NICOLAS NICOLAIDES

Nicolas Nicolaides, Professor of Mathematics (1826-1889)

Biography of Nicolaides by Theodoros Passas published in the Supplement of the Bulletin of the Greek Mathematical Society #4 (May - June 1931)

Biography of Nicolaides by Michael Kasiouras published in Euclid, Supplement of the Bulletin of the Greek Mathematical Society, New Series, Vol. 4, #3(Nov. 1970)

See short biographies of Nicolaides

Greek Wikipedia HERE

By Spyros Zervos HERE

2nd edition

YYYYYYYYYYYYYYYYYYYYYY

FEMININE GEOMETRY

In feminine geometry

all angles are tabooed;

But angling is woman's game,

when she is being wooed.

INCIRCLE

CIRCUMCIRCLE

A circumcircle with thousands of inscribed memories....

Perspective

Let ABC be a triangle, A1B1C1 the orthic triangle, A2B2C2 the circumcevian triangle of O with respect A1B1C1, A3B3C3 the antipodal triangle of A2B2C2 and A'B'C' the triangle bounded by the lines A1A3, B1B3, C1C3, resp..

The triangles A1B1C1 and A'B'C' are perspective.

Perspector?

APH, 19 March 2012

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

The perspector is X25.

Francisco Javier García Capitán
19 March 2012

HOW THEY DO GEOMETRY IN CHINA

how they do geometry in China

Perspective

Let ABC be a triangle, A'B'C' the cevian triangle of G and A"B"C" the circumcevian triangle of G with respect A'B'C'. The circles with diameters HA",HB",HC" intersect the NPC again at A1,B1,C1, resp.

The circles with diameters HA",HB",HC" intersect the NPC again at A1,B1,C1, resp.

The triangles ABC, A1B1C1 are perspective.

Perspector?

APH, 16 March 2012

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

Not in ETC. It is the isotomic conjugate of

(SA (b^4 + c^4-a^4), SB (a^4 - b^4 + c^4), SC (a^4 + b^4 - c^4))

************************************

ADDENDUM (12/9/19)
Perspector: X(13854)
Isotomic conjugate: X(34254)
************************************

The locus of P, instead of H, is trilinear polar of X648 (containing H) + NPC + a cubic:

