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

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

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

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

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

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