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Mapping Triangles to Points  
Two messages posted on the Math Forum Web site (geometryprecollege news group) long ago, followed by my response:  > Given equilateral triangle ABC and point P (in the plane containing  > Let us map the point P to the (abstract) triangle whose edgelengths  Neil Picciotto and I had a great time thinking about this, with the (essential!) help of Cabri. We ended up with a fair understanding of the geometry underlying the abstract correspondence suggested by John Conway. As it turns out, if you ignore symmetry, there are actually two points P and P' that correspond to each triangle shape. (With symmetry, there are 12 points.) Getting the triangle from P is straightforward. We found a way to:
Along the way, we found some interesting intermediary problems:
Finally, a couple of extensions:


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