- Dear Bernard!
>

My proof isn't purely synthetic, but the calculations are minimal. We know

>let W1, W2 be the Brocard points and X15, X16 the isodynamic points.

>

>P1 = W1 X15 /\ W2 X16

>P2 = W1 X16 /\ W2 X15

>

>KP1P2 is an equilateral triangle inscribed in the Brocard circle. (K =

>X6)

>

>any synthetic proof for this fact ?

>

that the inversion wrt the circumcicle of X(15) is X(16) and this inversion

of X(6) is the midpoint of X(15)X(16). Also it is known that the Brocard

points are in the circle with diameter OX(6) and the angles WiOX(6) are

equal to the Brocard angle $\phi$ (i=1,2). So using the expressions for

distance OX(6) and $\phi$ it is easy to obtain that KP1=P1P2.

Sincerely Alexey

Antivirus scanning: Symantec Mail Security for SMTP. - Dear friends

I notice the following fact certainly well known for a long time and very

easy to prove:

Call P any point in the plane of triangle ABC and L the trilinear polar of P

wrt ABC.

Call Gamma the inscribed conic in ABC of which P is the perspector or

Brianchon point, i.e : P is the perspector of ABC and the contact triangle.

Let M be any point on Gamma and call T the tangent in M at Gamma.

Then the trilinear pole O of T is on L.

From this fact can we find :

1°a simple projective construction of the tangent T of M at Gamma.

2°If O is any point on L, a simple projective construction of the contact

point M of the trilinear polar T of O wrt ABC with Gamma.

Thanks for your swift replies

François

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