barycentric coordinate system

Let $A\_0,\; \backslash ldots,\; A\_n$ be {{math|n + 1}} points in a Euclidean space, a flat or an affine space $\backslash mathbf\; A$ of dimension {{mvar|n}} that are affinely independent; this means that there is no affine subspace of dimension {{math|n − 1}} that contains all the points,Reventós Tarrida, Agustí. "Affine Maps, Euclidean Motions and Quadrics". Springer, 2011, {{ISBN|978-0-85729-709-9}}, page 11 or, equivalently that the points define a simplex. Given any point $P\backslash in\; \backslash mathbf\; A,$ there are scalars $a\_0,\; \backslash ldots,\; a\_n$ that are not all zero, such that

$$(\; a\_0\; +\; \backslash cdots\; +\; a\_n\; )\; \backslash overset\{\}\backslash overrightarrow\{OP\}\; =\; a\_0\; \backslash overset\{\}\backslash overrightarrow\; \{OA\_0\}\; +\; \backslash cdots\; +\; a\_n\; \backslash overset\{\}\backslash overrightarrow\; \{OA\_n\},$$

for any point {{mvar|O}}. (As usual, the notation $\backslash overset\{\}\backslash overrightarrow\; \{AB\}$ represents the translation vector or free vector that maps the point {{mvar|A}} to the point {{mvar|B}}.)

The elements of a {{math|(n + 1)}} tuple $(a\_0:\; \backslash dotsc:\; a\_n)$ that satisfies this equation are called barycentric coordinates of {{mvar|P}} with respect to $A\_0,\; \backslash ldots,\; A\_n.$ The use of colons in the notation of the tuple means that barycentric coordinates are a sort of homogeneous coordinates, that is, the point is not changed if all coordinates are multiplied by the same nonzero constant. Moreover, the barycentric coordinates are also not changed if the auxiliary point {{mvar|O}}, the origin, is changed.

The barycentric coordinates of a point are unique up to a scaling. That is, two tuples $(a\_0:\; \backslash dotsc:\; a\_n)$ and $(b\_0:\; \backslash dotsc:\; b\_n)$ are barycentric coordinates of the same point if and only if there is a nonzero scalar $\backslash lambda$ such that $b\_i=\backslash lambda\; a\_i$ for every {{mvar|i}}.

In some contexts, it is useful to constrain the barycentric coordinates of a point so that they are unique. This is usually achieved by imposing the condition

$$\backslash sum\; a\_i\; =\; 1,$$

or equivalently by dividing every $a\_i$ by the sum of all $a\_i.$ These specific barycentric coordinates are called normalized or absolute barycentric coordinates.Deaux, Roland. "Introduction to The Geometry of Complex Numbers". Dover Publications, Inc., Mineola, 2008, {{ISBN|978-0-486-46629-3}}, page 61 Sometimes, they are also called affine coordinates, although this term refers commonly to a slightly different concept.

Sometimes, it is the normalized barycentric coordinates that are called barycentric coordinates. In this case the above defined coordinates are called homogeneous barycentric coordinates.

With above notation, the homogeneous barycentric coordinates of {{mvar|A{{sub|i}}}} are all zero, except the one of index {{mvar|i}}. When working over the real numbers (the above definition is also used for affine spaces over an arbitrary field), the points whose all normalized barycentric coordinates are nonnegative form the convex hull of $\backslash \{A\_0,\; \backslash ldots,\; A\_n\backslash \},$ which is the simplex that has these points as its vertices.

With above notation, a tuple $(a\_1,\; \backslash ldots,\; a\_n)$ such that

$$\backslash sum\_\{i=0\}^n\; a\_i=0$$

does not define any point, but the vector

$$a\_0\; \backslash overset\{\}\backslash overrightarrow\; \{OA\_0\}\; +\; \backslash cdots\; +\; a\_n\; \backslash overset\{\}\backslash overrightarrow\; \{OA\_n\}$$

is independent from the origin {{mvar|O}}. As the direction of this vector is not changed if all $a\_i$ are multiplied by the same scalar, the homogeneous tuple $(a\_0:\; \backslash dotsc:\; a\_n)$ defines a direction of lines, that is a point at infinity. See below for more details.

Barycentric coordinates are strongly related to Cartesian coordinates and, more generally, affine coordinates. For a space of dimension {{mvar|n}}, these coordinate systems are defined relative to a point {{mvar|O}}, the origin, whose coordinates are zero, and {{mvar|n}} points $A\_1,\; \backslash ldots,\; A\_n,$ whose coordinates are zero except that of index {{mvar|i}} that equals one.

A point has coordinates

$$(x\_1,\; \backslash ldots,\; x\_n)$$

for such a coordinate system if and only if its normalized barycentric coordinates are

$$(1-x\_1-\backslash cdots\; -\; x\_n,x\_1,\; \backslash ldots,\; x\_n)$$

relatively to the points $O,\; A\_1,\; \backslash ldots,\; A\_n.$

The main advantage of barycentric coordinate systems is to be symmetric with respect to the {{math|n + 1}} defining points. They are therefore often useful for studying properties that are symmetric with respect to {{math|n + 1}} points. On the other hand, distances and angles are difficult to express in general barycentric coordinate systems, and when they are involved, it is generally simpler to use a Cartesian coordinate system.

Homogeneous barycentric coordinates are also strongly related with some projective coordinates. However this relationship is more subtle than in the case of affine coordinates, and, for being clearly understood, requires a coordinate-free definition of the projective completion of an affine space, and a definition of a projective frame.

