View source for Barycentric coordinate system

{{short description|Coordinate system that is defined by points instead of vectors}}
{{Distinguish|Barycentric coordinates (astronomy)}}
{{See also|Affine space#Barycentric coordinates}}
[[File:Barycentric_subdivision_of_a_3-simplex.png|thumb|right|A 3-simplex, with barycentric subdivisions of 1-faces (edges) 2-faces (triangles) and 3-faces (body).]]
In [[geometry]], the '''barycentric coordinate system''' is a [[coordinate system]] in which the location of a point of a [[simplex]] (a triangle, [[tetrahedron]], etc.) is specified as the [[center of mass]], or barycenter, of usually unequal [[mass]]es placed at its [[vertex (geometry)|vertices]]. Coordinates also extend outside the simplex, where one or more coordinates become [[negative number|negative]]. The system was introduced in 1827 by [[August Ferdinand Möbius]].<ref name=Hille>Hille, Einar. "Analytic Function Theory, Volume I", Second edition, fifth printing. Chelsea Publishing Company, New York, 1982,  {{ISBN|0-8284-0269-8}}, page 33, footnote 1</ref>

==Definition==

Let <math>\mathbf{x}_1, \ldots, \mathbf{x}_n</math> be the vertices of a [[simplex]] in an [[affine space]] ''A''. If, for some point <math>\mathbf{p}</math> in ''A'',

:<math> ( a_1 + \cdots + a_n ) \mathbf{p} = a_1 \, \mathbf{x}_1 + \cdots + a_n \, \mathbf{x}_n </math>

and at least one of <math>a_1, \ldots, a_n</math> does not vanish
then we say that the coefficients (<math>a_1, \ldots, a_n</math>) are ''barycentric coordinates'' of <math>\mathbf{p}</math> with respect to <math>\mathbf{x}_1, \ldots, \mathbf{x}_n</math>. The vertices themselves have the coordinates <math>\mathbf{x}_1=(1, 0, 0, \ldots, 0), \mathbf{x}_2=(0, 1, 0, \ldots, 0), \ldots, \mathbf{x}_n=(0, 0, 0, \ldots, 1)</math>. Barycentric coordinates are not unique: for any ''b'' not equal to zero, (<math>b a_1, \ldots, b a_n</math>) are also barycentric coordinates of ''p''.

When the coordinates are not negative, the point <math>\mathbf{p}</math> lies in the [[convex hull]] of <math>\mathbf{x}_1, \ldots, \mathbf{x}_n</math>, that is, in the simplex which has those points as its vertices.

Barycentric coordinates, as defined above, are a form of [[homogeneous coordinates]]: indeed, the "usual" homogeneous coordinates are the barycentric coordinates defined in the extended affine n-space on the simplex whose vertices are the points at infinity on the coordinate axes, plus the origin. Sometimes values of coordinates are restricted with a condition 
:<math>\sum a_i = 1</math>
which makes them [[unique (mathematics)|unique]]; then, they are [[affine coordinates]]. The classical terminology in this case is that of '''absolute barycentric coordinates'''.<ref name=Deaux>Deaux, Roland. "Introduction to The Geometry of Complex Numbers". Dover Publications, Inc., Mineola, 2008, {{ISBN|978-0-486-46629-3}}, page 61</ref>

==Barycentric coordinates on triangles==
{{Confusing section|small=no|reason=it is unnecessarily technical and complicated|date=December 2018 }}
{{see also|Ternary plot|Triangle center}}
[[Image:TriangleBarycentricCoordinates.svg|thumb|320px|Barycentric coordinates <math>(\lambda_{1}, \lambda_{2}, \lambda_{3})</math> on an equilateral triangle and on a right triangle.]]
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 [[triangulation (geometry)|triangular subdomains]]. These make analytic [[integral (mathematics)|integrals]] often easier to evaluate, and [[Gaussian quadrature]] tables are often presented in terms of area coordinates.

