This is the graph of the formula for an area of 25 solved for one of the variables. I noticed for most of the solution area the graph is pretty flat. So I approximated it with this:

I did that by putting a plane through the points:

These are valid points on the plane I'm looking for no matter the area A. That took a while to figure out.

Finding the plane that goes through those 3 points and simplifying yields:

It is linear in x,y,z for a constant A. That should make a few things more easily possible than with the original Heron's formula without losing much accuracy in the domain of possible triangles.

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