Τρίτη, 28 Φεβρουαρίου 2012

Concyclic Os


Let ABC be a triangle and P a point.


For which P's the circumcenters O,Oa,Ob,Oc of ABC, PBC, PCA, PAB, resp. are concyclic?

APH, 28 February 2012

Παρασκευή, 24 Φεβρουαρίου 2012

Oa,Ob,Oc


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


The reflection of BB' in AA' intersects AB in Ab
The reflection of CC' in AA' intersects AC in Ac

Let Oa be the circumcenter of A'AbAc

A'AbAc is the reflection of A'B'C' in AA' and Oa is the reflection
of the circumcenter of A'B'C' in AA'.

Similarly Ob, Oc.

Oa, Ob, Oc are the reflections of the circumcenter of the cevian triangle of I in the bisectors AA',BB',CC', resp. Therefore the lines AOa, BOb, COc concur at the isogonal conjugate of the circumcenter of A'B'C'.

Coordinates ?

Generalization:
For P instead of I:

We have two problems:

1. The locus of P such that ABC, OaObOc are perspective. (Oa,Ob,Oc as defined above)
2. The locus of P such that ABC, QaQbQc are perspective, where Qa is the circumcenter of the triangle which is the reflection of A'B'C' in AA' (ie Qa is the reflection of the circumcenter of A'B'C' in AA') and similarly Qb,Qc

APH, 24 February 2012

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Coordinates:

(a (a^5 b + a^4 b^2 - 2 a^3 b^3 - 2 a^2 b^4 + a b^5 + b^6 + a^5 c -
3 a^3 b^2 c - 2 a^2 b^3 c + 2 a b^4 c + 2 b^5 c + a^4 c^2 -
3 a^3 b c^2 - 2 a^2 b^2 c^2 - 3 a b^3 c^2 - b^4 c^2 - 2 a^3 c^3 -
2 a^2 b c^3 - 3 a b^2 c^3 - 4 b^3 c^3 - 2 a^2 c^4 + 2 a b c^4 -
b^2 c^4 + a c^5 + 2 b c^5 + c^6) : ...:...)

1. The locus is a 18 degree curve.

2. The locus is a 24 degree curve.

Nikos Dergiades, Hyacinthos #20870

Πέμπτη, 23 Φεβρουαρίου 2012

PP*


Let ABC be a triangle, P a point and A1B1C1 the pedal triangle of P.

Let Q be a point on the line PDP* [D = the center of the common pedal circle of P and its isog. conjugate P*] and Q1Q2Q3 the circumcevian triangle of Q with respect A1B1C1.

The triangles ABC, Q1Q2Q3 are perspective.

APH 23 February 2012


Τετάρτη, 22 Φεβρουαρίου 2012

PP* --> PP*


Let ABC be a triangle and A1B1C1, A2B2C2 the pedal triangles of two isogonal points P,P*, resp.


Let Q be a point on the line PP* and P1P2P3 the circumcevian triangle of Q with respect the triangle A1B1C1.

Conjecture:

The triangles A2B2C2, P1P2P3 are perspective. The perspector Q' lies on PP*.

APH, 22 February 2012

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The locus of Q such that P1P2P3 and A2B2C2 are perspective includes the line PP*

I find that the locus includes also the points on the circle with the midpoint
of PP* as center and and radius (AP BP CP)/(2 |OP^2 - R^2|).

Francisco Javier, Hyacinthos #20856.

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Let R1R2R3 be the reflection of the triangle P1P2P3 in the line PP*.

The triangles A2B2C2, R1R2R3 are perspective.

APH, 24 February 2012


Concurrent Euler Lines (generalization)


Problem 1:

Let L1,L2 be two lines intersected at A, and P a point. To draw line L intersecting L1,L2 at Ab,Ac, resp. such that:

P be

- the circumcenter of AAbAc


Ab, Ac are the other than A intersections of the circle (P,PA) with the lines L1,L2

- the orthocenter of AAbAc


The perpendiculars to L1,L2 through P intersect L2,L1 at Ac,Ab, resp.

