Formula for 2d coordinates of a Steiner Point of a triangle I've searched around a bunch, and I still haven't managed to find any clear-cut way to find the x and y coordinates of the Steiner point defined by the coordinates of exactly 3 points. Does anyone know of such a formula?
 A: Since the trilinear coordinates of the Steiner point are:
$$ \left[\frac{bc}{b^2-c^2};\frac{ca}{c^2-a^2};\frac{ab}{a^2-b^2}\right] $$
its barycentric coordinates are: $$\left[\frac{1}{b^2-c^2};\frac{1}{c^2-a^2};\frac{1}{a^2-b^2}\right]$$
and the Steiner point can be found as the following linear combination of the vertices $A,B,C$:
$$ S = \frac{1}{\mu}\cdot\left(\frac{A}{b^2-c^2}+\frac{B}{c^2-a^2}+\frac{C}{a^2-b^2}\right)$$
where $\mu = \frac{1}{b^2-c^2}+\frac{1}{c^2-a^2}+\frac{1}{a^2-b^2}$.
This approach works really fast for computing the coordinates of other ($O,H,N,K,\ldots$) centres, too, even if the vertices $A,B,C$ of the reference triangle lie in $\mathbb{R}^n$ with $n>2$.
A: I know that this is quite an old question, but this is just to extend Jack D'Aurizio's (perfectly correct) answer to more explicitly fit the question, which effectively was, given the coordinates of the vertices of a triangle, what are the coordinates of the associated Steiner point?
Once we have the Barycentric coordinates of any triangle centre (see the Encyclopedia of Triangle Centers for a convenient summary of triangle centres - the Steiner point is X(99) in the tome), finding the Cartesian coordinates of the centre becomes a straightforward weighted means problem; that is:
For an arbitrary triangle centre:
Given triangle $ABC$ with n-space vertex coordinates $A\ (a_1, a_2, \dots, a_n)$, $B\ (b_1, b_2, \dots, b_n)$ and $C\ (c_1, c_2, \dots, c_n)$, the triangle centre with Barycentric coordinates $(\mu_1; \mu_2; \mu_3)$ with $\sum \mu = \mu_1+\mu_2+\mu_3$ will have the coordinates $P\ (p_1, p_2, \dots, p_n)$ such that:
$$p_k = \frac{\mu_1 a_k+\mu_2 b_k + \mu_3 c_k}{\sum\mu}$$ 
All that is left, then, is to get an expression for the Barycentric coordinates. 
For the Steiner point specifically:
In the case of a Steiner point, as pointed out by Jack D'Aurizio, the Barycentric coordinates are $\left[\frac{1}{b^2-c^2};\frac{1}{c^2-a^2};\frac{1}{a^2-b^2}\right]$, where $a$, $b$ and $c$ are the lengths of the sides opposite the respectively named vertices, which can, of course, also be expressed in terms of the vertex coordinates given in the original problem.
