Loading ...
Sorry, an error occurred while loading the content.

Excircles problem

Expand Messages
  • xpolakis@otenet.gr
    The locus of the points whose the distances from two given circles are equal, are two hyperbolae (F.G.-M., paragraph 2148, of 5e edition) [ distance of a
    Message 1 of 2 , Sep 6, 2000
    • 0 Attachment
      The locus of the points whose the "distances" from two given
      circles are equal, are two hyperbolae (F.G.-M., paragraph 2148, of 5e edition)
      ["distance" of a point P from a circle (K) := PT, where T is the intersection
      of the circle and the line PK, or T := the point of tangency of the circle (K)
      and a circle centered at P]

      Now, which are the points (coordinates?) whose the "distances" from the
      three excircles are equal?

      APH
    • Paul Yiu
      Dear Antreas, ... edition) ... (K) ... This is the Apollonius problem for the excircles, and there are eight of them. Four of these are well known: the
      Message 2 of 2 , Sep 7, 2000
      • 0 Attachment
        Dear Antreas,

        At 05:58 PM 9/6/00 +0300, you wrote:

        >The locus of the points whose the "distances" from two given
        >circles are equal, are two hyperbolae (F.G.-M., paragraph 2148, of 5e
        edition)
        >["distance" of a point P from a circle (K) := PT, where T is the intersection
        >of the circle and the line PK, or T := the point of tangency of the circle
        (K)
        >and a circle centered at P]
        >
        >Now, which are the points (coordinates?) whose the "distances" from the
        >three excircles are equal?

        This is the Apollonius problem for the excircles, and there are eight of them.
        Four of these are well known: the nine-point circle and the three sides.
        The other four can be obtained by inverting these four with respect to
        the circle orthogonal to the excircles. This latter one has center the
        Spieker point (= inferior of incenter), and radius sqrt(r^2+s^2)/2.

        From these, one obtains, for example, the radius of the circle tangent to
        each of the excircles internally. This has radius

        (r^2+s^2)/(2Rr).

        The center of this circle has homogeneous barycentric coordinates

        (a^2(a^3(b+c)^2+a^2(b+c)(b^2+c^2)-a(b^4+2b^3c+2bc^3+c^4)-(b+c)(b^4+c^4)
        : ...
        : ...).

        This point does not appear in [ETC]. Clark has given the perspector of
        the triangle of points of tangency with the excircles as X(181). The
        line joining X(181) to the incenter contains this center, as does the
        line joining the nine-point center to the Spieker center.

        The other three circles, too, have a lot of interesting properties.

        Best regards.
        Sincerely,
        Paul
      Your message has been successfully submitted and would be delivered to recipients shortly.