Distance of a barycentric coordinate from a triangle vertex I have a triangle $ABC$ with side lengths $a,b,c$ (edges $BC, CA, AB$ respectively).
I have a point $p$ with barycentric coordinates $u:v:w$.
These are normalised: $u+v+w=1$.
$1:0:0$ corresponds to point $A$, $0:1:0$ is $B$ etc.
Is there a simple expression for the distance $d$ of the point $p$ from $A$ ?
(My initial naive guess based on $d(1:0:0)=0, d(0:1:0)=b, d(0:0:1)=c$ was that $d$ was linear $d(u,v,w)=v*b+w*c$ but this is clearly wrong as in the case of an equilateral triangle $a=b=c=1$ it returns $d=2/3$ for the centroid ($u:v:w = 1/3:1/3:1/3$), when the correct answer should be $1/\sqrt 3$ (the radius of the circumscribed circle)).
 A: Robin gives the correct formula, but not in the simplest form. Since $$a^2=b^2+c^2-2bc\cos\alpha,$$ the desired distance satisfies $$d^2=(bw)^2+(cv)^2+vw(b^2+c^2-a^2).$$ A little manipulation also yields $$d^2=(bw-cv)^2+vw((b+c)^2-a^2)$$
A: There isn't really a simple formula, but you can use vector methods.
Let $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ be position vectors
of the vertices. A point $P$ with normalized barycentric coordinates
$(u,v,w)$ has position vector $\mathbf{p}=u\mathbf{a}+v\mathbf{b}+w\mathbf{c}$.
Therefore $\mathbf{p}-\mathbf{a}=v(\mathbf{b}-\mathbf{a})+w(\mathbf{c}-\mathbf{a})$.
This leads to
$$|AP|^2=v^2|AB|^2+w^2|AC|^2+2vw|AB||AC|\cos\alpha$$
where $\alpha$ is the angle at $A$. Of course one can express
$|AB||AC|\cos\alpha$ in terms of the three side-lengths of the triangle
using the cosine rule.
This shows that neither $|AP|$ nor $|AP|^2$ is a linear function
of the barycentric coordinates (actually this is geometrically evident too).
But there is a simpler formula for the distance of $P$ to a given side
of the triangle.
A: There actually is a delightful formula -- see page 11 at 
http://www.mit.edu/~evanchen/handouts/bary/bary-full.pdf
Your displacement vector is $(u-1,v,w)$, giving $d^2=-(a^2vw+b^2w(u-1)+c^2v(u-1))$
A: So there is a paper A Hybrid GPU Rendering Pipeline for Alias-Free Hard Shadows
That claims to calculate the distance to a triangle $d(\omega,T)$ efficiently, they 

resort to some tricks based on the concepts of
  barycentric coordinates

they describe the squared distance between a point $w$ and some vertex $v_i$ of the triangle $T$ as
$d(\omega,v_i)^2=\left\Vert \omega-v_{i}\right\Vert=\lambda_{i-1}^{2}\left\Vert e_{i-1}\right\Vert ^{2}+\lambda_{i+1}^{2}\left\Vert e_{i+1}\right\Vert ^{2}-2\lambda_{i-1}\lambda_{i+1}\left(e_{i-1}\cdot e_{i}\right)$
There is a derivation in the paper.
I think using the notation you described it'd look something like this
$d(p,A)^2=\left\Vert p-A\right\Vert=w^2b^2+v^2a^2-2wv(CA\cdot AB)$
but I'd check the math just to be sure
A: Following from Benoît Kloeckner's comment above,
Place the points at $A=(0,0)$ at the origin, $B=(c,0)$ on the  x-axis with the distance $|AB|=c$, and $C=(x,y)$,
where we now want to satisfy $|AC|=b$ and $|BC|=a$.
Simple application of the Pythagorean theorem leads to
$x^2+y^2 = b^2$
 and $(x-c)^2+y^2 = a^2$
as the two constraints to be applied.
Expanding and subtracting the two equations:
$x^2-2cx+c^2+y^2=a^2$ and 
$x^2 + y^2 =b^2$
$2cx-c^2=b^2-a^2$
$2cx = (b^2-a^2+c^2)$
$x = \frac{b^2-a^2+c^2}{2c}$
Now you can define $y$ in terms of $x$.
Simply scale the points $\vec{A}=(0,0), \vec{B}=(0,c)$, and $\vec{C}=(x,y)$ by their respective $(u,v,w)$ barycentric coordinates to get $D=(x_D,y_D)$ as a function of $a,b,c,u,v,w$, apply the Pythagorean theorem again to get $d = |\vec{D}|$ = the square root of $(x_d)^2 + (y_d)^2$.  This last step shouldn't need to be spelled out for you, but $\vec{D}=u\vec{A}+v\vec{B}+w\vec{C}$