The projective completion of an affine space of dimension {{mvar|n}} is a projective space of the same dimension that contains the affine space as the complement of a hyperplane. The projective completion is unique up to an isomorphism. The hyperplane is called the hyperplane at infinity, and its points are the points at infinity of the affine space.{{Citation | last1=Berger | first1=Marcel | author1-link=Marcel Berger | title=Geometry I | publisher=Springer | location=Berlin | isbn= 3-540-11658-3 | year=1987}}

Given a projective space of dimension {{mvar|n}}, a projective frame is an ordered set of {{math|n + 2}} points that are not contained in the same hyperplane. A projective frame defines a projective coordinate system such that the coordinates of the {{math|(n + 2)}}th point of the frame are all equal, and, otherwise, all coordinates of the {{mvar|i}}th point are zero, except the {{mvar|i}}th one.

When constructing the projective completion from an affine coordinate system, one commonly defines it with respect to a projective frame consisting of the intersections with the hyperplane at infinity of the coordinate axes, the origin of the affine space, and the point that has all its affine coordinates equal to one. This implies that the points at infinity have their last coordinate equal to zero, and that the projective coordinates of a point of the affine space are obtained by completing its affine coordinates by one as {{math|(n + 1)}}th coordinate.

When one has {{math|n + 1}} points in an affine space that define a barycentric coordinate system, this is another projective frame of the projective completion that is convenient to choose. This frame consists of these points and their centroid, that is the point that has all its barycentric coordinates equal. In this case, the homogeneous barycentric coordinates of a point in the affine space are the same as the projective coordinates of this point. A point is at infinity if and only if the sum of its coordinates is zero. This point is in the direction of the vector defined at the end of {{slink||Definition}}.

{{see also|Ternary plot|Triangle center}}

In the context of a triangle, barycentric coordinates are also known as area coordinates or areal coordinates, because the coordinates of P with respect to triangle ABC are equivalent to the (signed) ratios of the areas of PBC, PCA and PAB to the area of the reference triangle ABC. Areal and trilinear coordinates are used for similar purposes in geometry.

Barycentric or areal coordinates are extremely useful in engineering applications involving triangular subdomains. These make analytic integrals often easier to evaluate, and Gaussian quadrature tables are often presented in terms of area coordinates.

Consider a triangle $ABC$ with vertices $A=(a\_1,a\_2)$, $B=(b\_1,b\_2)$, $C=(c\_1,c\_2)$ in the x,y-plane, $\backslash mathbb\{R\}^2$. One may regard points in $\backslash mathbb\{R\}^2$ as vectors, so it makes sense to add or subtract them and multiply them by scalars.

Each triangle $ABC$ has a signed area or sarea, which is plus or minus its area:

$\backslash operatorname\{sarea\}(ABC)\; =\; \backslash pm\; \backslash operatorname\{area\}(ABC).$

The sign is plus if the path from $A$ to $B$ to $C$ then back to $A$ goes around the triangle in a counterclockwise direction. The sign is minus if the path goes around in a clockwise direction.

Let $P$ be a point in the plane, and let $(\backslash lambda\_1,\backslash lambda\_2,\backslash lambda\_3)$ be its normalized barycentric coordinates with respect to the triangle $ABC$, so

$P\; =\; \backslash lambda\_1\; A\; +\; \backslash lambda\_2\; B\; +\; \backslash lambda\_3\; C$

and

$1\; =\; \backslash lambda\_1\; +\; \backslash lambda\_2\; +\; \backslash lambda\_3.$

Normalized barycentric coordinates $(\backslash lambda\_1,\backslash lambda\_2,\backslash lambda\_3)$ are also called areal coordinates because they represent ratios of signed areas of triangles:

\lambda_2 &= \operatorname{sarea}(APC)/\operatorname{sarea}(ABC)\\

\lambda_3 &= \operatorname{sarea}(ABP)/\operatorname{sarea}(ABC).\end{align}

One may prove these ratio formulas based on the facts that a triangle is half of a parallelogram, and the area of a parallelogram is easy to compute using a determinant.

Specifically, let

$D\; =\; -A+B+C.$

$ABCD$ is a parallelogram because its pairs of opposite sides, represented by the pairs of displacement vectors $D-C=B-A$, and $D-B=C-A$, are parallel and congruent.

Triangle $ABC$ is half of the parallelogram $ABDC$, so twice its signed area is equal to the signed area of the parallelogram, which is given by the $2\backslash times\; 2$ determinant $\backslash det(B-A,C-A)$ whose columns are the displacement vectors $B-A$ and $C-A$:

Expanding the determinant, using its alternating and multilinear properties, one obtains

&= \det(A,B)+\det(B,C)+\det(C,A) \end{align}

so

$2\; \backslash operatorname\{sarea\}(ABC)\; =\; \backslash det(A,B)+\backslash det(B,C)+\backslash det(C,A).$

Similarly,

$2\; \backslash operatorname\{sarea\}(PBC)\; =\; \backslash det(P,B)+\backslash det(B,C)+\backslash det(C,P)$,

To obtain the ratio of these signed areas, express $P$ in the second formula in terms of its barycentric coordinates:

$\backslash begin\{align\}2\; \backslash operatorname\{sarea\}(PBC)$

&= \det(\lambda_1 A + \lambda_2 B + \lambda_3 C, B) + \det(B,C)

+ \det(C,\lambda_1 A + \lambda_2 B + \lambda_3 C)\\

&= \lambda_1 \det(A,B) + \lambda_3 \det(C,B) + \det(B,C)

+ \lambda_1 \det(C,A) + \lambda_2 \det(C,B)\\

&= \lambda_1 \det(A,B) + \lambda_1 \det(C,A)

+ (1-\lambda_2 - \lambda_3) \det(B,C) \end{align}.

