Re: [math-fun] Hero[n]'s formula

If we label the triangle in the usual way (a is the side opposite to angle/vertex A, etc.), and if we choose A as the origin, and if we let b,c be _vectors_ (not just lengths), then

incenter = (|c|b+|b|c)/(2s), [s = (|a|+|b|+|c|)/2] = |b||c|/(2s) * (b/|b| + c/|c|)

(Sorry that this isn't symmetrical.)

It's equal to (aA+bB+cC)/(a+b+c) where a,b,c are lengths and A,B,C are vertices. I expect there's a not-too-painful proof of Hero[n] that goes via barycentric coordinates... -- g