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Re: [EMHL] Three collinear centers

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  • Barry Wolk
    A possible generalization led to something curious. Take all circles that are tangent to two of the sidelines BC, CA, AB.  Four of these circles pass through
    Message 1 of 37 , Jan 17, 2013
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      A possible generalization led to something curious. Take all circles that are tangent to two of the sidelines BC, CA, AB.  Four of these circles pass through the Feuerbach point, the incircle and 3 others. The centers of those 3 others are collinear.
       
      Does this work if the Feuerbach point is replaced with some other point on the incircle?
      --
      Barry Wolk
    • Antreas
      It is Another Seven Circles Theorem ! The Seven Circles Theorem states: Let a,b ,c, a ,b,c be a closed chain of six circles, all toouching a base circle
      Message 37 of 37 , Mar 21, 2013
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        It is Another "Seven Circles Theorem" !

        The "Seven Circles Theorem" states:

        Let a,b',c, a',b,c' be a closed chain of six circles,
        all toouching a base circle w[omega], and suppose that
        their points of contact with w are six distinct points
        A,B',C,A',B,C' respectively. Then, subject of a certain
        extra condition to be discussed below, AA',BB',CC' are concurrent.
        (C J A Evelyn, B G Money - Coutts, J A Tyrrell:
        The Seven Circles Theorem and other theorems. London 1974, p. 31)

        See also
        http://en.wikipedia.org/wiki/Seven_circles_theorem
        http://mathworld.wolfram.com/SevenCirclesTheorem.html

        APH


        --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
        >
        > Dear Antreas
        >
        > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
        > >
        > > Dear Alex
        > >
        > > Angel has tested the loci problems and it seems that
        > > for all P's the lines are concurrent for the cevian case,
        > > and for the pedal case as well except for the points P on the
        > > bisectors and on the circumcircle.
        > >
        > > Hmmmmmmm...... Maybe it is true in general ie for any
        > > circle intersecting the sidelines of the triangle!!
        > >
        >
        >
        > A check with GeoGebra:
        >
        >
        >
        > Let ABC be a triangle and P a point.
        >
        > Let (Q) be the CEVIAN circle of P and (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the points Tbc, Tba and Tca, Tcb.
        >
        > For all P (except for the points P on the bisectors) the lines TabTac, TbcTba, TcaTcb are concurrent.
        >
        > http://amontes.webs.ull.es/geogebra/Hyacinthos21421.html
        >
        > ----------------------
        >
        > Let (Q) be the PEDAL circle of P and (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the points Tbc, Tba and Tca, Tcb.
        >
        > For all P (except for the points P on the
        > bisectors and on the circumcircle) the lines TabTac, TbcTba, TcaTcb are concurrent.
        >
        > http://amontes.webs.ull.es/geogebra/Hyacinthos21421Pedal.html
        >
        > ------------------------------------------------
        >
        > Let (Q) be the circle Intersecting the sidelines of the triangle in D, E and F. (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the
        > points Tbc, Tba and Tca, Tcb.
        >
        > The lines TabTac, TbcTba, TcaTcb are concurrent.
        >
        > http://amontes.webs.ull.es/geogebra/Hyacinthos21428.html
        >
        > Best regards,
        > Angel M.
        >
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