5 a^12 x^3 - 28 a^10 b^2 x^3 + 53 a^8 b^4 x^3 - 40 a^6 b^6 x^3 +
7 a^4 b^8 x^3 + 4 a^2 b^10 x^3 - b^12 x^3 - 28 a^10 c^2 x^3 +
110 a^8 b^2 c^2 x^3 - 120 a^6 b^4 c^2 x^3 + 36 a^4 b^6 c^2 x^3 +
4 a^2 b^8 c^2 x^3 - 2 b^10 c^2 x^3 + 53 a^8 c^4 x^3 -
120 a^6 b^2 c^4 x^3 + 42 a^4 b^4 c^4 x^3 - 8 a^2 b^6 c^4 x^3 +
b^8 c^4 x^3 - 40 a^6 c^6 x^3 + 36 a^4 b^2 c^6 x^3 -
8 a^2 b^4 c^6 x^3 + 4 b^6 c^6 x^3 + 7 a^4 c^8 x^3 +
4 a^2 b^2 c^8 x^3 + b^4 c^8 x^3 + 4 a^2 c^10 x^3 - 2 b^2 c^10 x^3 -
c^12 x^3 + a^12 x^2 y - 4 a^10 b^2 x^2 y - 7 a^8 b^4 x^2 y +
40 a^6 b^6 x^2 y - 53 a^4 b^8 x^2 y + 28 a^2 b^10 x^2 y -
5 b^12 x^2 y - 2 a^10 c^2 x^2 y + 24 a^8 b^2 c^2 x^2 y -
44 a^6 b^4 c^2 x^2 y + 48 a^4 b^6 c^2 x^2 y - 34 a^2 b^8 c^2 x^2 y +
8 b^10 c^2 x^2 y - 17 a^8 c^4 x^2 y + 46 a^4 b^4 c^4 x^2 y +
3 b^8 c^4 x^2 y + 36 a^6 c^6 x^2 y - 24 a^4 b^2 c^6 x^2 y +
4 a^2 b^4 c^6 x^2 y - 8 b^6 c^6 x^2 y - 17 a^4 c^8 x^2 y +
4 a^2 b^2 c^8 x^2 y + b^4 c^8 x^2 y - 2 a^2 c^10 x^2 y +
c^12 x^2 y - 5 a^12 x y^2 + 28 a^10 b^2 x y^2 - 53 a^8 b^4 x y^2 +
40 a^6 b^6 x y^2 - 7 a^4 b^8 x y^2 - 4 a^2 b^10 x y^2 + b^12 x y^2 +
8 a^10 c^2 x y^2 - 34 a^8 b^2 c^2 x y^2 + 48 a^6 b^4 c^2 x y^2 -
44 a^4 b^6 c^2 x y^2 + 24 a^2 b^8 c^2 x y^2 - 2 b^10 c^2 x y^2 +
3 a^8 c^4 x y^2 + 46 a^4 b^4 c^4 x y^2 - 17 b^8 c^4 x y^2 -
8 a^6 c^6 x y^2 + 4 a^4 b^2 c^6 x y^2 - 24 a^2 b^4 c^6 x y^2 +
36 b^6 c^6 x y^2 + a^4 c^8 x y^2 + 4 a^2 b^2 c^8 x y^2 -
17 b^4 c^8 x y^2 - 2 b^2 c^10 x y^2 + c^12 x y^2 - a^12 y^3 +
4 a^10 b^2 y^3 + 7 a^8 b^4 y^3 - 40 a^6 b^6 y^3 + 53 a^4 b^8 y^3 -
28 a^2 b^10 y^3 + 5 b^12 y^3 - 2 a^10 c^2 y^3 + 4 a^8 b^2 c^2 y^3 +
36 a^6 b^4 c^2 y^3 - 120 a^4 b^6 c^2 y^3 + 110 a^2 b^8 c^2 y^3 -
28 b^10 c^2 y^3 + a^8 c^4 y^3 - 8 a^6 b^2 c^4 y^3 +
42 a^4 b^4 c^4 y^3 - 120 a^2 b^6 c^4 y^3 + 53 b^8 c^4 y^3 +
4 a^6 c^6 y^3 - 8 a^4 b^2 c^6 y^3 + 36 a^2 b^4 c^6 y^3 -
40 b^6 c^6 y^3 + a^4 c^8 y^3 + 4 a^2 b^2 c^8 y^3 + 7 b^4 c^8 y^3 -
2 a^2 c^10 y^3 + 4 b^2 c^10 y^3 - c^12 y^3 + a^12 x^2 z -
2 a^10 b^2 x^2 z - 17 a^8 b^4 x^2 z + 36 a^6 b^6 x^2 z -
17 a^4 b^8 x^2 z - 2 a^2 b^10 x^2 z + b^12 x^2 z -
4 a^10 c^2 x^2 z + 24 a^8 b^2 c^2 x^2 z - 24 a^4 b^6 c^2 x^2 z +
4 a^2 b^8 c^2 x^2 z - 7 a^8 c^4 x^2 z - 44 a^6 b^2 c^4 x^2 z +
46 a^4 b^4 c^4 x^2 z + 4 a^2 b^6 c^4 x^2 z + b^8 c^4 x^2 z +
40 a^6 c^6 x^2 z + 48 a^4 b^2 c^6 x^2 z - 8 b^6 c^6 x^2 z -
53 a^4 c^8 x^2 z - 34 a^2 b^2 c^8 x^2 z + 3 b^4 c^8 x^2 z +
28 a^2 c^10 x^2 z + 8 b^2 c^10 x^2 z - 5 c^12 x^2 z +
10 a^12 x y z - 36 a^10 b^2 x y z + 22 a^8 b^4 x y z +
8 a^6 b^6 x y z + 22 a^4 b^8 x y z - 36 a^2 b^10 x y z +
10 b^12 x y z - 36 a^10 c^2 x y z + 124 a^8 b^2 c^2 x y z -
88 a^6 b^4 c^2 x y z - 88 a^4 b^6 c^2 x y z +
124 a^2 b^8 c^2 x y z - 36 b^10 c^2 x y z + 22 a^8 c^4 x y z -
88 a^6 b^2 c^4 x y z + 132 a^4 b^4 c^4 x y z -
88 a^2 b^6 c^4 x y z + 22 b^8 c^4 x y z + 8 a^6 c^6 x y z -
88 a^4 b^2 c^6 x y z - 88 a^2 b^4 c^6 x y z + 8 b^6 c^6 x y z +
22 a^4 c^8 x y z + 124 a^2 b^2 c^8 x y z + 22 b^4 c^8 x y z -
36 a^2 c^10 x y z - 36 b^2 c^10 x y z + 10 c^12 x y z + a^12 y^2 z -
2 a^10 b^2 y^2 z - 17 a^8 b^4 y^2 z + 36 a^6 b^6 y^2 z -
17 a^4 b^8 y^2 z - 2 a^2 b^10 y^2 z + b^12 y^2 z +
4 a^8 b^2 c^2 y^2 z - 24 a^6 b^4 c^2 y^2 z + 24 a^2 b^8 c^2 y^2 z -
4 b^10 c^2 y^2 z + a^8 c^4 y^2 z + 4 a^6 b^2 c^4 y^2 z +
46 a^4 b^4 c^4 y^2 z - 44 a^2 b^6 c^4 y^2 z - 7 b^8 c^4 y^2 z -
8 a^6 c^6 y^2 z + 48 a^2 b^4 c^6 y^2 z + 40 b^6 c^6 y^2 z +
3 a^4 c^8 y^2 z - 34 a^2 b^2 c^8 y^2 z - 53 b^4 c^8 y^2 z +
8 a^2 c^10 y^2 z + 28 b^2 c^10 y^2 z - 5 c^12 y^2 z - 5 a^12 x z^2 +
8 a^10 b^2 x z^2 + 3 a^8 b^4 x z^2 - 8 a^6 b^6 x z^2 +
a^4 b^8 x z^2 + b^12 x z^2 + 28 a^10 c^2 x z^2 -
34 a^8 b^2 c^2 x z^2 + 4 a^4 b^6 c^2 x z^2 + 4 a^2 b^8 c^2 x z^2 -
2 b^10 c^2 x z^2 - 53 a^8 c^4 x z^2 + 48 a^6 b^2 c^4 x z^2 +
46 a^4 b^4 c^4 x z^2 - 24 a^2 b^6 c^4 x z^2 - 17 b^8 c^4 x z^2 +
40 a^6 c^6 x z^2 - 44 a^4 b^2 c^6 x z^2 + 36 b^6 c^6 x z^2 -
7 a^4 c^8 x z^2 + 24 a^2 b^2 c^8 x z^2 - 17 b^4 c^8 x z^2 -
4 a^2 c^10 x z^2 - 2 b^2 c^10 x z^2 + c^12 x z^2 + a^12 y z^2 +
a^8 b^4 y z^2 - 8 a^6 b^6 y z^2 + 3 a^4 b^8 y z^2 +
8 a^2 b^10 y z^2 - 5 b^12 y z^2 - 2 a^10 c^2 y z^2 +
4 a^8 b^2 c^2 y z^2 + 4 a^6 b^4 c^2 y z^2 - 34 a^2 b^8 c^2 y z^2 +
28 b^10 c^2 y z^2 - 17 a^8 c^4 y z^2 - 24 a^6 b^2 c^4 y z^2 +
46 a^4 b^4 c^4 y z^2 + 48 a^2 b^6 c^4 y z^2 - 53 b^8 c^4 y z^2 +
36 a^6 c^6 y z^2 - 44 a^2 b^4 c^6 y z^2 + 40 b^6 c^6 y z^2 -
17 a^4 c^8 y z^2 + 24 a^2 b^2 c^8 y z^2 - 7 b^4 c^8 y z^2 -
2 a^2 c^10 y z^2 - 4 b^2 c^10 y z^2 + c^12 y z^2 - a^12 z^3 -
2 a^10 b^2 z^3 + a^8 b^4 z^3 + 4 a^6 b^6 z^3 + a^4 b^8 z^3 -
2 a^2 b^10 z^3 - b^12 z^3 + 4 a^10 c^2 z^3 + 4 a^8 b^2 c^2 z^3 -
8 a^6 b^4 c^2 z^3 - 8 a^4 b^6 c^2 z^3 + 4 a^2 b^8 c^2 z^3 +
4 b^10 c^2 z^3 + 7 a^8 c^4 z^3 + 36 a^6 b^2 c^4 z^3 +
42 a^4 b^4 c^4 z^3 + 36 a^2 b^6 c^4 z^3 + 7 b^8 c^4 z^3 -
40 a^6 c^6 z^3 - 120 a^4 b^2 c^6 z^3 - 120 a^2 b^4 c^6 z^3 -
40 b^6 c^6 z^3 + 53 a^4 c^8 z^3 + 110 a^2 b^2 c^8 z^3 +
53 b^4 c^8 z^3 - 28 a^2 c^10 z^3 - 28 b^2 c^10 z^3 + 5 c^12 z^3

