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Lucas Circles

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  • xpolakis@otenet.gr
    In a triangle ABC we inscribe a square BaCaC aB a, whose the side BaCa lies on BC. If r_a is the circumradius of the triangle AB aC a, show that: Rh_1
    Message 1 of 2 , Apr 1, 2000
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      In a triangle ABC we inscribe a square BaCaC'aB'a, whose the side BaCa
      lies on BC. If r_a is the circumradius of the triangle AB'aC'a, show that:

      Rh_1 Rbc
      r_a = ------- = --------- [h_1 = a-altitude; R = circumradius of ABC]
      a + h_1 bc + 2aR

      If we consider the two other squares inscribed in ABC and the corresponding
      radii, then:

      Rh_2 Rca Rh_3 Rab
      r_b = ------- = ---------, r_c = ------- = ---------
      b + h_2 ca + 2bR c + h_3 ab + 2cR

      and 1/r_a + 1/r_b + 1/r_c = 3/R + 2(a^2 + b^2 + c^2)/abc.

      The three circumcircles are called Lucas Circles.

      (I. Panakis: Plane Trigonometry. Athens, 1973. Vol. II, p. 458, #4)

      Actually there are two such squares inscribed in the triangle corresponding
      to BC (ie whose one side lies on BC), since the vertices Ca, Ba can also lie
      on the extensions of AB, AC.

      These circles must have some interesting geometrical properties studied by
      Lucas. I don't believe that Lucas calculated just their radii and nothing else!
      (These three triangles are similar to reference triangle with similitude ratia:
      h_1/(a+h_1), h_2/(b+h_2), h_3/(c+h_3), and if I am not mistaken, this was
      already known to Hero of Alexandria).

      The next exercise in the book above (#5) is on de Longchamps Circle:

      A Triangle ABC is given, and we draw the circles (A,a), (B,b), (C,c),
      and the circle cutting orthogonally them. Prove that its radius is equal to:

      sqrt(2(8R^2 - (a^2+b^2+c^2))

      This circle is called de Longchamps Circle.

      G. de Longchamps studied the geometrical properties of this circle; not just
      its radius! (See the HM thread:

      http://forum.swarthmore.edu/epigone/historia/smildoibel/ )

      as also JHC did do for the Conway Circle.

      So, since the author does not give a reference, I am wondering which are
      their properties.
      (Probably, their radical center is a "notable" point, or the triangles
      bounded by their common tangents are in perspective with the reference
      triangle, or...)

    • Steve Sigur
      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
      Message 2 of 2 , 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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