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@...>

>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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