## 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

### Miscellaneous

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. 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?

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

• 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. 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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