## related to the Lucas circles

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• 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
Message 1 of 5 , Aug 5 7:26 AM
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

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
Message 2 of 5 , Aug 5 10:51 PM
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 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
>
>
> Thank you and best regards,
> Cosmin Pohoata
>
• ... No it not correct. 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
Message 3 of 5 , Aug 6 1:08 PM
> 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 ?

No it not correct.
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

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• Dear, Cosmin Pohoata, Nikos, and all, I was wrong, the perspector is X485, the outer Vecten point Best regards, ... points
Message 4 of 5 , Aug 7 3:13 AM
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:
>
> 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
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
> >
> >
> > Thank you and best regards,
> > Cosmin Pohoata
> >
>
• ... Dear Nikos, you are right, I misscalculated something. Thank you for pointing out the mistake.
Message 5 of 5 , Aug 7 7:09 AM
wrote:

> No it not correct.
> 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