The barycentric coordinates are normalized so $1\; =\; \backslash lambda\_1\; +\; \backslash lambda\_2\; +\; \backslash lambda\_3$, hence $\backslash lambda\_1\; =\; (1-\backslash lambda\_2\; -\; \backslash lambda\_3)$ . Plug that into the previous line to obtain

&= (\lambda_1)(2 \operatorname{sarea}(ABC)).\end{align}

Therefore

$\backslash lambda\_1\; =\; \backslash operatorname\{sarea\}(PBC)/\backslash operatorname\{sarea\}(ABC)$.

Similar calculations prove the other two formulas

$\backslash lambda\_2\; =\; \backslash operatorname\{sarea\}(APC)/\backslash operatorname\{sarea\}(ABC)$

$\backslash lambda\_3\; =\; \backslash operatorname\{sarea\}(ABP)/\backslash operatorname\{sarea\}(ABC)$.

Trilinear coordinates $(\backslash gamma\_1,\backslash gamma\_2,\backslash gamma\_3)$ of $P$ are signed distances from $P$ to the lines BC, AC, and AB, respectively. The sign of $\backslash gamma\_1$ is positive if $P$ and $A$ lie on the same side of BC, negative otherwise. The signs of $\backslash gamma\_2$ and $\backslash gamma\_3$ are assigned similarly. Let

$a\; =\; \backslash operatorname\{length\}(BC)$, $b\; =\; \backslash operatorname\{length\}(CA)$, $c\; =\; \backslash operatorname\{length\}(AB)$.

Then

\gamma_2 b &= \pm 2\operatorname{sarea}(APC)\\

\gamma_3 c &= \pm 2\operatorname{sarea}(ABP)\end{align}

where, as above, sarea stands for signed area. All three signs are plus if triangle ABC is positively oriented, minus otherwise. The relations between trilinear and barycentric coordinates are obtained by substituting these formulas into the above formulas that express barycentric coordinates as ratios of areas.

Switching back and forth between the barycentric coordinates and other coordinate systems makes some problems much easier to solve.

Given a point $\backslash mathbf\{r\}$ in a triangle's plane one can obtain the barycentric coordinates $\backslash lambda\_1$, $\backslash lambda\_2$ and $\backslash lambda\_3$ from the Cartesian coordinates $(x,\; y)$ or vice versa.

We can write the Cartesian coordinates of the point $\backslash mathbf\{r\}$ in terms of the Cartesian components of the triangle vertices $\backslash mathbf\{r\}\_1$, $\backslash mathbf\{r\}\_2$, $\backslash mathbf\{r\}\_3$ where $\backslash mathbf\{r\}\_i\; =\; (x\_i,\; y\_i)$ and in terms of the barycentric coordinates of $\backslash mathbf\{r\}$ as

$$\backslash begin\{align\}$$

x &= \lambda_1 x_1 + \lambda_2 x_2 + \lambda_3 x_3 \\[2pt]

y &= \lambda_1 y_1 + \lambda_2 y_2 + \lambda_3 y_3

\end{align}

That is, the Cartesian coordinates of any point are a weighted average of the Cartesian coordinates of the triangle's vertices, with the weights being the point's barycentric coordinates summing to unity.

To find the reverse transformation, from Cartesian coordinates to barycentric coordinates, we first substitute $\backslash lambda\_3\; =\; 1\; -\; \backslash lambda\_1\; -\; \backslash lambda\_2$ into the above to obtain

$$\backslash begin\{align\}$$

x &= \lambda_1 x_1 + \lambda_2 x_2 + (1 - \lambda_1 - \lambda_2) x_3 \\[2pt]

y &= \lambda_1 y_1 + \lambda_2 y_2 + (1 - \lambda_1 - \lambda_2) y_3

\end{align}

Rearranging, this is

$$\backslash begin\{align\}$$

\lambda_1(x_1 - x_3) + \lambda_2(x_2 - x_3) + x_3 - x &= 0 \\[2pt]

\lambda_1(y_1 - y_3) + \lambda_2(y_2 -\, y_3) + y_3 - \, y &= 0

\end{align}

This linear transformation may be written more succinctly as

\mathbf{T} \cdot \lambda = \mathbf{r}-\mathbf{r}_3

where $\backslash lambda$ is the vector of the first two barycentric coordinates, $\backslash mathbf\{r\}$ is the vector of Cartesian coordinates, and $\backslash mathbf\{T\}$ is a matrix given by

\mathbf{T} = \left(\begin{matrix}

x_1-x_3 & x_2-x_3 \\

y_1-y_3 & y_2-y_3

\end{matrix}\right)

Now the matrix $\backslash mathbf\{T\}$ is invertible, since $\backslash mathbf\{r\}\_1-\backslash mathbf\{r\}\_3$ and $\backslash mathbf\{r\}\_2-\backslash mathbf\{r\}\_3$ are linearly independent (if this were not the case, then $\backslash mathbf\{r\}\_1$, $\backslash mathbf\{r\}\_2$, and $\backslash mathbf\{r\}\_3$ would be collinear and would not form a triangle). Thus, we can rearrange the above equation to get

\left(\begin{matrix}\lambda_1 \\ \lambda_2\end{matrix}\right) = \mathbf{T}^{-1} ( \mathbf{r}-\mathbf{r}_3 )

Finding the barycentric coordinates has thus been reduced to finding the 2×2 inverse matrix of $\backslash mathbf\{T\}$, an easy problem.

