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Re: [EMHL] Three collinear centers

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  • Randy Hutson
    Some further results: Call the 3 circles Oa, Ob, Oc.  Let Sa be the insimilicenter of Ob and Oc, and define Sb, Sc cyclically. The lines ASa, BSb, CSc concur
    Message 1 of 37 , Jan 19, 2013
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      Some further results:

      Call the 3 circles Oa, Ob, Oc.  Let Sa be the insimilicenter of Ob and Oc, and define Sb, Sc cyclically.
      The lines ASa, BSb, CSc concur (non-ETC, search: 2.007236950111909).  Coordinates?

      Let A'B'C' be the reflection of the intouch triangle in X(11) (mentioned below).
      Let A"B"C" be the triangle formed by the radical axes of Oa and the A-excircle, Ob and the B-excircle, Oc and the C-excircle.
      Let A2B2C2 be the '2nd extouch triangle' (the triangle formed by lines through the pairs of external extouch points of each excircle).
      A'B'C', A"B"C", and A2B2C2 are homothetic, with common center of homothety being X(125) of A2B2C2 (non-ETC, search: 4.652666136903451).  Coordinates?

      It may be worth exploring the relationship between Oa, Ob, Oc and the mixtilinear in/excircles.

      Best regards,
      Randy Hutson





      >________________________________
      > From: Randy Hutson <rhutson2@...>
      >To: "Hyacinthos@yahoogroups.com" <Hyacinthos@yahoogroups.com>
      >Sent: Saturday, January 19, 2013 11:16 AM
      >Subject: Re: [EMHL] Three collinear centers
      >
      >

      >Some interesting results from this configuration:
      >
      >The line of centers is line X(484)X(516).
      >The radical axis of the 3 circles is line X(11)X(514).
      >
      >The lines through the contact points of the circles with the sidelines forms a triangle that is the reflection of the intouch triangle in X(11).
      >
      >I wonder if there is anything interesting about the centroid of the 3 centers.
      >
      >Randy Hutson
      >
      >>________________________________
      >> From: Barry Wolk wolkbarry@...>
      >>To: "Hyacinthos@yahoogroups.com" Hyacinthos@yahoogroups.com>
      >>Sent: Friday, January 18, 2013 2:50 PM
      >>Subject: Re: [EMHL] Three collinear centers
      >>
      >>
      >> 
      >>BW
      >>> 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.
      >>  
      >>
      >>These 3 circles meet at the Feuerbach point and at ( (b-c)^4 (b+c-a)^3, ... , ...)
      >> > Does this work if the Feuerbach point is replaced with some other point on the
      >>> incircle?
      >>
      >>I've solved this. If the Feuerbach point is changed to one of the 3 points where the incircle meets the sidelines, these are degenerate cases where there aren't 4 such circles. And the Feuerbach point is the only other point on the incircle where this collinearity holds.
      >>--
      >>Barry Wolk
      >>
      >>
      >>
      >>
      >
      >[Non-text portions of this message have been removed]
      >
      >
      >
      >
      >

      [Non-text portions of this message have been removed]
    • 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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