# 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)).

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This question seems not suitable for MO (not research level, this only uses Pythagore theorem), please read the FAQ. There is a formula, that you can devise by assuming $A$ is at the origin, $B$ is on the $x$-axis. The coordinates of $C$ are then easy to compute in terms of $a,b,c$, than you get the coordinates of your point. The distance is then given by the usual Euclidean formula. – Benoît Kloeckner Aug 25 '10 at 13:40
better yet, try it on math.stackexchange.com – Suresh Venkat Aug 25 '10 at 14:12
Sorry! Was completely unaware of math.stackexchange.com! Suspected I was a bit out of my depth here. Can it be migrated over ? – timday Aug 25 '10 at 14:30
not AFAIK. but it looks like you got an answer, so it's all good :) – Suresh Venkat Aug 25 '10 at 17:46

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.

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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 $$d^2=(bw-cv)^2+vw((b+c)^2-a^2)$$

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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 $\vec{D}=u\vec{A}+v\vec{B}+w\vec{C}$

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I would have added this as a comment, as Benoît Kloeckner did above; however, I did not have commenting ability at that time. Thus I replied as an answer. – sleepless in beantown Aug 26 '10 at 7:13

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

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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 $d^2=-(a^2vw+b^2w(u-1)+c^2v(u-1))$

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