Template:Short description Template:Distinguish
In geometry, a barycentric coordinate system is a coordinate system in which the location of a point is specified by reference to a simplex (a triangle for points in a plane, a tetrahedron for points in three-dimensional space, etc.). The barycentric coordinates of a point can be interpreted as masses placed at the vertices of the simplex, such that the point is the center of mass (or barycenter) of these masses. These masses can be zero or negative; they are all positive if and only if the point is inside the simplex.
Every point has barycentric coordinates, and their sum is never zero. Two tuples of barycentric coordinates specify the same point if and only if they are proportional; that is to say, if one tuple can be obtained by multiplying the elements of the other tuple by the same non-zero number. Therefore, barycentric coordinates are either considered to be defined up to multiplication by a nonzero constant, or normalized for summing to unity.
Barycentric coordinates were introduced by August Möbius in 1827.<ref>Template:Cite book
Reprinted in Template:Cite book</ref><ref>Max Koecher, Aloys Krieg: Ebene Geometrie. Springer-Verlag, Berlin 2007, Template:ISBN, S. 76.</ref><ref name=Hille>Hille, Einar. "Analytic Function Theory, Volume I", Second edition, fifth printing. Chelsea Publishing Company, New York, 1982, Template:ISBN, page 33, footnote 1</ref> They are special homogeneous coordinates. Barycentric coordinates are strongly related with Cartesian coordinates and, more generally, to affine coordinates (Template:Xref).
Barycentric coordinates are particularly useful in triangle geometry for studying properties that do not depend on the angles of the triangle, such as Ceva's theorem, Routh's theorem, and Menelaus's theorem. In computer-aided design, they are useful for defining some kinds of Bézier surfaces.<ref>Josef Hoschek, Dieter Lasser: Grundlagen der geometrischen Datenverarbeitung. Teubner-Verlag, 1989, Template:ISBN, S. 243.</ref><ref>Gerald Farin: Curves and Surfaces for Computer Aided Geometric Design. Academic Press, 1990, Template:ISBN, S. 20.</ref>
DefinitionEdit
Let <math>A_0, \ldots, A_n</math> be Template:Math points in a Euclidean space, a flat or an affine space <math>\mathbf A</math> of dimension Template:Mvar that are affinely independent; this means that there is no affine subspace of dimension Template:Math that contains all the points,<ref name="Reventós Tarrida">Reventós Tarrida, Agustí. "Affine Maps, Euclidean Motions and Quadrics". Springer, 2011, Template:ISBN, page 11</ref> or, equivalently that the points define a simplex. Given any point <math>P\in \mathbf A,</math> there are scalars <math>a_0, \ldots, a_n</math> that are not all zero, such that <math display=block> ( a_0 + \cdots + a_n ) \overset{}\overrightarrow{OP} = a_0 \overset{}\overrightarrow {OA_0} + \cdots + a_n \overset{}\overrightarrow {OA_n}, </math> for any point Template:Mvar. (As usual, the notation <math>\overset{}\overrightarrow {AB}</math> represents the translation vector or free vector that maps the point Template:Mvar to the point Template:Mvar.)
The elements of a Template:Math tuple <math>(a_0: \dotsc: a_n)</math> that satisfies this equation are called barycentric coordinates of Template:Mvar with respect to <math>A_0, \ldots, A_n.</math> 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 Template:Mvar, the origin, is changed.
The barycentric coordinates of a point are unique up to a scaling. That is, two tuples <math>(a_0: \dotsc: a_n)</math> and <math>(b_0: \dotsc: b_n)</math> are barycentric coordinates of the same point if and only if there is a nonzero scalar <math>\lambda</math> such that <math>b_i=\lambda a_i</math> for every Template:Mvar.
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 <math display=block>\sum a_i = 1,</math> or equivalently by dividing every <math>a_i</math> by the sum of all <math>a_i.</math> These specific barycentric coordinates are called normalized or absolute barycentric coordinates.<ref name=Deaux>Deaux, Roland. "Introduction to The Geometry of Complex Numbers". Dover Publications, Inc., Mineola, 2008, Template:ISBN, page 61</ref> 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 Template:Mvar are all zero, except the one of index Template:Mvar. 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 <math>\{A_0, \ldots, A_n\},</math> which is the simplex that has these points as its vertices.