Consider a triangle <math>T</math> defined by its three vertices, <math>\mathbf{r}_{1}</math>, <math>\mathbf{r}_{2}</math> and <math>\mathbf{r}_{3}</math>. Each point <math>\mathbf{r}</math> located inside this triangle can be written as a unique [[convex combination]] of the three vertices.  In other words, for each <math>\mathbf{r}</math> there is a unique sequence of three numbers, <math>\lambda_1,\lambda_2,\lambda_3\geq 0</math> such that <math>\lambda_1+\lambda_2+\lambda_3=1</math> and

:<math>\mathbf{r} = \lambda_{1} \mathbf{r}_{1} + \lambda_{2} \mathbf{r}_{2} + \lambda_{3} \mathbf{r}_{3},</math>

The three numbers <math>\lambda_1,\lambda_2,\lambda_3</math> indicate the "barycentric" or "area" coordinates of the point <math>\mathbf{r}</math> with respect to the triangle.  They are often denoted as <math>\alpha,\beta,\gamma</math> instead of <math>\lambda_1,\lambda_2,\lambda_3</math>.  Note that although there are three coordinates, there are only two [[Degrees of freedom (statistics)|degrees of freedom]], since <math>\lambda_1+\lambda_2+\lambda_3=1</math>.  Thus every point is uniquely defined by any two of the barycentric coordinates.

To explain why these coordinates are '''signed ratios of areas''', let us assume that we work in the [[euclidean space|Euclidean]] space <math>\mathbf{E}^{3}</math>. Here, consider the [[Cartesian coordinate system]] <math>Oxyz</math> and its associated [[standard basis|basis]], namely <math>\{\mathbf{i},\mathbf{j},\mathbf{k}\}</math>. Consider also the [[orientation (vector space)|positively oriented triangle]] <math>ABC</math> lying in the <math>Oxy</math> [[plane (geometry)|plane]]. It is known that for any [[basis (linear algebra)|basis]] <math>\{\mathbf{e},\mathbf{f},\mathbf{g}\}</math> of <math>\mathbf{E}^{3}</math> and any [[euclidean vector|free vector]] <math>\mathbf{h}</math> one has<ref name=Danby>Danby, J.M.A. "Fundamentals of Celestial Mechanics", Second edition, revised & enlarged, fifth printing. Willmann-Bell, Inc., Richmond, 2003,  {{ISBN|0-943396-20-4}}, page 26, problem 11</ref>

:<math>\mathbf{h}=\frac{1}{(\mathbf{e},\mathbf{f},\mathbf{g})}\cdot\left[(\mathbf{h},\mathbf{f},\mathbf{g})\mathbf{e}+(\mathbf{e},\mathbf{h},\mathbf{g})\mathbf{f}+(\mathbf{e},\mathbf{f},\mathbf{h})\mathbf{g}\right],</math>

where <math>(\mathbf{e},\mathbf{f},\mathbf{g})=(\mathbf{e}\times\mathbf{f})\cdot\mathbf{g}</math> stands for the [[triple product|mixed product]] of these three vectors.

Take <math>\mathbf{e}=\vec{AB},\,\mathbf{f}=\vec{AC},\,\mathbf{g}=\mathbf{k},\,\mathbf{h}=\vec{AP}</math>, where <math>P</math> is an arbitrary point in the plane <math>Oxy</math>, and remark that

:<math>(\mathbf{e},\mathbf{f},\mathbf{h})=(\vec{AB}\times\vec{AC})\cdot\vec{AP}=(\vert\vec{AB}\times\vec{AC}\vert\mathbf{k})\cdot\vec{AP} = 0.</math>

A subtle point regarding our choice of free vectors: <math>\mathbf{e}</math> is, in fact, the [[equipollence (geometry)|equipollence class]] of the [[euclidean vector|bound vector]] <math>\vec{AB}</math>.

