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# Mapping Triangles to Points

Two messages posted on the Math Forum Web site (geometry-precollege news group) long ago, followed by my response:

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> Given equilateral triangle ABC and point P (in the plane containing
> ABC), prove that PA, PB, and PC satisfy the triangle inequalities.
> That is: PA+PB >= PC PB+PC >= PA PC+PA >= PB.
> Determine the locus of P for which an equality can occur.
>
> Happy holidays to everyone!
>
> From,
> Ken

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> Let us map the point P to the (abstract) triangle whose edge-lengths
> are PA, PB, PC.
>
> Then this establishes a 1-1 correspondence between the points of
> the plane and the possible shapes of triangles.
>
> John Conway

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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:

1. Construct P and P' given the corresponding triangle. (This construction is the key to the proof that the correspondence is 1-1.)

Along the way, we found some interesting intermediary problems:

1. What is the locus of all the Ps that correspond to right triangles? (This is easier than the remaining questions.)
2. What is the locus of all Ps that correspond to triangles that include a 30 degree angle? Same question for 60, 120, 150.
3. What is the relationship between angles APB, BPC, and CPA and the angles in the corresponding triangle? (This was rather unexpected. The proof is a rather traditional Euclidean proof, and is the key to understanding this problem.)

Finally, a couple of extensions:

1. What is the relationship between P and P'? (I have a conjecture, but no proof.)
2. (This question was suggested by Jean-Marie Laborde.) What happens when the original triangle is not equilateral: for a given triangle ABC, for what points P do PA, PB, and PC satisfy the triangle inequality? for what points is there equality?
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