With above notation, a tuple <math>(a_1, \ldots, a_n)</math> such that <math display=block>\sum_{i=0}^n a_i=0</math> does not define any point, but the vector <math display=block> a_0 \overset{}\overrightarrow {OA_0} + \cdots + a_n \overset{}\overrightarrow {OA_n}</math> is independent from the origin Template:Mvar. As the direction of this vector is not changed if all <math>a_i</math> are multiplied by the same scalar, the homogeneous tuple <math>(a_0: \dotsc: a_n)</math> defines a direction of lines, that is a point at infinity. See below for more details.
Relationship with Cartesian or affine coordinatesEdit
Barycentric coordinates are strongly related to Cartesian coordinates and, more generally, affine coordinates. For a space of dimension Template:Mvar, these coordinate systems are defined relative to a point Template:Mvar, the origin, whose coordinates are zero, and Template:Mvar points <math>A_1, \ldots, A_n,</math> whose coordinates are zero except that of index Template:Mvar that equals one.
A point has coordinates <math display=block>(x_1, \ldots, x_n)</math> for such a coordinate system if and only if its normalized barycentric coordinates are <math display=block>(1-x_1-\cdots - x_n,x_1, \ldots, x_n)</math> relatively to the points <math>O, A_1, \ldots, A_n.</math>
The main advantage of barycentric coordinate systems is to be symmetric with respect to the Template:Math defining points. They are therefore often useful for studying properties that are symmetric with respect to Template:Math 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.
Relationship with projective coordinatesEdit
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 Template:Mvar 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.<ref name="Berger">Template:Citation</ref>
Given a projective space of dimension Template:Mvar, a projective frame is an ordered set of Template:Math points that are not contained in the same hyperplane. A projective frame defines a projective coordinate system such that the coordinates of the Template:Mathth point of the frame are all equal, and, otherwise, all coordinates of the Template:Mvarth point are zero, except the Template:Mvarth one.<ref name=Berger/>
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 Template:Mathth coordinate.
When one has Template:Math 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 Template:Slink.
Barycentric coordinates on trianglesEdit
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 <math>ABC</math> with vertices <math>A=(a_1,a_2)</math>, <math>B=(b_1,b_2)</math>, <math>C=(c_1,c_2)</math> in the x,y-plane, <math>\mathbb{R}^2</math>. One may regard points in <math>\mathbb{R}^2</math> as vectors, so it makes sense to add or subtract them and multiply them by scalars.
Each triangle <math>ABC</math> has a signed area or sarea, which is plus or minus its area:
<math>\operatorname{sarea}(ABC) = \pm \operatorname{area}(ABC).</math>
The sign is plus if the path from <math>A</math> to <math>B</math> to <math>C</math> then back to <math>A</math> goes around the triangle in a counterclockwise direction. The sign is minus if the path goes around in a clockwise direction.
Let <math>P</math> be a point in the plane, and let <math>(\lambda_1,\lambda_2,\lambda_3)</math> be its normalized barycentric coordinates with respect to the triangle <math>ABC</math>, so
<math>P = \lambda_1 A + \lambda_2 B + \lambda_3 C</math>
and
<math>1 = \lambda_1 + \lambda_2 + \lambda_3.</math>
Normalized barycentric coordinates <math>(\lambda_1,\lambda_2,\lambda_3)</math> are also called areal coordinates because they represent ratios of signed areas of triangles:
<math>\begin{align}\lambda_1 &= \operatorname{sarea}(PBC)/\operatorname{sarea}(ABC)\\
\lambda_2 &= \operatorname{sarea}(APC)/\operatorname{sarea}(ABC)\\
\lambda_3 &= \operatorname{sarea}(ABP)/\operatorname{sarea}(ABC).\end{align}</math>
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
<math>D = -A+B+C.</math>
<math>ABCD</math> is a parallelogram because its pairs of opposite sides, represented by the pairs of displacement vectors <math>D-C=B-A</math>, and <math>D-B=C-A</math>, are parallel and congruent.