In general, P be a fixed point on the Euler line of AAbAc.
(ie PO/PH = m/n, where m,n given numbers)

Problem 2:

Let ABC be a triangle and AaBbCc the orthic triangle. Let M1 be a line intersecting AB,AC at Ab,Ac resp. such that Aa is a fixed X(i) point on the Euler line of AAbAc. Similarly M2 a line intersecting BC,BA at Bc,Ba resp. such that Bb is X(i) point on the Euler line of BbBcBa, and M3 a line intersecting CA,CB at Ca,Cb such that Cc is X(i) point on the Euler line of CcCaCb.


For which points X(i) the Euler Lines of the triangles:

1. AAbAc, BBcBa, CCaCb

2. AaAbAc, BbBcBa, CcCaCb

are concurrent?

[For X(i) = O is the Problem Concurrent Euler Lines]

APH, 22 February 2012

Τρίτη, 21 Φεβρουαρίου 2012

Concurrent Euler Lines


Let ABC be a triangle and AaBbCc the orthic triangle.


The circle (Aa,AaA) intersects AB,AC at Ab,Ac, resp. (other than A)
The circle (Bb,BbB) intersects BC,BA at Bc,Ba, resp. (other than B)
The circle (Cc,CcC) intersects CA,CB at Ca,Cb, resp. (other than C)

The Euler lines of the triangles:

1. AAbAc, BBcBa, CCaCb

2. AaAbAc, BbBcBa, CcCaCb
[=perp. bisectors of AbAc, BcBa, CaCb, resp.]

are concurrent.

Points ?

APH, 21 February 2012

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1) (-s1*s2^2+2*s1^2*s3+s2*s3)/(2*s1*s3)
2) (3*s1*s3-s2^2)/(2*s3)
It is the orhocenter of the orthic triangle.

Where I use Moley'trick and s1=a+b+c, s2=ab+bc+ca and s3=abc

Etienne Rousee, Hyacinthos #20850

Τρίτη, 14 Φεβρουαρίου 2012

EROTOKRITOS CIRCLE


Let ABC be a triangle, A1B1C1 the circumcevian triangle of H and A2B2C2 the cevian triangle of G (medial triangle).


Denote:

A3 = the other than A1 intersection of A1A2 and the circumcircle.

B3 = the other than B1 intersection of B1B2 and the circumcircle.

C3 = the other than C1 intersection of C1C2 and the circumcircle.

N1 = The NPC center of A3BC

N2 = The NPC center of B3CA

N3 = The NPC center of C3AB.

The four NPC centers N, N1,N2,N3 are concyclic (??).

Center of the circle?

APH, 14 February 2012

Δευτέρα, 13 Φεβρουαρίου 2012

Euler Line


Let ABC be a triangle, Aa,Bb,Cc the orth. projections of A,B,C on the Euler line, resp., A1B1C1, A2B2C2 the medial, orthic triangles, resp. and P a point.


Let A'B'C', A"B"C" be the circumcevian triangles of P with respect the triangles A1B1C1, A2B2C2, resp.

Which is the locus of P such that the lines:

1. A'Aa, B'Bb, C'Cc

2. A"Aa, B"Bb, C"Cc

are concurrent?

The Euler line + ??

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

Generalization:

P,P* = two isogonal conjugate points.

A1B1C1, A2B2C2 = the pedal triangles of P,P*, resp.

Aa, Bb, Cc = the orth. projections of A,B,C on PP*, resp.

A'B'C', A"B"C" = the circumcevian triangles of a point Q with respect A1B1C1, A2B2C2, resp.

Which is the locus of Q such that the lines:

1. A'Aa, B'Bb, C'Cc

2. A"Aa, B"Bb, C"Cc

are concurrent?

Is it the line PP* + ??

Locus of point of concurrence? Common Circumcircle of A1B1C1 and A2B2C2 + ??

APH, 13 February 2012

Παρασκευή, 10 Φεβρουαρίου 2012

Perspective


Let ABC be a triangle, A'B'C' the orthic triangle, A1B1C1 the cevian triangle of G and A2B2C2 the circumcevian triangle of G with respect A1B1C1.


Denote:

A* = A2O /\ A'N

B* = B2O /\ B'N

C* = C2O /\ C'N

The triangles ABC, A*B*C* are perspective (?)
(perspector on the Euler line?)

