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## Re: a construction in the Ricardo problem

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• There is some typos in my last message, you must read: On Fri, 31 Dec 2004 20:55:32 +0100, Francois Rideau
Message 1 of 22 , Dec 31, 2004
There is some typos in my last message, you must read:

On Fri, 31 Dec 2004 20:55:32 +0100, Francois Rideau
<francois.rideau@...> wrote:
> On Tue, 28 Dec 2004 08:44:01 -0000, jpehrmfr
> <Jean-Pierre.Ehrmann.70@...> wrote:
> >
> >
> > Dear Ricardo
> > > To construct ABC knowing:
> > > G,
> > > a point N on AC,
> > > a point L on AB
> > > and the straight r in which it is BC
> >
> > r intersects GL at P and GN at Q.
> > N' - or L' - is the homothetic in (G,-1/2) of the harmonic conjugate
> > of G wrt (L,P) - or (N,Q) -
> > Then NN' is the line AC and LL' is the line AB.
> > I wish a very good year 2005 to every Hyacinthist.
> > Jean-Pierre
>
> About the construction of N' in the Jean-Pierre solution, it's better
> to use the intersection P' of the line LG with the line called "s"
> in my own solution, ("s" is the image of "r" by the central dilatation
> of center G and ratio -2); "s" is the locus of the vertex A. Since we
> have an harmonic pencil of lines (As,AG,AB,AC), we have an harmonic
> cross ratio (P',G,L,N') = -1.
> The points L and N' are related by an involutive collineation of
> center G and axis "s".
>
> Happy New Year
>
> >
> >
> >
> >
> >
> >
> > Yahoo! Groups Links
>
• Dear All, while playing around in GSP with various kinds of hexagons, I noticed that if a hexagon ABCDEF is circumscribed by a conic section (i.e. A, B, C, D,
Message 2 of 22 , Jan 8, 2005
Dear All,

while playing around in GSP with various kinds of hexagons, I noticed
that if a hexagon ABCDEF is circumscribed by a conic section (i.e. A, B,
C, D, E, F all lie on the same conic section), then the six "short"
diagonals AC, BD, CE, DF, EA, FB always seem to form a hexagon that is
inscribed by a conic, i.e. the diagonals are all six tangent lines to
one and the same conic. A dual version of this hypothesis is easily
formulated and seems to be true as well.

I am not familiar with this result, at least not in this form. I
imagine, however, it is something that can be found in the literature or
can be trivially derived from better known results. Anyone has a
reference or quick proof?

best,
Eisso

--

========================================
Eisso J. Atzema, Ph.D.
Department of Mathematics & Statistics
University of Maine
Orono, ME 04469
Tel.: (207) 581-3928 (office)
(207) 866-3871 (home)
Fax.: (207) 581-3902
E-mail: atzema@...
========================================
• Dear all, Happy New Year. ... This problem can be put another way: Let ABC be a triangle and A1, B1, C1 be points on the arcs BC, CA, AB of the ABC
Message 3 of 22 , Jan 9, 2005
Dear all,
Happy New Year.
Eisso wrote:
> while playing around in GSP with various kinds of
> hexagons, I noticed
> that if a hexagon ABCDEF is circumscribed by a conic
> section (i.e. A, B,
> C, D, E, F all lie on the same conic section), then
> the six "short"
> diagonals AC, BD, CE, DF, EA, FB always seem to form
> a hexagon that is
> inscribed by a conic, i.e. the diagonals are all six
> tangent lines to
> one and the same conic.

This problem can be put another way:
Let ABC be a triangle and A1, B1, C1 be points on the
arcs
BC, CA, AB of the ABC circumcircle. Let the distances
of A from the points A1,B1,C1 be a1,a2,a3
of B from the points A1,B1,C1 be b1,b2,b3
of C from the points A1,B1,C1 be c1,c2,c3

Ptolemaeus' theorem gives
aa1 = bb1 + cc1 (1)
bb2 = cc2 + aa2 (2)
cc3 = aa3 + bb3 (3)

If Ab is the intersection (BC, B1C1)= (opp side of A,
opp side of B1)
then from BAb/CAb = area(BA1C1)/area(CA1C1) =
b1b3/c1c3
we conclude that the barycentrics of Ab are
Ab = (0 : c1c3 : b1b3) and similarly
Ac = (0 : c1c2 : b1b2)
Bc = (c1c2 : 0 : a1a2)
Ba = (c2c3 : 0 : a2a3)
Ca = (b2b3 : a2a3 : 0 )
Cb = (b1b3 : a1a3 : 0 )

Finding the equations of the lines BcCb, CaAc, AbBa
and using (1), (2), (3) we prove that these diagonals
of your hexagon are concurrent. Hence by the inverse
Brianchon's theorem we conclude that this hexagon
has an inconic.