Explicitly, the formulae for the barycentric coordinates of point $\backslash mathbf\{r\}$ in terms of its Cartesian coordinates (x, y) and in terms of the Cartesian coordinates of the triangle's vertices are:

$$\backslash begin\{align\}$$

\lambda_1 =&\ \frac{(y_2-y_3)(x-x_3) + (x_3-x_2)(y-y_3)}{\det(\mathbf T)} \\[4pt]

&= \frac{(y_2-y_3)(x-x_3) + (x_3-x_2)(y-y_3)}{(y_2-y_3)(x_1-x_3) + (x_3-x_2)(y_1-y_3)} \\[4pt]

&= \frac{(\mathbf{r}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})} \\[12pt]

\lambda_2 =&\ \frac{(y_3-y_1)(x-x_3) + (x_1-x_3)(y-y_3)}{\det(\mathbf T)} \\[4pt]

&= \frac{(y_3-y_1)(x-x_3) + (x_1-x_3)(y-y_3)}{(y_2-y_3)(x_1-x_3) + (x_3-x_2)(y_1-y_3)} \\[4pt]

&= \frac{(\mathbf{r}-\mathbf{r_3})\times(\mathbf{r_3}-\mathbf{r_1})}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})} \\[12pt]

\lambda_3 =&\ 1 - \lambda_1 - \lambda_2 \\[4pt]

&= 1-\frac{(\mathbf{r}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_1})}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})} \\[4pt]

&= \frac{(\mathbf{r}-\mathbf{r_1})\times(\mathbf{r_1}-\mathbf{r_2 })}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})}

\end{align}When understanding the last line of equation, note the identity $(\backslash mathbf\{r\_1\}-\backslash mathbf\{r\_3\})\backslash times(\backslash mathbf\{r\_2\}-\backslash mathbf\{r\_3\})=(\backslash mathbf\{r\_3\}-\backslash mathbf\{r\_1\})\backslash times(\backslash mathbf\{r\_1\}-\backslash mathbf\{r\_2\})$.

Another way to solve the conversion from Cartesian to barycentric coordinates is to write the relation in the matrix form $$$$

\mathbf{R} \boldsymbol{\lambda} = \mathbf{r}with $\backslash mathbf\{R\}\; =\; \backslash left(\backslash ,\; \backslash mathbf\{r\}\_1\; \backslash ,|\backslash ,\; \backslash mathbf\{r\}\_2\; \backslash ,|\backslash ,\; \backslash mathbf\{r\}\_3\; \backslash right)$ and $\backslash boldsymbol\{\backslash lambda\}\; =\; \backslash left(\backslash lambda\_1,\backslash lambda\_2,\backslash lambda\_3\backslash right)^\backslash top,$ i.e.$$$$

\begin{pmatrix}

x_1 & x_2 & x_3\\

y_1 & y_2 & y_3

\end{pmatrix}

\begin{pmatrix}

\lambda_1 \\ \lambda_2 \\ \lambda_3

\end{pmatrix} =

\begin{pmatrix}x\\y\end{pmatrix}

To get the unique normalized solution we need to add the condition $\backslash lambda\_1\; +\; \backslash lambda\_2\; +\; \backslash lambda\_3\; =\; 1$. The barycentric coordinates are thus the solution of the linear system$$$$

\left(\begin{matrix}

1 & 1 & 1 \\

x_1 & x_2 & x_3\\

y_1 & y_2 & y_3

\end{matrix}\right)

\begin{pmatrix}

\lambda_1 \\ \lambda_2 \\ \lambda_3

\end{pmatrix} =

\left(\begin{matrix}

1\\x\\y

\end{matrix}\right)

which is$$$$

\begin{pmatrix}

\lambda_1 \\ \lambda_2 \\ \lambda_3

\end{pmatrix} = \frac{1}{2A}

\begin{pmatrix}

x_2y_3-x_3y_2 & y_2-y_3 & x_3-x_2 \\

x_3y_1-x_1y_3 & y_3-y_1 & x_1-x_3 \\

x_1y_2-x_2y_1 & y_1-y_2 & x_2-x_1

\end{pmatrix}\begin{pmatrix}

1\\x\\y

\end{pmatrix}

where $$$$

2A = \det(1|R) = x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)is twice the signed area of the triangle. The area interpretation of the barycentric coordinates can be recovered by applying Cramer's rule to this linear system.

A point with trilinear coordinates x : y : z has barycentric coordinates ax : by : cz where a, b, c are the side lengths of the triangle. Conversely, a point with barycentrics $\backslash lambda\_1\; :\; \backslash lambda\_2\; :\; \backslash lambda\_3$ has trilinears $\backslash lambda\_1/a:\backslash lambda\_2/b:\backslash lambda\_3/c.$

The three sides a, b, c respectively have equations

$$\backslash lambda\_1=0,\; \backslash quad\; \backslash lambda\_2=0,\; \backslash quad\; \backslash lambda\_3=0.$$

The equation of a triangle's Euler line is

Using the previously given conversion between barycentric and trilinear coordinates, the various other equations given in Trilinear coordinates#Formulas can be rewritten in terms of barycentric coordinates.

The displacement vector of two normalized points $P=(p\_1,p\_2,p\_3)$ and $Q=(q\_1,q\_2,q\_3)$ is

$$\backslash overset\{\}\backslash overrightarrow\{P\; Q\}=(p\_1-q\_1,p\_2-q\_2,p\_3-q\_3).$$

The distance {{mvar|d}} between {{mvar|P}} and {{mvar|Q}}, or the length of the displacement vector $\backslash overset\{\}\backslash overrightarrow\{P\; Q\}=(x,y,z),$ is

$$\backslash begin\{align\}$$

d^2 &= |PQ|^2 \\[2pt]

&= -a^2yz - b^2zx - c^2xy \\[4pt]

&= \frac{1}{2} \left[x^2(b^2+c^2-a^2) + y^2(c^2+a^2-b^2) + z^2(a^2+b^2-c^2)\right].

