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  • jhen
    Good Day Everyone, I just would like to ask if you know some of the properties of Morley s Triangle? I really have limited resources. =) Thanks, and have a
    Message 1 of 2 , Jan 31, 2011
      Good Day Everyone,

      I just would like to ask if you know some of the properties of Morley's Triangle? I really have limited resources. =)

      Thanks, and have a great day!
    • Den Roussel
      ... Dear Jenny and Hyacinthos; There are many sites on the web with much information pertaining to Morley s triangle. Hopefully you have sufficient internet
      Message 2 of 2 , Feb 10, 2011
        --- In Hyacinthos@yahoogroups.com, "jhen" <pretty_jenny88@...> wrote:
        >
        > Good Day Everyone,
        >
        > I just would like to ask if you know some of the properties of Morley's Triangle? I really have limited resources. =)
        >
        > Thanks, and have a great day!
        >

        Dear Jenny and Hyacinthos;

        There are many sites on the web with much information pertaining to Morley's triangle. Hopefully you have sufficient internet access to explore these. A couple of properties little known, (if not new), are included in this note.

        I have put a Geometer's Sketchpad document in the file section entitled
        'Isogonal Configuration'

        Consider a triangle A1A2A3 and pairs of isogonal lines p1,q1; p2,q2; p3,q3 emanating from the vertices. These 6 lines intersect in 15 points.

        Now consider a contiguous hexagonal non-coincident path from A1 with the form
        [vertex - intersection - vertex - intersection - vertex - intersection].
        Each path is a hexagon and traverses a segment of each isogonal line.

        e.g. from the first vertex A1, proceed along an isogonal line, say p1, to the first intersection, say p1p2. The second vertex A2 is entered along the p2 isogonal line so must leave by the q2 isogonal line to intersect at say, q2p3. The third vertex A3 is entered on the p3 line so must exit on the q3 line to the intersection of the q1 line at q3q1 and from there back to the first vertex A1 completing the hexagonal path.

        There are 8 distinct paths and the 3 intersection points of each determine an 'isogonal triangle' which is perspective with the reference triangle. A Conway message of Nov. 28, 1997 spoke about these perspectors.

        The 8 perspectors of the 8 isogonal triangles lie in pairs on the 4 lines thru each vertex V which pass thru the 4 intersection points which are not on either of the isogonal lines of V. In other words, each of the 4 lines thru a vertex contains 2 perspectors and an isogonal intersection.

        Some metric properties with respect to the hexagonal path.
        Let x1 represent the length of the segment from A1 along the isogonal line p1 to any of the 4 intersection points on p1, i.e., p1p2, p1q2, p1p3, p1q3.
        Let y1 represent the length of the segment from A1 along the isogonal line q1 to any of the 4 intersection points on q1, i.e., q1p2, q1q2, q1p3, q1q3.
        Likewise for the other 4 isogonal lines.

        For the 8 paths, the product of 3 non adjacent segments of the hexagon is equal to the product of the other 3 segments.

        x1x2x3 = y1y2y3

        As an example, by definition the Morley triangle is an isogonal triangle. It's vertex path is A1 > M3 > A2 > M1 > A3 > M2 or writing the path as isogonal intersections

        q1p1 > p1q2 > q2p2 > p2q3 > q3p3 > p3q1
        A1 M3 A2 M1 A3 M2

        or writing the path as segments
        x1 > y2 > x2 > y3 > x3 > y1

        A curious equality occurs when the isogonal lines are trisectors. In that case, 2 of the 8 hexagons corresponding to the so called '1st Morley triangle' and the 'adjunct Morley triangle' have the property that the sum of the non adjacent segments is equal to the sum of the other segments.

        x1+x2+x3 = y1+y2+y3

        A little diversion:
        With compass and straight edge, construct a triangle whose Morley triangle has a side parallel to a given line.

        Fondly, Den Roussel
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