Best regards
Nikolaos Dergiades

____________________________________________________________
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• Dear Hyacinthists ... noticed ... A, B, ... six short ... that is ... to ... easily ... literature or ... I don t know who discovered the fact that if two
Message 4 of 22 , Jan 9, 2005
Dear Hyacinthists
--- In Hyacinthos@yahoogroups.com, "Eisso J. Atzema" <atzema@m...>
wrote:

> while playing around in GSP with various kinds of hexagons, I
noticed
> that if a hexagon ABCDEF is circumscribed by a conic section (i.e.
A, B,
> C, D, E, F all lie on the same conic section), then the
six "short"
> diagonals AC, BD, CE, DF, EA, FB always seem to form a hexagon
that is
> inscribed by a conic, i.e. the diagonals are all six tangent lines
to
> one and the same conic. A dual version of this hypothesis is
easily
> formulated and seems to be true as well.
>
> I am not familiar with this result, at least not in this form. I
> imagine, however, it is something that can be found in the
literature or
> can be trivially derived from better known results. Anyone has a
> reference or quick proof?

I don't know who discovered the fact that if two triangles (ACE and
BDF with your notations) are inscribed in the same conic, their 6
sidelines touch a conic (von Staudt? Steiner? Brianchon?...)
In any case, it happened probably about 1850.
Does anyone have a reference about the first occurence of this
result?
Friendly. Jean-Pierre
• Dear Jean-Pierre, your hunch is right. I found a reference to the theorem in Coolidge s History of Conic Sections. According to Coolidge, the theorem can be
Message 5 of 22 , Jan 10, 2005
Dear Jean-Pierre,

your hunch is right. I found a reference to the theorem in Coolidge's
History of Conic Sections. According to Coolidge, the theorem can be
found in Steiner's Systematische Entwickelung (...) of 1832. (see p.59
of the Dover edition of Coolidge's book).

Eisso

jpehrmfr wrote:

>
>
>I don't know who discovered the fact that if two triangles (ACE and
>BDF with your notations) are inscribed in the same conic, their 6
>sidelines touch a conic (von Staudt? Steiner? Brianchon?...)
>In any case, it happened probably about 1850.
>Does anyone have a reference about the first occurence of this
>result?
>Friendly. Jean-Pierre
>
>
>
>
>
>
>
>Yahoo! Groups Links
>
>
>
>
>
>
>
>

--

========================================
Eisso J. Atzema, Ph.D.
Department of Mathematics & Statistics
University of Maine
Orono, ME 04469
Tel.: (207) 581-3928 (office)
(207) 866-3871 (home)
Fax.: (207) 581-3902
E-mail: atzema@...
========================================
• Miquel Locus: Let B,C be two points on a line, B ,C two points on an other line and A a point on the line BC. Let M be the 2nd intersection of the circles
Message 6 of 22 , Aug 28, 2008
Miquel Locus:

Let B,C be two points on a line, B',C' two points on an other line
and A' a point on the line BC.
Let M be the 2nd intersection of the circles A'B'B, A'C'C.

As A' moves on BC, which is the locus of M?
If P := BB' /\ CC', the locus is the circle PB'C'.
[ The circles A'B'B, A'C'C, PB'C' are concurrent ]

Now, let ABC be a triangle, P a point, and A'B'C' the cevian
triangle of P.

Ma := The point of concurrence of the circles A'B'B, A'C'C, PB'C'

Mb := The point of concurrence of the circles B'C'C, B'A'A, PC'A'

Mc := The point of concurrence of the circles C'A'A, C'B'B, PA'B'

Are the triangles ABC, MaMbMc perspective??
(if no, which is the locus of P such that they are)

Antreas
• ... Another application: Let ABC be a triangle, and A1B1C1, A2B2C2 the cevian triangles of two points P,Q, resp. Denote: A := B1C1 / B2C2 B := C1A1 / C2A2
Message 7 of 22 , Aug 29, 2008
--- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
>
>
> Miquel Locus:
>
> Let B,C be two points on a line, B',C' two points on an other line
> and A' a point on the line BC.
> Let M be the 2nd intersection of the circles A'B'B, A'C'C.
>
> As A' moves on BC, which is the locus of M?
> If P := BB' /\ CC', the locus is the circle PB'C'.
> [ The circles A'B'B, A'C'C, PB'C' are concurrent ]

Another application:

Let ABC be a triangle, and A1B1C1, A2B2C2 the cevian triangles
of two points P,Q, resp.

Denote:

A' := B1C1 /\ B2C2

B' := C1A1 /\ C2A2

C' := A1B1 /\ A2B2

The circles A'BC, B'CA, C'AB are concurrent.
[Notice that A,B',C' - B,C',A' - C,A',B' are collinear triads of
points]

Special Case:

Q = P* (ie P, Q are Isogonal Conjugate points)

Which is the point of concurrence?