Francisco Javier García Capitán
17 March 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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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!

Point on the Euler Line

Let ABC be a triangle, A1B1C1 the pedal triangle of O = cevian triangle of G, G1G2G3 the circumcevian triangle of G with respect A1B1C1 (ie G1 is the other than A1 intersection of the NPC and AA1 etc), O1O2O3 the circumcevian triangle of O with respect A1A2A3 (ie O1 is the other than A1 intersection of A1O and the NPC etc) and GaGbGc the circumcevian triangle of O with respect G1G2G3 (ie Ga is the other than G1 intersection of OG1 and the NPC etc).

The triangles O1O2O3 and GaGbGc are perspective.

Perspector (on the Euler line of ABC) ?

APH 17 January 2012

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

The perspector Q lies on the OH line and satisfies OQ:QH = -2 cosA cosB cosC

The coordinates are:

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

This is not in ETC.

ADDENDUM (9/9/19)

Peter Moses:
X(18531)

Generalization for point P instead of O:

P a point (instead of O)
A1B1C1 = the cevian triangle of G (medial triangle)
P1P2P3 = the circumcevian triangle of P with respect A1B1C1
G1G2G3 = the circumcevian triangle of G with respect A1B1C1
GaGbGc = the circumcevian triangle of P with respect G1G2G3

P1P2P3 and GaGbGc are perspective.
The perspector S is on line GP and satisfies the ratio:

k = GS:SP = (PN^2 - R^2/4) (a^2x+b^2 y+ c^2z)/(x+y+z), where N is the nine point center (and R/2 is the radius of the nine point circle)

Francisco Javier García Capitán
18 January 2012

PICASSO POINT Generalized

Generalization of 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.

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

Which is the center, General Picasso Point, of the circle ?

APH, 17 January 2012

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

This is the midpoint of DH.

Francisco Javier García Capitán
17 January 2012