16 x^8 y^8 - 6 c^14 x^7 y^7 (b^2 x + a^2 y) z -
c^12 x^6 y^6 (14 b^4 x^2 + 30 a^2 b^2 x y + 17 a^4 y^2) z^2 -
c^10 x^5 y^5 (14 b^6 x^3 + 54 a^2 b^4 x^2 y + 69 a^4 b^2 x y^2 +
30 a^6 y^3) z^3 -
3 a^2 c^8 x^4 y^5 (10 b^6 x^3 + 31 a^2 b^4 x^2 y +
33 a^4 b^2 x y^2 + 12 a^6 y^3) z^4 +
c^6 x^3 y^3 (14 b^10 x^5 + 30 a^2 b^8 x^4 y - 21 a^4 b^6 x^3 y^2 -
107 a^6 b^4 x^2 y^3 - 99 a^8 b^2 x y^4 - 30 a^10 y^5) z^5 +
c^4 x^2 y^2 (b^2 x + a^2 y)^3 (14 b^6 x^3 + 12 a^2 b^4 x^2 y -
18 a^4 b^2 x y^2 - 17 a^6 y^3) z^6 +
6 c^2 x y (b^2 x - a^2 y) (b^2 x + a^2 y)^6 z^7 + (b^2 x -
a^2 y) (b^2 x + a^2 y)^7 z^8),
c^2 (c^16 x^8 y^8 + 6 c^14 x^7 y^7 (b^2 x + a^2 y) z +
2 c^12 x^6 y^6 (7 b^4 x^2 + 15 a^2 b^2 x y + 7 a^4 y^2) z^2 +
2 c^10 x^5 y^5 (7 b^6 x^3 + 27 a^2 b^4 x^2 y + 27 a^4 b^2 x y^2 +
7 a^6 y^3) z^3 + 30 c^8 x^5 y^5 (a b^3 x + a^3 b y)^2 z^4 -
c^6 x^3 y^3 (14 b^10 x^5 + 30 a^2 b^8 x^4 y +
21 a^4 b^6 x^3 y^2 + 21 a^6 b^4 x^2 y^3 + 30 a^8 b^2 x y^4 +
14 a^10 y^5) z^5 -
c^4 x^2 y^2 (14 b^12 x^6 + 54 a^2 b^10 x^5 y +
93 a^4 b^8 x^4 y^2 + 107 a^6 b^6 x^3 y^3 +
93 a^8 b^4 x^2 y^4 + 54 a^10 b^2 x y^5 + 14 a^12 y^6) z^6 -
3 c^2 x y (b^2 x + a^2 y)^3 (2 b^8 x^4 + 4 a^2 b^6 x^3 y +
5 a^4 b^4 x^2 y^2 + 4 a^6 b^2 x y^3 + 2 a^8 y^4) z^7 - (b^2 x +
a^2 y)^4 (b^4 x^2 + a^2 b^2 x y + a^4 y^2)^2 z^8)}

For 3. the equation of the 12th curve is

c^12 x^6 y^6 + 5 b^2 c^10 x^6 y^5 z + 5 a^2 c^10 x^5 y^6 z +
11 b^4 c^8 x^6 y^4 z^2 + 23 a^2 b^2 c^8 x^5 y^5 z^2 +
11 a^4 c^8 x^4 y^6 z^2 + 14 b^6 c^6 x^6 y^3 z^3 +
44 a^2 b^4 c^6 x^5 y^4 z^3 + 44 a^4 b^2 c^6 x^4 y^5 z^3 +
14 a^6 c^6 x^3 y^6 z^3 + 11 b^8 c^4 x^6 y^2 z^4 +
44 a^2 b^6 c^4 x^5 y^3 z^4 + 65 a^4 b^4 c^4 x^4 y^4 z^4 +
44 a^6 b^2 c^4 x^3 y^5 z^4 + 11 a^8 c^4 x^2 y^6 z^4 +
5 b^10 c^2 x^6 y z^5 + 23 a^2 b^8 c^2 x^5 y^2 z^5 +
44 a^4 b^6 c^2 x^4 y^3 z^5 + 44 a^6 b^4 c^2 x^3 y^4 z^5 +
23 a^8 b^2 c^2 x^2 y^5 z^5 + 5 a^10 c^2 x y^6 z^5 + b^12 x^6 z^6 +
5 a^2 b^10 x^5 y z^6 + 11 a^4 b^8 x^4 y^2 z^6 +
14 a^6 b^6 x^3 y^3 z^6 + 11 a^8 b^4 x^2 y^4 z^6 +
5 a^10 b^2 x y^5 z^6 + a^12 y^6 z^6 = 0.

Francisco Javier García Capitán
27 January 2012

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

Variation:

Let ABC be triangle, P a point, A1B1C1 the pedal (instead of cevian) triangle of P and A2B2C2 the circumcevian triangle of P. etc

APH

LOCUS

Let ABC be a triangle, A'B'C' the cevian triangle of G, A"B"C" the circumcevian triangle of G with respect the triangle A'B'C' and O1,O2,O3 the circumcenters of GB"C",GC"A",GA"B", resp.

The triangles ABC, O1O2O3 are perspective.

Perspector?

APH, 25 January 2012

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

Generalization:

Let ABC be a triangle, A'B'C' the cevian triangle of P, A"B"C" the circumcevian triangle of P with respect the triangle A'B'C' and O1,O2,O3 the circumcenters of PB"C", PC"A", PA"B", resp.
The triangles ABC, O1O2O3 are perspective gives as locus the Yiu quintic and an octic through G and H.