Triangle <math>ABC</math> is half of the parallelogram <math>ABDC</math>, so twice its signed area is equal to the signed area of the parallelogram, which is given by the <math>2\times 2</math> determinant <math>\det(B-A,C-A)</math> whose columns are the displacement vectors <math>B-A</math> and <math>C-A</math>:
<math display=block>\operatorname{sarea}(ABCD)=\det\begin{pmatrix}b_1-a_1 & c_1-a_1 \\ b_2-a_2 & c_2-a_2\end{pmatrix}</math>
Expanding the determinant, using its alternating and multilinear properties, one obtains
<math>\begin{align}\det(B-A,C-A) &= \det(B,C)-\det(A,C)-\det(B,A)+\det(A,A) \\
&= \det(A,B)+\det(B,C)+\det(C,A) \end{align}</math>
so
<math>2 \operatorname{sarea}(ABC) = \det(A,B)+\det(B,C)+\det(C,A).</math>
Similarly,
<math>2 \operatorname{sarea}(PBC) = \det(P,B)+\det(B,C)+\det(C,P) </math>,
To obtain the ratio of these signed areas, express <math>P</math> in the second formula in terms of its barycentric coordinates:
<math>\begin{align}2 \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}.</math>
The barycentric coordinates are normalized so <math>1 = \lambda_1 + \lambda_2 + \lambda_3</math>, hence <math>\lambda_1 = (1-\lambda_2 - \lambda_3)</math> . Plug that into the previous line to obtain
<math>\begin{align}2 \operatorname{sarea}(PBC) &= \lambda_1 (\det(A,B)+\det(B,C)+\det(C,A)) \\
&= (\lambda_1)(2 \operatorname{sarea}(ABC)).\end{align}</math>
Therefore
<math>\lambda_1 = \operatorname{sarea}(PBC)/\operatorname{sarea}(ABC)</math>.
Similar calculations prove the other two formulas
<math>\lambda_2 = \operatorname{sarea}(APC)/\operatorname{sarea}(ABC)</math>
<math>\lambda_3 = \operatorname{sarea}(ABP)/\operatorname{sarea}(ABC)</math>.
Trilinear coordinates <math>(\gamma_1,\gamma_2,\gamma_3)</math> of <math>P</math> are signed distances from <math>P</math> to the lines BC, AC, and AB, respectively. The sign of <math>\gamma_1</math> is positive if <math>P</math> and <math>A</math> lie on the same side of BC, negative otherwise. The signs of <math>\gamma_2</math> and <math>\gamma_3</math> are assigned similarly. Let
<math>a = \operatorname{length}(BC)</math>, <math>b = \operatorname{length}(CA)</math>, <math>c = \operatorname{length}(AB)</math>.
Then
<math>\begin{align}\gamma_1 a &= \pm 2\operatorname{sarea}(PBC)\\
\gamma_2 b &= \pm 2\operatorname{sarea}(APC)\\
\gamma_3 c &= \pm 2\operatorname{sarea}(ABP)\end{align}</math>
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.
Conversion between barycentric and Cartesian coordinatesEdit
Edge approachEdit
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 display=block>\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}</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 display=block>\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}</math>
Rearranging, this is
<math display=block>\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}</math>
This linear transformation may be written more succinctly as
<math display=block> \mathbf{T} \cdot \lambda = \mathbf{r}-\mathbf{r}_3 </math>
where <math>\lambda</math> is the vector of the first two barycentric coordinates, <math>\mathbf{r}</math> is the vector of Cartesian coordinates, and <math>\mathbf{T}</math> is a matrix given by
<math display=block> \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, 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 collinear and would not form a triangle). Thus, we can rearrange the above equation to get
<math display=block> \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 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 display="block">\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}</math>When understanding the last line of equation, note the identity <math>(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})=(\mathbf{r_3}-\mathbf{r_1})\times(\mathbf{r_1}-\mathbf{r_2})</math>.