\end{align}

where a, b, c are the sidelengths of the triangle. The equivalence of the last two expressions follows from $x+y+z=0,$ which holds because

$$\backslash begin\{align\}$$

x+y+z &= (p_1-q_1) + (p_2-q_2) + (p_3-q_3) \\[2pt]

&= (p_1+p_2+p_3) - (q_1+q_2+q_3) \\[2pt]

&= 1 - 1 = 0.

\end{align}

The barycentric coordinates of a point can be calculated based on distances d_i to the three triangle vertices by solving the equation

\left(\begin{matrix}

-c^2 & c^2 & b^2-a^2 \\

-b^2 & c^2-a^2 & b^2 \\

1 & 1 & 1

\end{matrix}\right)\boldsymbol{\lambda} = \left(\begin{matrix}

d^2_A - d^2_B \\

d^2_A - d^2_C \\

1

\end{matrix}\right).

File:3_jugs_puzzle_barycentric_plot.svg using a barycentric plot. The yellow area denotes combinations achievable with the jugs. The solid red and dashed blue paths show pourable transitions. When a vertex lands on the dotted triangle, 4 L has been measured.]]

Although barycentric coordinates are most commonly used to handle points inside a triangle, they can also be used to describe a point outside the triangle. If the point is not inside the triangle, then we can still use the formulas above to compute the barycentric coordinates. However, since the point is outside the triangle, at least one of the coordinates will violate our original assumption that $\backslash lambda\_\{1...3\}\backslash geq\; 0$. In fact, given any point in cartesian coordinates, we can use this fact to determine where this point is with respect to a triangle.

If a point lies in the interior of the triangle, all of the Barycentric coordinates lie in the open interval $(0,1).$ If a point lies on an edge of the triangle but not at a vertex, one of the area coordinates $\backslash lambda\_\{1...3\}$ (the one associated with the opposite vertex) is zero, while the other two lie in the open interval $(0,1).$ If the point lies on a vertex, the coordinate associated with that vertex equals 1 and the others equal zero. Finally, if the point lies outside the triangle at least one coordinate is negative.

Summarizing,

:Point $\backslash mathbf\{r\}$ lies inside the triangle if and only if .

$$\backslash mathbf\{r\}$$ lies on the edge or corner of the triangle if $0\; \backslash leq\; \backslash lambda\_i\; \backslash leq\; 1\; \backslash ;\backslash forall\backslash ;\; i\; \backslash text\{\; in\; \}\; \{1,2,3\}$ and $\backslash lambda\_i\; =\; 0\backslash ;\; \backslash text\; \{,\; for\; some\; i\; in\; \}\; \{1,\; 2,\; 3\}$.

:Otherwise, $\backslash mathbf\{r\}$ lies outside the triangle.

In particular, if a point lies on the far side of a line the barycentric coordinate of the point in the triangle that is not on the line will have a negative value.

File:Piecewise linear function2D.svg

If $f(\backslash mathbf\{r\}\_1),f(\backslash mathbf\{r\}\_2),f(\backslash mathbf\{r\}\_3)$ are known quantities, but the values of {{mvar|f}} inside the triangle defined by $\backslash mathbf\{r\}\_1,\backslash mathbf\{r\}\_2,\backslash mathbf\{r\}\_3$ is unknown, they can be approximated using linear interpolation. Barycentric coordinates provide a convenient way to compute this interpolation. If $\backslash mathbf\{r\}$ is a point inside the triangle with barycentric coordinates $\backslash lambda\_1$, $\backslash lambda\_2$, $\backslash lambda\_3$, then

$$f(\backslash mathbf\{r\})\; \backslash approx\; \backslash lambda\_1\; f(\backslash mathbf\{r\}\_1)\; +\; \backslash lambda\_2\; f(\backslash mathbf\{r\}\_2)\; +\; \backslash lambda\_3\; f(\backslash mathbf\{r\}\_3)$$

In general, given any unstructured grid or polygon mesh, this kind of technique can be used to approximate the value of {{mvar|f}} at all points, as long as the function's value is known at all vertices of the mesh. In this case, we have many triangles, each corresponding to a different part of the space. To interpolate a function {{mvar|f}} at a point $\backslash mathbf\{r\}$, first a triangle must be found that contains $\backslash mathbf\{r\}$. To do so, $\backslash mathbf\{r\}$ is transformed into the barycentric coordinates of each triangle. If some triangle is found such that the coordinates satisfy $0\; \backslash leq\; \backslash lambda\_i\; \backslash leq\; 1\; \backslash ;\backslash forall\backslash ;\; i\; \backslash text\{\; in\; \}\; 1,2,3$, then the point lies in that triangle or on its edge (explained in the previous section). Then the value of $f(\backslash mathbf\{r\})$ can be interpolated as described above.

These methods have many applications, such as the finite element method (FEM).

