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726The Orthogonal circles and some others

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  • Steve Sigur
    Apr 2, 2000
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      Hello Hyacynthians,

      Here I continue to attempt to understand via inversive techniques.

      In Conway's Global View of inversive triangle geometry it is often
      easy to see the existence and properties of circles related to the
      triangle. In this global viewpoint we choose any point P on a sphere,
      and three equally spaced points on the equator and perform a
      stereographic projection, which is inversive and hence angle
      preserving, onto the tangent plane on the opposite side of the sphere
      from P. By appropriate choice of P we can obtain any triangle, so
      that many properties of the triangle can be worked out by observing a
      highly symmetric situation on a sphere.

      The equator projects to the circumcircle of ABC and the meridian
      lines through A, B, and C, which intersect at the north and south
      poles, become the Appolonian circles, intersecting in the isodynamic

      Aside: The pedal triangle of a point on a Appolonian circle is isosceles.

      We see immediately that the Appolonian circles are perpendicular to
      the circumcircle. Better yet we can immediately see two more triples
      of circles and some of their properties.

      On the sphere consider the six points formed by ABC and their
      antipodal points A'B'C'. These last points are the second
      intersections of the Appolonian meridians with the equator, which
      project into the intersection of the Appolonian circles with the

      Now consider on the sphere a circle centered at A' going through B
      and C. This circle is obviously tangent to the meridians through B
      and C. Similar circles can be found at B' and C'. When projected onto
      the plane, there are three circles that are each orthogonal to the
      circumcircle, go through two vertices of ABC, and are tangent to the
      Appolonian circles. Conway calls these the orthogonal circles. More
      about them below.

      Aside: The pedal triangles of points on orthogonal circles are right.

      Now consider on the sphere a circle centered at A and going through
      B' and C'. This circle is twice orthogonal to the meridian through A
      and tangent to the meridians at B' and C'. Projected into the plane
      this means that there is a circle orthogonal to the a-Appolonian
      circle and tangent to the other two. This circle is not centered at A
      in the plane, because the center of a circle is not preserved under
      inversion. There are two other such circles. I do not know the names
      of these circles, if indeed they have names.

      The orthogonal circles

      We know that the exsymmedian lines are the tangents to the
      circumcircle at the vertices. Hence the vertices of this triangle
      A^K, B^K, C^K are the centers of circles orthogonal to the
      circumcircle at two vertices and so are centers of the orthogonal
      circles. By symmetry AB^K = CB^K, and since the exsymmedians are
      antiparallel the the opposites sides, we conclude that the base angle
      of the isosceles triangle on AC is angle B.

      Going back to the sphere we find a relevant concurrence in the plane.
      Take the meridian great circle through AA' and tilt it so that it
      goes through A, A', and P. Note that this goes through the antipodal
      point to P. This is also true for the circles PBB' and PCC'. Since
      any circle through P projects to a line in the plane, we find that,
      in the plane, the the lines AA', BB', and CC' concur. By other
      orguments we can determine that the point of concurrence is K, the
      symmedian point. Hence as shown in the diagram, B', the intersection
      of the b-Appolonian circle with the circumcircle, is on the symmedian
      through B.

      B^K, an exsymmedian point
      center of
      /\ circle
      / |\
      / | \
      / | \
      / | \
      / | \
      / | \
      / | \
      / B' \
      / | \
      / <B | <B \
      A /-----------|----------\ C
      | symmedian

      This isoseceles triangle lets us compute the radius CB^K of the
      b-orthogonal circle. It equals b/2 . sec B = abc/ (2SB) where SB is
      Conway notation for (cc+aa-bb)/2. Calling this radius ra we have a
      formula for the three radii of the orthogonal circles

      1/ra + 1/rb + 1/rc = 2/abc (SA+SB+SC) = 2S/abc where S = (aa+bb+cc)/2.

      The new circles

      I am calling them "new" because they are new to me.

      The centers of the orthogonal circles were found using the tangent
      lines to the circumcircle at A, B, C. The centers of these new
      circles are found using the tangents at A', B', C', the three second
      intersections of the Appolonian circles with the circumcircle.

      These centers are again on the symmedian lines.

      This means that ABC, A'B'C', the centers of the orthogonal circles,
      and the centers of the new circles are all perspective at K.

      For what it is worth some coordinates are

      B' = (aa: -bb/2 : cc)
      center of orthogonal circle = B^K = (aa : - bb : cc)
      center of new circle = (aa : -5bb : cc)

      Enough for now.

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