Barycentric Coordinates | ELEMENTARY GEOMETRY

Coordinates relative to a triangle

Fix a nondegenerate triangle $ABC$ , and write

a=BC,\qquad b=CA,\qquad c=AB.

Homogeneous barycentric coordinates are a nonzero triple $(x:y:z)$ considered up to multiplication by a common nonzero scalar. Thus $(1:2:3)$ and $(2:4:6)$ represent the same coordinates. When $\sigma=x+y+z\ne0$ , the corresponding finite point has position vector

\mathbf p=\frac{x\mathbf A+y\mathbf B+z\mathbf C}{\sigma},

where the vectors on the right are taken from any common origin. Dividing by $\sigma$ gives normalized coordinates whose sum is $1$ . This normalization is useful for calculations involving small displacements; homogeneous coordinates are useful when only ratios matter. A triple with sum zero does not represent a finite point in this formula and will not be used here.

The vertices are $A=(1:0:0)$ , $B=(0:1:0)$ , and $C=(0:0:1)$ . The midpoint of $BC$ is $(0:1:1)$ . More generally, a point $D=(0:y:z)$ inside the segment $BC$ satisfies $BD:DC=z:y$ . The reversal follows directly from

\mathbf D=\frac{y\mathbf B+z\mathbf C}{y+z}.

If $AD$ is the internal angle bisector, then $BD:DC=c:b$ by the angle bisector theorem, hence $D=(0:b:c)$ .

Coordinates need not be positive. For example, $(-1:1:1)$ represents the point with position vector $-\mathbf A+\mathbf B+\mathbf C$ , outside the triangle. The algebra below therefore applies beyond its interior.

Once normalized, the three coordinates are uniquely determined: the vectors $\mathbf B-\mathbf A$ and $\mathbf C-\mathbf A$ are independent, and $\mathbf p-\mathbf A=y(\mathbf B-\mathbf A)+z(\mathbf C-\mathbf A)$ . Thus each finite point has one normalized triple, while its homogeneous representatives form a family of scalar multiples. In particular, the coordinates are weights attached to the three vertices, not distances measured along three axes.

Lines and the circumcircle

A line has a homogeneous linear equation

\ell x+my+nz=0.

Substitution tests whether a point lies on it. Conversely, such an equation defines a Euclidean line when $\ell,m,n$ are not all equal: on the normalization plane $x+y+z=1$ , it is a nonconstant affine linear equation. In particular, $BC$ has equation $x=0$ .

Multiplying all three coefficients by the same nonzero constant leaves the line unchanged. To find the line through two known points, substitute their coordinates and solve the resulting two linear equations for $\ell,m,n$ , again only up to scale.

To derive the circumcircle equation, take its center $O$ as origin and its radius as $R$ . Then

|\mathbf A|^2=|\mathbf B|^2=|\mathbf C|^2=R^2.

Expanding $|\mathbf B-\mathbf C|^2=a^2$ gives $2\mathbf B\cdot\mathbf C=2R^2-a^2$ ; the other two pairs give the corresponding formulas with $b$ and $c$ . Therefore

\begin{aligned} |x\mathbf A+y\mathbf B+z\mathbf C|^2 &=R^2(x^2+y^2+z^2)\\ &\quad +(2R^2-c^2)xy\\ &\quad +(2R^2-a^2)yz\\ &\quad +(2R^2-b^2)zx\\ &=R^2\sigma^2-(a^2yz+b^2zx+c^2xy). \end{aligned}

After division by $\sigma^2$ ,

\begin{aligned} |\mathbf p|^2&=R^2-\frac{F(x,y,z)}{\sigma^2},\\ F(x,y,z)&=a^2yz+b^2zx+c^2xy. \end{aligned}

A finite point lies on the circumcircle exactly when $|\mathbf p|^2=R^2$ . Its equation is consequently $F(x,y,z)=0$ . The derivation uses no positivity assumption on the coordinates.

Tangents from the linear terms

Near $B$ , use normalized coordinates $(u,1-u-v,v)$ . Substitution gives the exact expansion

\begin{aligned} F(u,1-u-v,v) &=c^2u+a^2v-c^2u^2-a^2v^2\\ &\quad +(b^2-a^2-c^2)uv. \end{aligned}

The terms of first degree describe the tangent at $u=v=0$ . Its equation is $c^2u+a^2v=0$ , or, in homogeneous coordinates,

c^2x+a^2z=0.

At $C$ , substitute $(u,v,1-u-v)$ . The linear terms are $b^2u+a^2v$ , giving the tangent $b^2x+a^2y=0$ .

More generally, for a differentiable plane curve $G(U)=0$ at a regular point $q$ , where $\nabla G(q)\ne0$ , the tangent equation is

\nabla G(q)\cdot(U-q)=0.

The gradient is evaluated at the contact point and applied to a displacement. This is not a rule that an arbitrary partial derivative, left as a function of the variable point, may be set equal to zero.

For our homogeneous quadratic $F$ , write $X=(x,y,z)$ , with $q$ and $X$ initially normalized. The tangent can be written $\nabla F(q)\cdot X=0$ : the displacement form loses its constant term because $\nabla F(q)\cdot q=2F(q)=0$ . The resulting equation is homogeneous and thus applies to every representative of $X$ . This quadratic also satisfies the symmetry

\nabla F(q)\cdot X=\nabla F(X)\cdot q.

With $q=B=(0,1,0)$ , the right-hand side happens to be $F_y(X)=a^2z+c^2x$ . Thus setting $F_y(X)=0$ works here because of this symmetry and the particular coordinates of $B$ , not as a general differentiation rule for tangents.

Intersecting two tangents

Assume that the tangents at $B$ and $C$ meet at a finite point $P$ . Their equations imply

y=-\frac{b^2}{a^2}x,\qquad z=-\frac{c^2}{a^2}x.

Here $x$ cannot vanish, since that would force all three coordinates to vanish. Choosing $x=-a^2$ gives

P=(-a^2:b^2:c^2).

The finite-intersection assumption means $-a^2+b^2+c^2\ne0$ . These coordinates and $D=(0:b:c)$ are used in Tangents, Angle Bisectors, and a Common Point.

Tangents, Angle Bisectors, and a Common Point

Coordinates relative to a triangle

Lines and the circumcircle

Tangents from the linear terms

Intersecting two tangents

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