The integral of a function over the domain of the triangle can be annoying to compute in a cartesian coordinate system. One generally has to split the triangle up into two halves, and great messiness follows. Instead, it is often easier to make a change of variables to any two barycentric coordinates, e.g. $\backslash lambda\_1,\backslash lambda\_2$. Under this change of variables,

\int_{T} f(\mathbf{r}) \ d\mathbf{r} = 2A \int_{0}^{1} \int_{0}^{1 - \lambda_2} f(\lambda_1 \mathbf{r}_1 + \lambda_2 \mathbf{r}_2 +

(1 - \lambda_1 - \lambda_2) \mathbf{r}_3) \ d\lambda_1 \ d\lambda_2

where {{mvar|A}} is the area of the triangle. This result follows from the fact that a rectangle in barycentric coordinates corresponds to a quadrilateral in cartesian coordinates, and the ratio of the areas of the corresponding shapes in the corresponding coordinate systems is given by $2A$. Similarly, for integration over a tetrahedron, instead of breaking up the integral into two or three separate pieces, one could switch to 3D tetrahedral coordinates under the change of variables

\int\int_{T} f(\mathbf{r}) \ d\mathbf{r}

= 6V \int_{0}^{1} \int_{0}^{1 - \lambda_3} \int_ {0}^{1-\lambda_2-\lambda_3}

f(\lambda_1\mathbf{r}_1 + \lambda_2\mathbf{r}_2 +

\lambda_3\mathbf{r}_3 + (1-\lambda_1-\lambda_2-\lambda_3)\mathbf{r}_4)

\ d\lambda_1 \ d\lambda_2 \ d\lambda_3

where {{mvar|V}} is the volume of the tetrahedron.

In the homogeneous barycentric coordinate system defined with respect to a triangle $ABC$, the following statements about special points of $ABC$ hold.

The three vertices {{mvar|A}}, {{mvar|B}}, and {{mvar|C}} have coordinatesScott, J. A. "Some examples of the use of areal coordinates in triangle geometry", Mathematical Gazette 83, November 1999, 472–477.

$$\backslash begin\{array\}\{rccccc\}$$

A = & 1 &:& 0 &:& 0 \\

B = & 0 &:& 1 &:& 0 \\

C = & 0 &:& 0 &:& 1

\end{array}

The centroid has coordinates $1:1:1.$

If {{mvar|a}}, {{mvar|b}}, {{mvar|c}} are the edge lengths $BC$, $CA$, $AB$ respectively, $\backslash alpha$, $\backslash beta$, $\backslash gamma$ are the angle measures $\backslash angle\; CAB$, $\backslash angle\; ABC$, and $\backslash angle\; BCA$ respectively, and {{mvar|s}} is the semiperimeter of $ABC$, then the following statements about special points of $ABC$ hold in addition.

The circumcenter has coordinates{{cite web|last1=Schindler|first1=Max|last2=Chen|first2=Evan|title=Barycentric Coordinates in Olympiad Geometry|url=https://www.mit.edu/~evanchen/handouts/bary/bary-full.pdf|access-date=14 January 2016|date=July 13, 2012}}Clark Kimberling's Encyclopedia of Triangles {{cite web |url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |title=Encyclopedia of Triangle Centers |access-date=2012-06-02 |url-status=dead |archive-url=https://web.archive.org/web/20120419171900/http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |archive-date=2012-04-19 }}[http://mathworld.wolfram.com/BarycentricCoordinates.html Wolfram page on barycentric coordinates]

$$\backslash begin\{array\}\{rccccc\}$$

& \sin 2\alpha &:& \sin 2\beta &:& \sin 2\gamma \\[2pt]

=& 1-\cot\beta\cot\gamma &:& 1-\cot\gamma\cot\alpha &:& 1-\cot\alpha\cot\beta \\[2pt]

=& a^2(-a^2+b^2+c^2) &:& b^2(a^2-b^2+c^2) &:& c^2(a^2+b^2-c^2)

\end{array}

The orthocenter has coordinates

$$\backslash begin\{array\}\{rccccc\}$$

& \tan\alpha &:& \tan\beta &:& \tan\gamma \\[2pt]

=& a\cos\beta\cos\gamma &:& b\cos\gamma\cos\alpha &:& c\cos\alpha\cos\beta \\[2pt]

=& (a^2+b^2-c^2)(a^2-b^2+c^2) &:& (-a^2+b^2+c^2)(a^2+b^2-c^2) &:& (a^2-b^2+c^2)(-a^2+b^2+c^2)

\end{array}

The incenter has coordinates $a:b:c=\backslash sin\; \backslash alpha:\backslash sin\; \backslash beta:\backslash sin\; \backslash gamma.$Dasari Naga, Vijay Krishna, "On the Feuerbach triangle",

Forum Geometricorum 17 (2017), 289–300: p. 289. http://forumgeom.fau.edu/FG2017volume17/FG201731.pdf

The excenters have coordinates

$$\backslash begin\{array\}\{rrcrcr\}$$

J_A = & -a &:& b &:& c \\

J_B = & a &:& -b &:& c \\

J_C = & a &:& b &:& -c

\end{array}

The nine-point center has coordinates

$$\backslash begin\{array\}\{rccccc\}$$

& a\cos(\beta-\gamma) &:& b\cos(\gamma-\alpha) &:& c\cos(\alpha-\beta) \\[4pt]

=& 1+\cot\beta\cot\gamma &:& 1+\cot\gamma\cot\alpha &:& 1+\cot\alpha\cot\beta \\[4pt]

=& a^2(b^2+c^2) - (b^2-c^2)^2 &:& b^2(c^2+a^2) - (c^2-a^2)^2 &:& c^2(a^2+b^2) - (a^2-b^2)^2

\end{array}

The Gergonne point has coordinates $(s-b)(s-c):(s-c)(s-a):(s-a)(s-b)$.

The Nagel point has coordinates $s-a:s-b:s-c$.