We have obtained that

:<math>\vec{AP}=m_B\cdot\vec{AB}+m_C\cdot\vec{AC},\,\mbox{ where }\,m_B=\frac{(\vec{AP},\vec{AC},\mathbf{k})}{(\vec{AB},\vec{AC},\mathbf{k})},\,m_C=\frac{(\vec{AB},\vec{AP},\mathbf{k})}{(\vec{AB},\vec{AC},\mathbf{k})}.</math>

Given the positive ([[clockwise|counterclockwise]]) orientation of triangle <math>ABC</math>, the [[fraction (mathematics)|denominator]] of both <math>m_B</math> and <math>m_C</math> is precisely the double of the [[triangle|area of the triangle]] <math>ABC</math>. Also,

:<math>(\vec{AP},\vec{AC},\mathbf{k})=(\vec{PC},\vec{PA},\mathbf{k})\,\mbox{ and }\,(\vec{AB},\vec{AP},\mathbf{k})=(\vec{PA},\vec{PB},\mathbf{k})</math>

and so the [[fraction (mathematics)|numerators]] of <math>m_B</math> and <math>m_C</math> are the doubles of the '''signed areas''' of triangles <math>APC</math> and respectively <math>ABP</math>.

Further, we deduce that

:<math>\vec{OP}=(1-m_B-m_C)\cdot\vec{OA}+m_B\cdot\vec{OB}+m_C\cdot\vec{OC}</math>

which means that the numbers <math>1-m_B-m_C</math>, <math>m_B</math> and <math>m_C</math> are the barycentric coordinates of <math>P</math>. Similarly, the third barycentric coordinate reads as

:<math>m_A = 1 - m_B - m_C = \frac{(\vec{PB},\vec{PC},\mathbf{k})}{(\vec{AB},\vec{AC},\mathbf{k})}.</math>

This <math>m</math>-letter notation of the barycentric coordinates comes from the fact that the point <math>P</math> may be interpreted as the [[center of mass]] for the masses <math>m_A</math>, <math>m_B</math>, <math>m_C</math> which are located in <math>A</math>, <math>B</math> and <math>C</math>.

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

=== Conversion between barycentric and Cartesian coordinates ===

Given a point <math>\mathbf{r}</math> in a triangle's plane one can obtain the barycentric coordinates <math>\lambda_{1}</math>, <math>\lambda_{2}</math> and <math>\lambda_{3}</math> from the [[Cartesian coordinates]] <math>(x, y)</math> or vice versa.

We can write the Cartesian coordinates of the point <math>\mathbf{r}</math> in terms of the Cartesian components of the triangle vertices <math>\mathbf{r}_1</math>, <math>\mathbf{r}_2</math>, <math>\mathbf{r}_3</math> where <math>\mathbf{r}_i = (x_i, y_i)</math> and in terms of the barycentric coordinates of <math>\mathbf{r}</math> as

:<math>
\begin{matrix}
x = \lambda_{1} x_{1} +  \lambda_{2} x_{2} +  \lambda_{3} x_{3} \\
y = \lambda_{1} y_{1} +  \lambda_{2} y_{2} +  \lambda_{3} y_{3} \\
\end{matrix}
</math>

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 <math>\lambda_{3} = 1 - \lambda_{1} - \lambda_{2}</math> into the above to obtain

:<math>
\begin{matrix}
x = \lambda_{1} x_{1} +  \lambda_{2} x_{2} + (1 - \lambda_{1} - \lambda_{2}) x_{3} \\
y = \lambda_{1} y_{1} +  \lambda_{2} y_{2} + (1 - \lambda_{1} - \lambda_{2}) y_{3} \\
\end{matrix}
</math>

Rearranging, this is

:<math>
\begin{matrix}
\lambda_{1}(x_{1} - x_{3}) + \lambda_{2}(x_{2} - x_{3}) + x_{3} - x = 0 \\
\lambda_{1}(y_{1} - y_{3}) + \lambda_{2}(y_{2} - y_{3}) + y_{3} - y = 0 \\
\end{matrix}
</math>

This [[linear transformation]] may be written more succinctly as

:<math>
\mathbf{T} \cdot \lambda = \mathbf{r}-\mathbf{r}_3
</math>

where <math>\lambda</math> is the [[coordinate space|vector]] of the first two barycentric coordinates, <math>\mathbf{r}</math> is the [[Euclidean vector|vector]] of [[Cartesian coordinates]], and <math>\mathbf{T}</math> is a [[matrix (mathematics)|matrix]] given by

:<math>
\mathbf{T} = \left(\begin{matrix}
x_1-x_3 & x_2-x_3 \\
y_1-y_3 & y_2-y_3 \\
\end{matrix}\right)
</math>

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

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

Finding the barycentric coordinates has thus been reduced to finding the [[Matrix inversion#Inversion of 2×2 matrices|2×2 inverse matrix]] of <math>\mathbf{T}</math>, an easy problem.