Vertex approachEdit
Another way to solve the conversion from Cartesian to barycentric coordinates is to write the relation in the matrix form <math display="block"> \mathbf{R} \boldsymbol{\lambda} = \mathbf{r}</math>with <math>\mathbf{R} = \left(\, \mathbf{r}_1 \,|\, \mathbf{r}_2 \,|\, \mathbf{r}_3 \right)</math> and <math>\boldsymbol{\lambda} = \left(\lambda_1,\lambda_2,\lambda_3\right)^\top,</math> i.e.<math display="block"> \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} </math>To get the unique normalized solution we need to add the condition <math>\lambda_1 + \lambda_2 + \lambda_3 = 1</math>. The barycentric coordinates are thus the solution of the linear system<math display="block"> \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) </math>which is<math display="block"> \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} </math>where <math display="block"> 2A = \det(1|R) = x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)</math>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.
Conversion between barycentric and trilinear coordinatesEdit
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 coordinatesEdit
The three sides a, b, c respectively have equations<ref name=Scott/>
<math display=block>\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 display=block> \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 pointsEdit
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 display=block>\overset{}\overrightarrow{P Q}=(p_1-q_1,p_2-q_2,p_3-q_3).</math>
The distance Template:Mvar between Template:Mvar and Template:Mvar, or the length of the displacement vector <math>\overset{}\overrightarrow{P Q}=(x,y,z),</math> is<ref name=Scott/><ref name=Olympiad/>
<math display=block>\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}</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 display=block>\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}</math>
The barycentric coordinates of a point can be calculated based on distances di to the three triangle vertices by solving the equation <math display=block> \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>
ApplicationsEdit
Determining location with respect to a triangleEdit
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 display=block>\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 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.
Interpolation on a triangular unstructured gridEdit
If <math>f(\mathbf{r}_1),f(\mathbf{r}_2),f(\mathbf{r}_3)</math> are known quantities, but the values of Template:Mvar 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 display=block>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 Template:Mvar 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 Template:Mvar 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 or tetrahedronEdit
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. <math>\lambda_1,\lambda_2</math>. Under this change of variables,
<math display=block> \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 Template:Mvar 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 <math>2A</math>. 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
<math display="block"> \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 </math>where Template:Mvar is the volume of the tetrahedron.
Examples of special pointsEdit
In the homogeneous barycentric coordinate system defined with respect to a triangle <math>ABC</math>, the following statements about special points of <math>ABC</math> hold.
The three vertices Template:Mvar, Template:Mvar, and Template:Mvar have coordinates<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>
<math display=block>\begin{array}{rccccc}
A = & 1 &:& 0 &:& 0 \\ B = & 0 &:& 1 &:& 0 \\ C = & 0 &:& 0 &:& 1
\end{array}</math>
The centroid has coordinates <math>1:1:1.</math><ref name=Scott/>
If Template:Mvar, Template:Mvar, Template:Mvar are the edge lengths <math>BC</math>, <math>CA</math>, <math>AB</math> respectively, <math>\alpha</math>, <math>\beta</math>, <math>\gamma</math> are the angle measures <math>\angle CAB</math>, <math>\angle ABC</math>, and <math>\angle BCA</math> respectively, and Template:Mvar is the semiperimeter of <math>ABC</math>, then the following statements about special points of <math>ABC</math> hold in addition.
The circumcenter has coordinates<ref name=Scott/><ref name=Olympiad>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref><ref name=ck>Clark Kimberling's Encyclopedia of Triangles {{#invoke:citation/CS1|citation |CitationClass=web }}</ref><ref name=":0">Wolfram page on barycentric coordinates</ref>
<math display=block>\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}</math>
The orthocenter has coordinates<ref name=Scott/><ref name=Olympiad/>
<math display=block>\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}</math>
The incenter has coordinates <math>a:b:c=\sin \alpha:\sin \beta:\sin \gamma.</math><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>
The excenters have coordinates<ref name=NK/>
<math display=block>\begin{array}{rrcrcr}
J_A = & -a &:& b &:& c \\ J_B = & a &:& -b &:& c \\ J_C = & a &:& b &:& -c
\end{array}</math>
The nine-point center has coordinates<ref name=Scott/><ref name=NK/>
<math display=block>\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}</math>
The Gergonne point has coordinates <math>(s-b)(s-c):(s-c)(s-a):(s-a)(s-b)</math>.