The symmedian point has coordinates $a^2:b^2:c^2$.

Barycentric coordinates may be easily extended to three dimensions. The 3D simplex is a tetrahedron, a polyhedron having four triangular faces and four vertices. Once again, the four barycentric coordinates are defined so that the first vertex $\backslash mathbf\{r\}\_1$ maps to barycentric coordinates $\backslash lambda\; =\; (1,0,0,0)$, $\backslash mathbf\{r\}\_2\; \backslash to\; (0,1,0,0)$, etc.

This is again a linear transformation, and we may extend the above procedure for triangles to find the barycentric coordinates of a point $\backslash mathbf\{r\}$ with respect to a tetrahedron:

\left(\begin{matrix}\lambda_1 \\ \lambda_2 \\ \lambda_3\end{matrix}\right) = \mathbf{T}^{-1} ( \mathbf{r}-\mathbf{r}_4 )

where $\backslash mathbf\{T\}$ is now a 3×3 matrix:

\mathbf{T} = \left(\begin{matrix}

x_1-x_4 & x_2-x_4 & x_3-x_4\\

y_1-y_4 & y_2-y_4 & y_3-y_4\\

z_1-z_4 & z_2-z_4 & z_3-z_4

\end{matrix}\right)

and $\backslash lambda\_4\; =\; 1\; -\; \backslash lambda\_1\; -\; \backslash lambda\_2\; -\; \backslash lambda\_3$with the corresponding Cartesian coordinates:$$\backslash begin\{align\}$$

x &= \lambda_1 x_1 + \lambda_2 x_2 + \lambda_3 x_3 + (1-\lambda_1-\lambda_2-\lambda_3)x_4 \\

y &= \lambda_1 y_1 + \,\lambda_2 y_2 + \lambda_3 y_3 + (1-\lambda_1-\lambda_2-\lambda_3)y_4 \\

z &= \lambda_1 z_1 + \,\lambda_2 z_2 + \lambda_3 z_3 + (1-\lambda_1-\lambda_2-\lambda_3)z_4

\end{align}Once again, the problem of finding the barycentric coordinates has been reduced to inverting a 3×3 matrix.

3D barycentric coordinates may be used to decide if a point lies inside a tetrahedral volume, and to interpolate a function within a tetrahedral mesh, in an analogous manner to the 2D procedure. Tetrahedral meshes are often used in finite element analysis because the use of barycentric coordinates can greatly simplify 3D interpolation.

Barycentric coordinates $(\backslash lambda\_1,\; \backslash lambda\_2,\; ...,\; \backslash lambda\_k)$ of a point $p\; \backslash in\; \backslash mathbb\{R\}^n$ that are defined with respect to a finite set of k points $x\_1,\; x\_2,\; ...,\; x\_k\; \backslash in\; \backslash mathbb\{R\}^n$ instead of a simplex are called generalized barycentric coordinates. For these, the equation

$$(\backslash lambda\_1\; +\; \backslash lambda\_2\; +\; \backslash cdots\; +\; \backslash lambda\_k)p\; =\; \backslash lambda\_1\; x\_1\; +\; \backslash lambda\_2\; x\_2\; +\; \backslash cdots\; +\; \backslash lambda\_k\; x\_k$$

is still required to hold.{{Cite journal |last1=Meyer |first1=Mark |last2=Barr |first2=Alan |last3=Lee |first3=Haeyoung |last4=Desbrun |first4=Mathieu |date=6 April 2012 |title=Generalized Barycentric Coordinates on Irregular Polygons |url=http://www.geometry.caltech.edu/pubs/MHBD02.pdf |journal=Journal of Graphics Tools |volume=7 |pages=13–22 |doi=10.1080/10867651.2002.10487551|s2cid=13370238 }} Usually one uses normalized coordinates, $\backslash lambda\_1\; +\; \backslash lambda\_2\; +\; \backslash cdots\; +\; \backslash lambda\_k\; =\; 1$. As for the case of a simplex, the points with nonnegative normalized generalized coordinates ($0\; \backslash le\; \backslash lambda\_i\; \backslash le\; 1$) form the convex hull of {{math|x₁, ..., x_n}}. If there are more points than in a full simplex ($k\; >\; n\; +\; 1$) the generalized barycentric coordinates of a point are not unique, as the defining linear system (here for n=2)$$$$

\left(\begin{matrix}

1 & 1 & 1 & ... \\

x_1 & x_2 & x_3 & ... \\

y_1 & y_2 & y_3 & ...

\end{matrix}\right)

\begin{pmatrix}

\lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \vdots

\end{pmatrix} =

\left(\begin{matrix}

1\\x\\y

\end{matrix}\right)

is underdetermined. The simplest example is a quadrilateral in the plane. Various kinds of additional restrictions can be used to define unique barycentric coordinates.{{Cite journal|last=Floater|first=Michael S.|date=2015|title=Generalized barycentric coordinates and applications *|url=https://www.mn.uio.no/math/english/people/aca/michaelf/papers/gbc.pdf|journal=Acta Numerica|language=en|volume=24|pages=161–214|doi=10.1017/S0962492914000129|s2cid=62811364 |issn=0962-4929}}

More abstractly, generalized barycentric coordinates express a convex polytope with n vertices, regardless of dimension, as the image of the standard $(n-1)$-simplex, which has n vertices – the map is onto: $\backslash Delta^\{n-1\}\; \backslash twoheadrightarrow\; P.$ The map is one-to-one if and only if the polytope is a simplex, in which case the map is an isomorphism; this corresponds to a point not having unique generalized barycentric coordinates except when P is a simplex.