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

:<math>\lambda_1=\frac{(y_2-y_3)(x-x_3)+(x_3-x_2)(y-y_3)}{\det(T)}=\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)}\, ,</math>
:<math>\lambda_2=\frac{(y_3-y_1)(x-x_3)+(x_1-x_3)(y-y_3)}{\det(T)}=\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)}\, ,</math>
:<math>\lambda_3=1-\lambda_1-\lambda_2\, .</math>

Another way to solve the conversion from Cartesian to barycentric coordinates is to rewrite the problem in matrix form so that

:<math>
\mathbf{r} = \mathbf{R} \boldsymbol{\lambda}</math>

with <math>\mathbf{R} = \left(\begin{matrix}
\mathbf{r}_1 |
\mathbf{r}_2 |
\mathbf{r}_3
\end{matrix}\right)</math>
and
<math>\boldsymbol{\lambda} = \left(\lambda_1,\lambda_2,\lambda_3\right)^\top</math>.
Then, the condition <math>\lambda_1 + \lambda_2 + \lambda_3 = 1</math>
reads <math>\left(1,1,1\right) \boldsymbol{\lambda} = 1</math>
and the barycentric coordinates can be solved as the solution of
the linear system

:<math>
\left(\begin{matrix}
x_1 & x_2 & x_3\\
y_1 & y_2 & y_3\\
1 & 1 & 1
\end{matrix}\right)
\boldsymbol{\lambda} =
\left(\begin{matrix}
x\\
y\\
1
\end{matrix}\right)
</math>

===Conversion between barycentric and trilinear coordinates===
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 <math>\lambda_1 : \lambda_2 : \lambda_3</math> has trilinears <math>\lambda_1/a:\lambda_2/b:\lambda_3/c.</math>

===Equations in barycentric coordinates===

The sides ''a, b, c'' respectively have equations<ref name=Scott/>

:<math>\lambda_1=0, \quad \lambda_2=0, \quad \lambda_3=0.</math>

The equation of a triangle's [[Euler line]] is<ref name=Scott/>

:<math> \begin{vmatrix} \lambda_1 & \lambda_2 & \lambda_3 \\1 & 1 & 1\\\tan A & \tan B & \tan C \end{vmatrix} =0.</math>

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.

===Distance between points===
The displacement vector of two normalized points <math>P=(p_1,p_2,p_3)</math> and <math>Q=(q_1,q_2,q_3)</math> is<ref name=Olympiad/>

:<math>\overrightarrow{P Q}=(p_1-q_1,p_2-q_2,p_3-q_3).</math>

The distance <math>d</math> between <math>P</math> and <math>Q</math>, or the length of the displacement vector <math>\overrightarrow{P Q}=(x,y,z),</math> is<ref name=Scott/><ref name=Olympiad/>

:<math>d^2=\left | P Q \right |^2 = -a^2yz - b^2zx - c^2xy =\frac{1}{2}[x^2(b^2+c^2-a^2)+y^2(c^2+a^2-b^2)+z^2(a^2+b^2-c^2)].</math>

where ''a, b, c'' are the sidelengths of the triangle. The equivalence of the last two expressions follows from <math>x+y+z=0,</math> which holds because <math>x+y+z=(p_1-q_1)+(p_2-q_2)+(p_3-q_3)=(p_1+p_2+p_3)-(q_1+q_2+q_3)=1-1=0.</math>

The barycentric coordinates of a point can be calculated based on distances ''d''<sub>''i''</sub> to the three triangle vertices by solving the equation
:<math>
\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).
</math>

=== Applications===

====Determining location with respect to a triangle ====

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 <math>\lambda_{1...3}\geq 0</math>.  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]] <math>(0,1).</math>  If a point lies on an edge of the triangle but not at a vertex, one of the area coordinates <math>\lambda_{1...3}</math> (the one associated with the opposite vertex) is zero, while the other two lie in the open interval <math>(0,1).</math> 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 <math>\mathbf{r}</math> lies inside the triangle [[if and only if]] <math>0 < \lambda_i < 1 \;\forall\; i \text{ in } {1,2,3}</math>.
:<math>\mathbf{r}</math> lies on the edge or corner of the triangle if <math>0 \leq \lambda_i \leq 1 \;\forall\; i \text{ in } {1,2,3}</math> and <math>\lambda_i = 0\; \text {, for some i in } {1, 2, 3}</math>.

