- Dear Francois,

> 1° In case of (Co) beeing the circumcircle, I think

Yes you are right if the areas are signed.

> the Tuan locus is also

> the locus of the points P such that OP = OQ, where O

> is the circumcenter and

> Q the isogonal conjugate of P.

The circumcevian triangle A1B1C1 of P is

similar to the pedal(P) and

the circumcevian triangle A2B2C2 of Q is

similar to the pedal(Q) and

Since P and Q are isogonal conjugates

pedal(P) and pedal(Q) have the same circumcircle.

the ratio of areas pedal(P)/A1B1C1 = pedal(Q)/A2B2C2

The triangles A1B1C1, A2B2C2 have equal areas

iff pedal(P), pedal(Q) have equal areas

but it is known that

(pedal(P))/(ABC) = |R^2-OP^2|/(4RR)

Hence |R^2-OP^2| = |R^2-OQ^2|.

If the areas are signed then then

we get OP = OQ as you mentioned

but if the areas are not signed then we can have

OP^2 + OQ^2 = 2R^2.

Best regards

Nikos Dergiades

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

Thank you very much for your hints. I think they are complete proof already and not only normal proof. It is special proof: proof by picture.

By your instructions, we can draw four special lines: AA'//BB'//CC' and one line passing through midpoints of these segments, center of circumconic and is the collinear line of Pascal theorem for our six points. These four lines combine with sidelines of two triangles to make some equal area pieces so we can see the result right after drawing.

I love this simple picture!

Thank you again and best regards,

Bui Quang Tuan

Francois Rideau <francois.rideau@...> wrote: .> This fact is:

Here some hint : We start with the triangle ABC, the circumconic {Gamma} and

> If through any point X on circumconic we draw three parallel lines with

> three sidelines of reference triangle then the triangle bounded by three

> second intesections (other than X) of three these lines with circumconic has

> area value the same as area value of reference triangle.

>

> any point M on {Gamma}. We get a new in-triangle A'B'C' such that MA' is

.

> parallel to BC, MB' is parallel to CA, MC' is parallel to AB. Then lines

> AA', BB', CC' are parallel and triangle A'B'C' is the image of ABC by some

> affine symmetry f, preserving {Gamma} of which the axis is some diameter of

> {Gamma} and the direction AA', BB' or CC'. As f change the signed area in

> its opposite, we are done! Friendly Francois

>

>

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>

>

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