Variation:

A** = A2N /\ A'O

B** = B2N /\ B'O

C** = C2N /\ C'O

Are the triangles:

ABC, A**B**C**

A*B*C*, A**B**C**

perspective?

APH, 10 February 2012

EL GRECO CIRCLE

Continued from X3542

Let N, N1, N2, N3 be the NPC centers of ABC, A2BC, B2CA, C2AB.

Are  N,N1,N2,N3 lying on a circle (with center on the HO line [Euler Line], and for a point P instead of O, on the HP line)?

APH, 10 February 2012

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It is true for P = O (see the proof in the comment).
I name the circle as EL CRECO circle in honor of the great Cretan - Spanish painter DOMINICOS THEOTOKOPOULOS known as EL GRECO

APH




Τετάρτη, 8 Φεβρουαρίου 2012

X3542


Let ABC be a triangle, A'B'C' the cevian triangle of H (orthic) and A"B"C" the circumcevian triangle of O with respect A'B'C'.


The line A"H intersects the circumcircle of ABC at A1,A2 with |HA1| < |HA2|. Similarly the points B1, B2 with |HB1| < |HB2| and C1, C2 with |HC1| < |HC2|.

Are the triangles ABC, A1B1C1 perspective ?

APH,  8 February 2012

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Yes, they're perspective with perspector Q=X3542, on the Euler line.
This point satisfies, HQ:QG = - 6 cosA cosB cosC

Francisco Javier García Capitán
9 February 2012

A2B2C2 and ABC perspective if and only P is  lies on the NPC or in the trilinear
polar of X2052.

 Francisco Javier García Capitán Hyacinthos #20815


_______________________________________________


Generalization: Is it true if we replace O with a point P?
(The perspector on the HP line)
APH, Hyacinthos #20809

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Yes, it is true even if you replace O by P,
It follows from the following problem:-
It follows from the following problem which is a generalisation of a
problem you posted earlier:- Given a triangle ABC, A1B1C1 be the
circumcevian triangle of a point R and A2B2C2 be the circumcevian triangle
of a point Q. If A3B3C3 is the circumcevian triangle of R wrt A2B2C2, then
A1A3,B1B3,C1C3 concur on RQ.
Now take R= reflection of H on P and Q=H and the result follows.

CHANDAN, Hyacinthos #20810

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Κυριακή, 5 Φεβρουαρίου 2012

PICASSO CIRCLES


PICASSO POINT.

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:

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

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

See Picasso Point Generalized.

Call the circle (N,N1,N2,N3) as Picasso Circle with respect ABC and D and let P0 be its center (ie the center of the Picasso circle with respect ABC and D)

Now, omit the vertex A and replace it with A" and denote Pa = the center of the Picasso circle with respect A"BC and D.

Similarly denote Pb = the center of the Picasso circle with respect AB"C and Pc = the center of the Picasso circle with respect ABC" and D.


Are the following true?

1. The four circles (P0), (Pa), (Pb), (Pc) are concurrent at a point Pp.

2. The four centers P0, Pa, Pb, Pc and the point D lie on a circle with center Pp.

APH, 5 February 2012

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1. It is true. The point Pp is the reflection of O on the midpoint of DN, that is DONPp is a parallelogram, and DPp is always parallel to Euler line.

2. The four centers P0, Pa, Pb, Pc, BUT NOT THE POINT D lie on a circle with center Pp. The radius of this circle is HALF the distance between D and O.

Francisco Javier García Capitán
6 February 2012

Παρασκευή, 3 Φεβρουαρίου 2012

X275


Let ABC be a triangle, P a point, P1P2P3 the cevian triangle of P and PaPbPc the cevian triangle of P with respect the triangle P1P2P3.


Denote

R1 = the radical axis of the circles (BPbP3) and (CPcP2)

R2 = the radical axis of the circles (CPcP1) and (APaP3)

R3 = the radical axis of the circles (APaP2) and (BPbP1)

For which P's the lines R1,R2,R3 are concurrent ?

APH, 3 February 2012

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It is a 15th degree locus through H. For P=H, the intersection point is X275.

Francisco Javier García Capitán
3 February 2012

 

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