- Greetings,

I see that in Paul Yiu, Introduction to the geometry of the triangle,

it is mentioned that the perspector of triangle formed by the points

of tangency of the Lucas circles and ABC is X371, the isogonal of the

Vecten point.

Isn't it true that this point is the radical center of the circles

circumscribed the three squares inscribed in ABC ?

Sorry if it's a dumb question, I didn't see it mentioned. In my sketch

the collinearity of the circumcenter of ABC, radical center of the

Lucas circles and X371 followed from the above remark, so that's why I

ask.

Thank you and best regards,

Cosmin Pohoata - Hello,

For now, what I have seen is that the centers of these squares are

perspective with ABC, having as perspector X486, the inner Vecten

point.

I don't know also if this is new.

--- In Hyacinthos@yahoogroups.com, "pohoata_cosmin2000"

<pohoata_cosmin2000@...> wrote:>

triangle,

> Greetings,

> I see that in Paul Yiu, Introduction to the geometry of the

> it is mentioned that the perspector of triangle formed by the points

the

> of tangency of the Lucas circles and ABC is X371, the isogonal of

> Vecten point.

sketch

>

> Isn't it true that this point is the radical center of the circles

> circumscribed the three squares inscribed in ABC ?

>

> Sorry if it's a dumb question, I didn't see it mentioned. In my

> the collinearity of the circumcenter of ABC, radical center of the

why I

> Lucas circles and X371 followed from the above remark, so that's

> ask.

>

>

> Thank you and best regards,

> Cosmin Pohoata

> > I see that in Paul Yiu, Introduction to the geometry

No it not correct.

> of the triangle,

> it is mentioned that the perspector of triangle

> formed by the points

> of tangency of the Lucas circles and ABC is X371,

> the isogonal of the

> Vecten point.

>

> Isn't it true that this point is the radical center

> of the circles

> circumscribed the three squares inscribed in ABC ?

I think you conclude this by a sketch.

If we call these circes C1, C2, C3

then the radical axis of C2, C3 seems to pass

through X371 but if you calculate the distance

of X371 from the above axis you'll see it is not zero.

The barycentrics of the radical center are

sinAsinAcosA /(sinA+cosA), . . , . .

while the barycentrics of X371 are

sinA(sinA+cosA), . . , . .

Best regards

Nikos Dergiades

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr- Dear, Cosmin Pohoata, Nikos, and all,

I was wrong, the perspector is X485, the outer Vecten point

Best regards,

--- In Hyacinthos@yahoogroups.com, "garciacapitan"

<garciacapitan@...> wrote:>

points

> Hello,

>

> For now, what I have seen is that the centers of these squares are

> perspective with ABC, having as perspector X486, the inner Vecten

> point.

>

> I don't know also if this is new.

>

>

> --- In Hyacinthos@yahoogroups.com, "pohoata_cosmin2000"

> <pohoata_cosmin2000@> wrote:

> >

> > Greetings,

> > I see that in Paul Yiu, Introduction to the geometry of the

> triangle,

> > it is mentioned that the perspector of triangle formed by the

> > of tangency of the Lucas circles and ABC is X371, the isogonal of

> the

> > Vecten point.

> >

> > Isn't it true that this point is the radical center of the circles

> > circumscribed the three squares inscribed in ABC ?

> >

> > Sorry if it's a dumb question, I didn't see it mentioned. In my

> sketch

> > the collinearity of the circumcenter of ABC, radical center of the

> > Lucas circles and X371 followed from the above remark, so that's

> why I

> > ask.

> >

> >

> > Thank you and best regards,

> > Cosmin Pohoata

> >

> - --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...>

wrote:

> No it not correct.

Dear Nikos, you are right, I misscalculated something. Thank you for

> I think you conclude this by a sketch.

> If we call these circes C1, C2, C3

> then the radical axis of C2, C3 seems to pass

> through X371 but if you calculate the distance

> of X371 from the above axis you'll see it is not zero.

> The barycentrics of the radical center are

> sinAsinAcosA /(sinA+cosA), . . , . .

> while the barycentrics of X371 are

> sinA(sinA+cosA), . . , . .

>

> Best regards

> Nikos Dergiades

pointing out the mistake.