The Nagel point has coordinates <math>s-a:s-b:s-c</math>.
The symmedian point has coordinates <math>a^2:b^2:c^2</math>.<ref name=":0" />
Barycentric coordinates on tetrahedraEdit
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 <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 display=block> \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 display=block> \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>
and <math>\lambda_4 = 1 - \lambda_1 - \lambda_2 - \lambda_3</math>with the corresponding Cartesian coordinates:<math display="block">\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}</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 coordinatesEdit
Barycentric coordinates <math>(\lambda_1, \lambda_2, ..., \lambda_k)</math> of a point <math>p \in \mathbb{R}^n</math> that are defined with respect to a finite set of k points <math>x_1, x_2, ..., x_k \in \mathbb{R}^n</math> instead of a simplex are called generalized barycentric coordinates. For these, the equation
<math display=block>(\lambda_1 + \lambda_2 + \cdots + \lambda_k)p = \lambda_1 x_1 + \lambda_2 x_2 + \cdots + \lambda_k x_k</math>
is still required to hold.<ref>Template:Cite journal</ref> Usually one uses normalized coordinates, <math>\lambda_1 + \lambda_2 + \cdots + \lambda_k = 1</math>. As for the case of a simplex, the points with nonnegative normalized generalized coordinates (<math>0 \le \lambda_i \le 1</math>) form the convex hull of Template:Math. If there are more points than in a full simplex (<math>k > n + 1</math>) the generalized barycentric coordinates of a point are not unique, as the defining linear system (here for n=2)<math display="block"> \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) </math>is underdetermined. The simplest example is a quadrilateral in the plane. Various kinds of additional restrictions can be used to define unique barycentric coordinates.<ref>Template:Cite journal</ref>
AbstractionEdit
More abstractly, generalized barycentric coordinates express a convex 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 to generalized barycentric coordinates are slack variables, 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.
ApplicationsEdit
Generalized barycentric coordinates have applications in computer graphics and more specifically in geometric modelling.<ref>Template:Cite journal</ref> 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 alsoEdit
ReferencesEdit
<references/>
- 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.
- Template:Cite book
- Template:Cite book
- Barycentric Calculus In Euclidean And Hyperbolic Geometry: A Comparative Introduction, Abraham Ungar, World Scientific, 2010
- Hyperbolic Barycentric Coordinates, Abraham A. Ungar, The Australian Journal of Mathematical Analysis and Applications, Vol.6, No.1, Article 18, pp. 1–35, 2009
- {{#invoke:Template wrapper|{{#if:|list|wrap}}|_template=cite web
|_exclude=urlname, _debug, id |url = https://mathworld.wolfram.com/{{#if:ArealCoordinates%7CArealCoordinates.html}} |title = Areal Coordinates |author = Weisstein, Eric W. |website = MathWorld |access-date = |ref = Template:SfnRef }}
- {{#invoke:Template wrapper|{{#if:|list|wrap}}|_template=cite web
|_exclude=urlname, _debug, id |url = https://mathworld.wolfram.com/{{#if:BarycentricCoordinates%7CBarycentricCoordinates.html}} |title = Barycentric Coordinates |author = Weisstein, Eric W. |website = MathWorld |access-date = |ref = Template:SfnRef }}
- Barycentric coordinates computation in homogeneous coordinates, Vaclav Skala, Computers and Graphics, Vol.32, No.1, pp. 120–127, 2008
External linksEdit
- Law of the lever
- The uses of homogeneous barycentric coordinates in plane euclidean geometry
- Barycentric Coordinates – a collection of scientific papers about (generalized) barycentric coordinates
- Barycentric coordinates: A Curious Application (solving the "three glasses" problem) at cut-the-knot
- Accurate point in triangle test
- Barycentric Coordinates in Olympiad Geometry by Evan Chen and Max Schindler
- Barycenter command and TriangleCurve command at Geogebra.