- Dear Alexei and Jean-Pierre

another approach that needs an elementary

construction is the following:

but I can not give a synthetic proof.

If X, Y are points on the sides AB, AC

and the perpendicular bisector of XY meets

the radical axis of the excircles (Ib), (Ic)

or the radical axis of (I), (Ia) at the point Z

then the circumcircle of XYZ is tangent to

(Ib), (Ic) or to (I), (Ia) if and only if

XY is parallel of antiparallel to BC.

Jean-Pierre what is the general statement

of the property of hyperbola you used?

Best regards

Nikos Dergiades

> [Alexei]

___________________________________________________________

> > Let the points X, Y be on the sidelines AC, BC of

> the triangle ABC

> > and XY is parallel to AB. Then there exists a

> circle passing through

> > X, Y and touching two excircles of the triangle.

>

>[JPE] More precisely :

> - there exists a circle through X, Y touching the

> A-excircle and the B-

> excircle

> - there exists a circle through X, Y touching the

> incircle and the C-

> excircle

> We have the same result if XY is antiparallel to AB

>

> Here are some remarks (I've changed the notations

> and I didn't find a

> synthetic proof)

> Let I,Ia,Ib,Ic be the incenter and excenters; U,U'

> the feet on BC of

> the internal and external A-bisectors

> Then there exists

> - an hyperbola h with center U going through the 4

> projections of I and

> Ia upon AB and AC (the asymptots are BC and the

> other common tangent

> through U to the incircle and the A-excircle)

> - an hyperbola h' with center U' going through the 4

> projections of Ib

> and Ic upon AB and AC (the asymptots are BC and the

> other common

> tangent through U' to the B- and C-excircles)

>

> Consider a point M lying on h; the parallel and

> antiparallel to BC

> through M intersect AB and AC at four points lying

> on a circle, and

> this circle touches the incircle and the A-excircle

> (and the line

> through the contact points goes through U)

> Consider a point M' lying on h'; the parallel and

> antiparallel to BC

> through M' intersect AB and AC at four points lying

> on a circle, and

> this circle touches the B-excircle and the

> C-excircle (and the line

> through the contact points goes through U')

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr - Dear Hyacinthists,

Dear François,

It seems the messages are leaving/arriving normally again.

I see now your solution is to draw a circle through two

given points and tangent to a line. This is classical.

One can have two solutions. These are the circumcenters

of the triangles ABC given <h_a,d_a,m_b>.

Barukh Ziv some 5 years ago sent me a message showing

that O belongs to a parabola <P=(F,d)> and to h'_a,

reflection of h_a on d_a.

The directrix <d> is parallel to (B,C) and distant m_b^2/4h_a

from it. The focus <F> is distant \ell/4 from (A,H_a) and

9h_a/8 from <d>. Here \ell=\sqrt{m_b^2-(h_a/2)^2}.

Data to have 2 solutions (from Die Konstruktion von

Dreiecken, Kurt Herterich).

<h_a=6 cm>, <m_b=5 cm>, and <d_a=6.5 cm>;

<h_a=3 cm>, <m_b=3.9 cm>, and <d_a=5 cm>;

Best regards,

Luis

>From: "Francois Rideau" <francois.rideau@...>

[clip]

>Reply-To: Hyacinthos@yahoogroups.com

>To: Hyacinthos@yahoogroups.com

>Subject: Re: [EMHL] line/conic intersection

>Date: Mon, 2 Jul 2007 20:21:59 +0200

>

>Yes, nice solution and now I remember that I learned it more than fifty

>years ago.

>Of course the most important is the discussion which must follows on the

>number of solutions.

>About the teaching of the theory of conics, I think the best book was

>written by the great Henri Lebesgue himself: Les coniques, published in

>1942

>at Gauthier-Villars, just after his death, during the nazi occupation of

>France.

>Poor Henri, nowadays, the only conic still in use in France is the circle

>on

>roadsigns just before runabouts!

>Friendly

>Francois

>

>

>On 7/2/07, Luís Lopes <qed_texte@...> wrote:

_________________________________________________________________

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