:Otherwise, <math>\mathbf{r}</math> lies outside the triangle.

In particular, if a point lies on the opposite side of a sideline from the vertex opposite that sideline, then that point's barycentric coordinate corresponding to that vertex is negative.

====Interpolation on a triangular unstructured grid ====
[[File:Piecewise linear function2D.svg|thumb|300px|Surface (upper part) obtained from linear interpolation over a given triangular grid (lower part) in the ''x'',''y'' plane. The surface approximates a function ''z''=''f''(''x'',''y''), given only the values of ''f'' on the grid's vertices.]]
If <math>f(\mathbf{r}_1),f(\mathbf{r}_2),f(\mathbf{r}_3)</math> are known quantities, but the values of <math>f</math> inside the triangle defined by <math>\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3</math> is unknown, they can be approximated using [[linear interpolation]].  Barycentric coordinates provide a convenient way to compute this interpolation.  If <math>\mathbf{r}</math> is a point inside the triangle with barycentric coordinates <math>\lambda_{1}</math>, <math>\lambda_{2}</math>, <math>\lambda_{3}</math>, then

:<math>f(\mathbf{r}) \approx \lambda_1 f(\mathbf{r}_1) + \lambda_2 f(\mathbf{r}_2) + \lambda_3 f(\mathbf{r}_3)</math>

In general, given any [[unstructured grid]] or [[polygon mesh]], this kind of technique can be used to approximate the value of <math>f</math> 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 <math>f</math> at a point <math>\mathbf{r}</math>, first a triangle must be found that contains <math>\mathbf{r}</math>.  To do so, <math>\mathbf{r}</math> is transformed into the barycentric coordinates of each triangle.  If some triangle is found such that the coordinates satisfy <math>0 \leq \lambda_i \leq 1 \;\forall\; i \text{ in } 1,2,3</math>, then the point lies in that triangle or on its edge (explained in the previous section). Then the value of <math>f(\mathbf{r})</math> can be interpolated as described above.

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

====Integration over a triangle ====

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 [[Integration by substitution|change of variables]] to any two barycentric coordinates, e.g. <math>\lambda_1,\lambda_2</math>.  Under this change of variables,

:<math>
\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}
</math>

where <math>A</math> is the [[Triangle#Using coordinates|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 <math>2A</math>.

===Examples of special points===

The three [[vertex (geometry)|vertices]] of a triangle have barycentric coordinates <math>1:0:0, \quad 0:1:0, \quad \text{and} \quad 0:0:1.</math><ref name=Scott>Scott, J. A. "Some examples of the use of areal coordinates in triangle geometry", ''[[Mathematical Gazette]]'' 83, November 1999, 472–477.</ref>

The [[centroid]] has barycentrics <math>1:1:1.</math><ref name=Scott/>

The [[circumcenter]] of a triangle ''ABC'' has barycentric coordinates<ref name=Scott/><ref name=Olympiad>{{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|accessdate=14 January 2016|date=July 13, 2012}}</ref><ref name=ck>Clark Kimberling's Encyclopedia of Triangles {{cite web |url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |title=Encyclopedia of Triangle Centers |accessdate=2012-06-02 |url-status=dead |archiveurl=https://web.archive.org/web/20120419171900/http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |archivedate=2012-04-19 }}</ref><ref>[http://mathworld.wolfram.com/BarycentricCoordinates.html Wolfram page on barycentric coordinates]</ref>

:<math> a^2(-a^2 + b^2 + c^2):\;b^2(a^2 - b^2 + c^2):\;c^2(a^2 + b^2 - c^2)</math>

:<math>=\sin 2A :\sin 2B:\sin 2C=(1-\cos B\cos C ):(1-\cos C\cos A):(1-\cos A\cos B).</math>

where {{nowrap|''a'', ''b'', ''c''}} are edge lengths {{nowrap|''BC'', ''CA'', ''AB''}} respectively of the triangle.

