Three circles

This is one of my favorite problems in Euclidean Geometry. It goes like this:

Consider three circles c_1, c_2, c_3 that intersect in a common point O. (see figure below) Circles c_1 and c_2 intersect also at point A. Circles c_2 and c_3 intersect also at point B. Circles c_1 and c_3 intersect also at a point C. Consider any point P \in c_1, and trace a line through P and A. This line intersects circle c_2 at a second point P'. Trace a line through P' and B. This line intersects circle c_3 at a second point P''. We want to prove that the points P, P'' and C are collinear


See the post Using tikz as an IGSE for an explanation on how to produce interactive geometric diagrams ala Cabri or Geometer's Sketchpad

  1. Daniel Maxin
    December 12, 2010 at 9:00 am

    OCP is the suplementary angle of OAP hence OCP=OAP’
    OCP” is the suplementary agle of OBP” hence OCP”=OBP’
    But OAP’+OBP’=180 degrees hence OCP+OCP”=180

    • December 13, 2010 at 5:40 pm

      Daniel, thanks for your proof. I just have one question: Why are those angles supplementary?

      \begin{array}{c}\angle OCP + \angle OAP = \pi?\\ \angle OCP'' + \angle OBP'' = \pi? \end{array}

      • Daniel Maxin
        December 13, 2010 at 8:39 pm

        This is known to happen in a quadrilateral inscribed in a circle, i.e. the sum of the opposite angles is 180. This is because an angle inscribed in a circle has a measure equal to half the radian measure of the subtended arc. When we add the opposite angles mentioned above we obtain the full circle, half of which is 180 degrees.

        • December 13, 2010 at 8:52 pm

          A beautiful proof, Daniel. Thanks again.

  2. Alison
    December 12, 2010 at 9:20 am

    If instead you find the point O, then you’re doing exactly what’s done to find Earthquakes. In fact it’s one of my favourite outreach exercises.

    • December 12, 2010 at 9:32 am

      Neat! Would you care to contribute with a post explaining how?

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