Dual to generalized barycentric coordinates are slack variables, which measure by how much margin a point satisfies the linear constraints, and gives an embedding $P\; \backslash hookrightarrow\; (\backslash mathbf\{R\}\_\{\backslash geq\; 0\})^f$ into the f-orthant, where f is the number of faces (dual to the vertices). This map is one-to-one (slack variables are uniquely determined) but not onto (not all combinations can be realized).

This use of the standard $(n-1)$-simplex and f-orthant as standard objects that map to a polytope or that a polytope maps into should be contrasted with the use of the standard vector space $K^n$ as the standard object for vector spaces, and the standard affine hyperplane $\backslash \{(x\_0,\backslash ldots,x\_n)\; \backslash mid\; \backslash sum\; x\_i\; =\; 1\backslash \}\; \backslash subset\; K^\{n+1\}$ as the standard object for affine spaces, where in each case choosing a linear basis or affine basis provides an isomorphism, allowing all vector spaces and affine spaces to be thought of in terms of these standard spaces, rather than an onto or one-to-one map (not every polytope is a simplex). Further, the n-orthant is the standard object that maps to cones.

File:Barycentric RGB.svg

Generalized barycentric coordinates have applications in computer graphics and more specifically in geometric modelling.{{Cite journal |last=Floater |first=Michael S. |date=2003 |title=Mean value coordinates |url=https://linkinghub.elsevier.com/retrieve/pii/S0167839603000025 |journal=Computer Aided Geometric Design |language=en |volume=20 |issue=1 |pages=19–27 |doi=10.1016/S0167-8396(03)00002-5}} Often, a three-dimensional model can be approximated by a polyhedron such that the generalized barycentric coordinates with respect to that polyhedron have a geometric meaning. In this way, the processing of the model can be simplified by using these meaningful coordinates. Barycentric coordinates are also used in geophysics.ONUFRIEV, VG; DENISIK, SA; FERRONSKY, VI, BARICENTRIC MODELS IN ISOTOPE STUDIES OF NATURAL-WATERS. NUCLEAR GEOPHYSICS, 4, 111-117 (1990)

• Ternary plot
• Convex combination
• Water pouring puzzle
• Homogeneous coordinates

• Scott, J. A. Some examples of the use of areal coordinates in triangle geometry, Mathematical Gazette 83, November 1999, 472–477.
• Schindler, Max; Chen, Evan (July 13, 2012). Barycentric Coordinates in Olympiad Geometry (PDF). Retrieved 14 January 2016.
• Clark Kimberling's Encyclopedia of Triangles Encyclopedia of Triangle Centers. Archived from the original on 2012-04-19. Retrieved 2012-06-02.
• {{cite book |last=Bradley |first=Christopher J. |title=The Algebra of Geometry: Cartesian, Areal and Projective Co-ordinates |year=2007 |publisher=Highperception |location=Bath |isbn=978-1-906338-00-8 }}
• {{cite book | last=Coxeter | first=H.S.M. | author-link=Harold Scott MacDonald Coxeter | title=Introduction to geometry | url=https://archive.org/details/introductiontoge00coxe | url-access=limited | edition=2nd | publisher=John Wiley and Sons | year=1969 | isbn=978-0-471-50458-0 | pages=[https://archive.org/details/introductiontoge00coxe/page/n233 216]–221 | zbl=0181.48101 }}
• [https://web.archive.org/web/20120519111614/http://www.worldscibooks.com/mathematics/7740.html Barycentric Calculus In Euclidean And Hyperbolic Geometry: A Comparative Introduction], Abraham Ungar, World Scientific, 2010
• [http://ajmaa.org/searchroot/files/pdf/v6n1/v6i1p18.pdf Hyperbolic Barycentric Coordinates], Abraham A. Ungar, The Australian Journal of Mathematical Analysis and Applications, Vol.6, No.1, Article 18, pp. 1–35, 2009
• {{MathWorld|title=Areal Coordinates|urlname=ArealCoordinates}}
• {{MathWorld|title=Barycentric Coordinates|urlname=BarycentricCoordinates}}
• [http://www.sciencedirect.com/science/article/pii/S009784930700180X Barycentric coordinates computation in homogeneous coordinates], Vaclav Skala, Computers and Graphics, Vol.32, No.1, pp. 120–127, 2008

• [https://www.math.nyu.edu/~crorres/Archimedes/Lever/LeverLaw.html Law of the lever]
• [http://www.matematicas.unam.mx/gfgf/ga20071/data/material/barycentricpaper.pdf The uses of homogeneous barycentric coordinates in plane euclidean geometry]
• [http://www.inf.usi.ch/hormann/barycentric/ Barycentric Coordinates] – a collection of scientific papers about (generalized) barycentric coordinates
• [http://www.cut-the-knot.org/triangle/glasses.shtml Barycentric coordinates: A Curious Application] (solving the "three glasses" problem) at cut-the-knot
• [http://totologic.blogspot.fr/2014/01/accurate-point-in-triangle-test.html Accurate point in triangle test]
• [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=721&t=475427 Barycentric Coordinates in Olympiad Geometry] by Evan Chen and Max Schindler
• [https://www.geogebra.org/wiki/en/Barycenter_Command Barycenter command] and [https://www.geogebra.org/wiki/en/TriangleCurve_Command TriangleCurve command] at Geogebra.

{{DEFAULTSORT:Barycentric Coordinate System}}

Category:Linear algebra

Category:Affine geometry

Category:Triangle geometry

Category:Coordinate systems

Category:Two-dimensional coordinate systems