The [[orthocenter]] has barycentric coordinates<ref name=Scott/><ref name=Olympiad/>

:<math> (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)</math>

:<math>=\tan A:\tan B:\tan C=a\cos B\cos C:b\cos C\cos A:c\cos A\cos B.</math>

The [[incenter]] has barycentric coordinates<ref name=Olympiad/><ref name=NK>Dasari Naga, Vijay Krishna, "On the Feuerbach triangle",
''Forum Geometricorum'' 17 (2017), 289–300: p. 289. http://forumgeom.fau.edu/FG2017volume17/FG201731.pdf</ref>

:<math>a:b:c=\sin A:\sin B:\sin C.</math>

The [[excenter]]s' barycentrics are<ref name=NK/>

:<math>-a:b:c \quad \quad a:-b:c \quad \quad a:b:-c.</math>

The [[nine-point center]] has barycentric coordinates<ref name=Scott/><ref name=NK/>

:<math>a\cos(B-C):b\cos (C-A):c\cos (A-B)=(1+\cos B\cos C):(1+\cos C\cos A):(1+\cos A\cos B)</math>
::<math>=[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].</math>

==Barycentric coordinates on tetrahedra==

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

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

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

where <math>\mathbf{T}</math> is now a 3×3 matrix:

:<math>
\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)
</math>

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.

== Generalized barycentric coordinates ==
{{unreferenced section|date=December 2010}}

Barycentric coordinates (''a''<sub>1</sub>, ..., ''a''<sub>''n''</sub>) that are defined with respect to a [[polytope]] instead of a simplex are called ''generalized barycentric coordinates''. For these, the equation

:<math>(a_1 + \cdots + a_n) p = a_1 x_1 + \cdots + a_n x_n</math>

is still required to hold where ''x''<sub>1</sub>, ..., ''x''<sub>''n''</sub> are the vertices of the given polytope. Thus, the definition is formally unchanged but while a simplex with ''n'' vertices needs to be embedded in a vector space of dimension of at least ''n-1'', a polytope may be embedded in a vector space of lower dimension. The simplest example is a quadrilateral in the plane. Consequently, even normalized generalized barycentric coordinates (i.e. coordinates such that the sum of the coefficients is 1) are in general not uniquely determined anymore while this is the case for normalized barycentric coordinates with respect to a simplex.

More abstractly, generalized barycentric coordinates express a polytope with ''n'' vertices, regardless of dimension, as the ''image'' of the standard <math>(n-1)</math>-simplex, which has ''n'' vertices – the map is onto: <math>\Delta^{n-1} \twoheadrightarrow P.</math> 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 linear program|Dual]] to generalized barycentric coordinates are [[slack variable]]s, which measure by how much margin a point satisfies the linear constraints, and gives an [[embedding]] <math>P \hookrightarrow (\mathbf{R}_{\geq 0})^f</math> 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 <math>(n-1)</math>-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 <math>K^n</math> as the standard object for vector spaces, and the standard [[affine hyperplane]] <math>\{(x_0,\ldots,x_n) \mid \sum x_i = 1\} \subset K^{n+1}</math> 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.

=== Applications ===
Generalized barycentric coordinates have applications in [[computer graphics]] and more specifically in [[geometric model]]ling. 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]] <ref>ONUFRIEV, VG; DENISIK, SA; FERRONSKY, VI,  BARICENTRIC MODELS IN ISOTOPE STUDIES OF NATURAL-WATERS. NUCLEAR GEOPHYSICS, 4, 111-117 (1990)</ref>

==See also==
*[[Ternary plot]]
*[[Convex combination]]

==References==
{{reflist}}
* {{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. | authorlink=Harold Scott MacDonald Coxeter | title=Introduction to geometry | edition=2nd | publisher=John Wiley and Sons | year=1969 | isbn=978-0-471-50458-0 | pages=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.&nbsp;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.&nbsp;120–127, 2008

==External links==
* [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]].

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[[Category:Linear algebra]]
[[Category:Affine geometry]]
[[Category:Triangle geometry]]
[[Category:Coordinate systems]]