Antreas
• Dear Antreas, Triangles ABC, MaMbMc are perspective and if P = (x:y:z) then first barycentric of perspector is: (y + z)* (a^2*(x + y)*(x + z) - x*(b^2*(x + y)
Message 8 of 22 , Aug 29, 2008
Dear Antreas,

Triangles ABC, MaMbMc are perspective and if P = (x:y:z) then first barycentric of perspector is:

(y + z)*
(a^2*(x + y)*(x + z) - x*(b^2*(x + y) + c^2*(x + z)))*
(a^2*y^2*z^2 + b^2*z^2*x^2 - c^2*x^2*y^2 + 2*x*y*z*SC*(x + y + z))*
(a^2*y^2*z^2 - b^2*z^2*x^2 + c^2*x^2*y^2 + 2*x*y*z*SB*(x + y + z))

(For easy seeing I put some parts in separate line)

Best regards,
Bui Quang Tuan

--- On Fri, 8/29/08, xpolakis <xpolakis@...> wrote:

> From: xpolakis <xpolakis@...>
> Subject: [EMHL] Miquel
> To: Hyacinthos@yahoogroups.com
> Date: Friday, August 29, 2008, 4:13 AM
>
> Now, let ABC be a triangle, P a point, and
> A'B'C' the cevian
> triangle of P.
>
> Ma := The point of concurrence of the circles
> A'B'B, A'C'C, PB'C'
>
> Mb := The point of concurrence of the circles
> B'C'C, B'A'A, PC'A'
>
> Mc := The point of concurrence of the circles
> C'A'A, C'B'B, PA'B'
>
> Are the triangles ABC, MaMbMc perspective??
> (if no, which is the locus of P such that they are)
>
> Antreas
• Dear Antreas, If P = (p:q:r) and Q = (u:v:w) then first barycentric of point of concurrence is: (p*u*(q*w - r*v)*(p*v*w*(q - r) + q*u*w*(p + r) - r*u*v*(p +
Message 9 of 22 , Aug 29, 2008
Dear Antreas,
If P = (p:q:r) and Q = (u:v:w) then first barycentric of point of concurrence is:

(p*u*(q*w - r*v)*(p*v*w*(q - r) + q*u*w*(p + r) - r*u*v*(p + q)))/
(p*q*r*(a^2*v*w*(-p*v*w + q*w*u + r*u*v) - b^2*w*u*(-q*w*u + r*u*v + p*v*w) - c^2*u*v*(-r*u*v + p*v*w + q*w*u)) - u*v*w*(a^2*q^2*r^2*u - b^2*r^2*p^2*v - c^2*p^2*q^2*w))

Best regards,
Bui Quang Tuan

--- On Fri, 8/29/08, xpolakis <xpolakis@...> wrote:

> From: xpolakis <xpolakis@...>
> Subject: [EMHL] Re: Miquel
> To: Hyacinthos@yahoogroups.com
> Date: Friday, August 29, 2008, 4:56 PM
> --- In Hyacinthos@yahoogroups.com, "xpolakis"
> <xpolakis@...> wrote:
> >
>
> Another application:
>
> Let ABC be a triangle, and A1B1C1, A2B2C2 the cevian
> triangles
> of two points P,Q, resp.
>
> Denote:
>
> A' := B1C1 /\ B2C2
>
> B' := C1A1 /\ C2A2
>
> C' := A1B1 /\ A2B2
>
> The circles A'BC, B'CA, C'AB are concurrent.
> [Notice that A,B',C' - B,C',A' -
> C,A',B' are collinear triads of
> points]
>
> Special Case:
>
> Q = P* (ie P, Q are Isogonal Conjugate points)
>
> Which is the point of concurrence?
>
>
> Antreas
>
• Dear Antreas [Antreas] ... As Quang Tuan Bui told us, the answer is yes. I can add that A B C and MaMbMc are perspective with perspector the isogonal
Message 10 of 22 , Aug 29, 2008
Dear Antreas
[Antreas]
> Now, let ABC be a triangle, P a point, and A'B'C' the cevian
> triangle of P.
>
> Ma := The point of concurrence of the circles A'B'B, A'C'C, PB'C'
>
> Mb := The point of concurrence of the circles B'C'C, B'A'A, PC'A'
>
> Mc := The point of concurrence of the circles C'A'A, C'B'B, PA'B'
>
> Are the triangles ABC, MaMbMc perspective??

As Quang Tuan Bui told us, the answer is yes.
I can add that A'B'C' and MaMbMc are perspective with perspector the
isogonal conjugate of the complement of P.
Friendly. Jean-Pierre
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