The perspector for H is X381 and that for G is the isotomic conjugate of the point {3 a^4 - 4 a^2 b^2 + b^4 - 4 a^2 c^2 - 6 b^2 c^2 + c^4,
a^4 - 4 a^2 b^2 + 3 b^4 - 6 a^2 c^2 - 4 b^2 c^2 + c^4,
a^4 - 6 a^2 b^2 + b^4 - 4 a^2 c^2 - 4 b^2 c^2 + 3 c^4}, not in ETC.

Francisco Javier García Capitán
26 January 2012

ADDENDUM (10/9/19)

Perspector for P = G: X(14494)
Its isotomic conjugate: X(34229)

ZIG ZAG LOCUS

Let ABC be a triangle, P a point (not on the circumcircle), A1B1C1 the circumcevian triangle of P, A2B2C2 the circumcevian triangle of H with respect the triangle A1B1C1 and A3B3C3 the circumcevian triangle of O with respect the triangle A2B2C2 (antipodal triangle of A2B2C2).

Which is the locus of P such that the triangles ABC, A3B3C3 are perspective? And the locus of the perspectors?
Generalization: Replace H with a point Q.

APH, 25 January 2012

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

The locus is the Euler line.
For an arbitrary Q, instead of H, the locus is the line OQ, and the locus of perspectors P' is the same line, so we have a map P -> P', although the formula is not simple.

I found the following formula showing the relationship between some P on line OQ such that OP:PQ = k, and the perspector P', also on line OQ such that OP':P'Q = k': k' = R^2 / (k OQ^2 - (k+2) R^2)

Francisco Javier García Capitán
25 January 2012

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

Variation:

A2B2C2 = the circumcevian triangle of P* (= the isogonal or isotomic conjugate of P, instead of H) with respect the triangle A1B1C1

APH

SEGOVIA POINT Continued

The antipodal triangle A'B'C' of O (= circumcevian of O) is the reflection of ABC in O. The triangle A2B2C2 is the reflection of the orthic A1B1C1 in N. Consider now the triangle A'2B'2C'2 = the reflection of A2B2C2 in O.

The triangles A'B'C', A2B2C2 are perspective at SEGOVIA Point of ABC.

Similarly the triangles ABC, A'2B'2C'2 are perspective at the SEGOVIA point of A'B'C'.

Denote O1O2O3 = the pedal triangle of O (=medial triangle).

We will work in an acute triangle ABC (similarly if ABC is not acute. Simply we have to change some signs).

ABC, A'2B'2C'2 are perspective <==>

[cotB + cot(O1BA'2)] / [cotC + cot(O1CA'2] * Cyclically = 1

We have:

cot(O1BA'2) = O1B / O1A'2 = (BC/2)/[OA'2 - OO1] = (BC/2)/[HA1 - OO1] =

= sinA /(2cosBcosC - cosA)

Therefore:

[cotB + cot(O1BA'2)] = (cosB/sinB) - [sinA /(2cosBcosC - cosA)] =

cosC[1 + 2cos^2B] / sinB(2cosBcosC - cosA)

and

[cotB + cot(O1BA'2)] / [cotC + cot(O1CA'2] = (cosC/cosB)*(sinC/sinB)*[(1+2cos^2B)/(1+2cos^2C)] (*)

and similarly the other two ratia.

Multiplicating them we get 1, therefore the triangles are perspective.

As for the coordinates of the perspector:

We get it in barycentrics from (*) and are:

(cosA * sinA * (1/(1+2cos^2A)) ::)

These are the barycentrics of the Segovia Point of the circumcevian triangle of O (if I did not make some computational error!)

BROCARD PRIZE

Let ABC be a triangle and Ka, Kb, Kc the Brocard axes of
the triangles GBC, GCA, GAB, resp.

Let A'B'C' be a triangle homothetic and sharing the same
centroid G with ABC.

Conjecture:

The reflections La,Lb,Lc of Ka,Kb,Kc in the sidelines B'C', C'A',
A'B' of A'B'C' resp. are concurrent.

See: ANOPOLIS list, Message 137

The first who will send a solution (synthetic or not) to list HYACINTHOS will win the book: F.G.-M.: Exercices d' Algebre (1198 pages)

Good Luck!

PAUL PRIZE

Paul Erdos was a great mathematician who used to pose problems in mathematics periodicals (American Mathematical Monthly and others) with money prizes. I will do the same for a problem, but instead of money, the prize will be a classical book of geometry. It is a good coincidence that I have named the Point of the problem by the name of another great Paul: Pablo Picasso, and also today (18 January) is the birthday of my son Paul.

The Problem:

Let ABC be a triangle, A'B'C' the antipodal triangle of ABC (=circumcevian triangle of O) and D a point. Let A"B"C" be the circumcevian triangle of D with respect A'B'C' (ie A" is the other than A' intersection of A'D with the circumcircle etc)

Denote:

N,N1,N2,N3 the NPC centers of the triangles ABC, A"BC, B"CA, C"AB.

The four centers N,N1,N2,N3 are concyclic.

The center of the circle is General Picasso Point [See HERE]

Terms of Proofs:

The proof must be Euclidean synthetic (ie with no use of algebra or coordinate geometry)

The Prize:

The first, who will post a proof to the list Hyacinthos, will win the book:

F.G.-M.: Exercices de Geometrie. Huitieme Edition.

The book has 1302 pages + a 36 pages supplement

From page 1130 to page 1259: Geometrie du triangle:

Good Luck!