I manually created my own version of the Wikipedia math article for barycentric coordinates, which are related to triangle geometry including trilinear coordinates.
I made some edits with distance functions in barycentric coordinates and added geometric scenes. I wonder if there are higher dimensional versions of the centroid, orthocenter, Gergonne point, circumcenter, etc. I typed out the Wikipedia article for the Euler line and adding it as an attachment to this post. I have almost finished typing out the article for triangle center.

Computational exploration of Barycentric coordinate system​
​by Peter C . Burbery

In geometry, a barycentric coordinate system is a coordinate 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 an only if the point is inside the simplex.

Every point has barycentric coordinates, and their sum is not 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 Ferdinand Möbius in 1827. They are special homogeneous coordinates. Barycentric coordinates are strongly related with Cartesian coordinates, and more generally, to affine coordinates (see Affine Space § Relationship between barycentric and affine coordinates).

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. In computer-aided design, they useful for defining some kinds of Bezier surfaces.

Conversion between barycentric and Cartesian coordinates

In[]:=

Edge Approach

Given a point r in a triangle's plane one can obtain the barycentric coordinates

λ

1

,

λ

2

,

and

λ

3

fromtheCartesiancoordinates(x,y)

or vice versa.

We can write the Cartesian coordinates of the point r in terms of the Cartesian components of the triangle vertices

r

1

,

r

2

,

r

3

where

r

i

=(

x

i

,

y

i

)

and in terms of the barycentric coordinates of r as:

x= λ 1 x 1 + λ 2 x 2 + λ 3 x 3

y= λ 1 y 1 + λ 2 y 2 + λ 3 y 3

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

λ

3

=1-

λ

1

-

λ

2

into the above to obtain

x= λ 1 x 1 + λ 2 x 2 +(1- λ 1 - λ 2 ) x 3

y= λ 1 y 1 + λ 2 y 2 +(1- λ 1 - λ 2 ) y 3

Rearranging this is

λ 1 ( x 1 - x 3 )+ λ 2 ( x 2 - x 3 )+ x 3 -x=0

λ 1 ( y 1 - y 3 )+ λ 2 ( y 2 - y 3 )+ y 3 -y=0

This linear transformation may be written more succinctly as

T·λ=r-

r

3

where λ is the vector of the first two barycentric coordinates, r is the vector of Cartesian coordinates and T is the matrix given by

T=

x 1 - x 3	x 2 - x 3
y 1 - y 3	y 2 - y 3



In[]:=

(*Comment.IwanttomaketheTnotitalicasavector*)

Now the matrix T is invertible since

r

1

-

r

3

and

r

2

-

r

3

are linearly independent (if this were not the case, then

r

1

,

r

2

, and

r

3

would be collinear and would not form a triangle). Thus, we can rearrange the above equation to get:

λ 1

λ 2

=

-1

T

(r-

r

3

)

Finding the barycentric coordinates has thus been reduced to finding the 2x2 inverse matrix of T, an easy problem.

Explicitly, the formulae for the barycentric coordinates of point r in terms of its Cartesian coordinates

(x,y)

and in terms of the Cartesian coordinates of the triangles vertices are:

λ 1 = ( y 2 - y 3 )(x- x 3 )+( x 3 - y 3 ) det(T) = ( y 2 - y 3 )(x- x 3 )+( x 3 - y 3 ) ( y 1 - y 3 )( x 1 - x 3 )+( x 3 - x 2 )( y 1 - y 3 )

λ 2 = ( y 3 - y 1 )(x- x 3 )+( x 1- x 3 )(y- y 3 ) det(T) = ( 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 ) λ 3 =1- λ 1 - λ 2

In[]:=

Turning Barycentric Coordinates into Cartesian coordinates

We will convert the barycentric coordinates (2,7,1). I am using the digits of  as the example values by the way.

The triangle vectors will be:

r

1

=(8,1)​​

r

2

=(8,2)​​

r

3

=(4,5)

Set up the numbers

In[]:=

barycentricCoordinates={2,7,1};​​r1={8,1};​​r2={8,2};​​r3={4,5};

We can convert this to Cartesian coordinates with the dot product

x= λ 1 x 1 + λ 2 x 2 + λ 3 x 3 =( λ 1 , λ 2 , λ 3 ).( x 1 , x 2 , x 3 )

y= λ 1 y 1 + λ 2 y 2 + λ 3 y 3 =( λ 1 , λ 2 , λ 3 ).( y 1 , y 2 , y 3 )

Evaluate the Cartesian coordinate with the dot product

In[]:=

x=barycentricCoordinates.{r1,r2,r3}[[All,1]]

Out[]=

76

In[]:=

y=barycentricCoordinates.{r1,r2,r3}[[All,2]]

Out[]=

21

The Cartesian coordinates of the barycentric coordinates (2,7,1) in a triangle with vertex vectors at (8, 1), (8, 2), and (4,5) is 76 and 21.

In[]:=

Turning Cartesian coordinates into Barycentric Coordinates

We can find the inverse to double check our work

T=

x 1 - x 3	x 2 - x 3
y 1 - y 3	y 2 - y 3



This time we are going from (8,1), (8,2) and (4,5) and the Cartesian coordinates (76,21) to barycentric coordinates with an inverse matrix

We set the vector r to (x,y)

In[]:=

r={76,21};

In[]:=

MatrixFormT=

r1[[1]]-r3[[1]]	r2[[1]]-r3[[1]]
r1[[2]]-r3[[2]]	r2[[2]]-r3[[2]]



Out[]//MatrixForm=



4	4
-4	-3



We will use this to find

λ

1

and

λ

2

.

λ 1

λ 2

=

-1

T

(r-

r

3

)

In[]:=

{

λ

1

,

λ

2

}=Inverse[T].(r-r3);

λ

3

is related to

λ

1

and

λ

2

by the sum of weights which is unity or

λ

1

+

λ

2

+

λ

3

=1

. Therefore we can calculate

λ

3

with

1-

λ

1

-

λ

2

In[]:=

λ

3

=1-

λ

1

-

λ

2

;

We can now display the three barycentric coordinates

In[]:=

MatrixForm

λ 1

λ 2

λ 3



Out[]//MatrixForm=

-70

88

-17

We can double check our work with the dot product.

In[]:=

{

λ

1

,

λ

2

,

λ

3

}.{r1,r2,r3}[[All,1]]​​{

λ

1

,

λ

2

,

λ

3

}.{r1,r2,r3}[[All,2]]​​{

λ

1

,

λ

2

,

λ

3

}.{r1,r2,r3}[[All,1]]==r[[1]]​​{

λ

1

,

λ

2

,

λ

3

}.{r1,r2,r3}[[All,2]]==r[[2]]

Out[]=

76

Out[]=

21

Out[]=

True

Out[]=

True

In[]:=

Using the explicit relationship

Explicitly, the formulae for the barycentric coordinates of point r in terms of its Cartesian coordinates

(x,y)

and in terms of the Cartesian coordinates of the triangles vertices are:

λ 1 = ( y 2 - y 3 )(x- x 3 )+( x 3 - y 3 ) det(T) = ( 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 )

λ 2 = ( y 3 - y 1 )(x- x 3 )+( x 1- x 3 )(y- y 3 ) det(T) = ( 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 ) λ 3 =1- λ 1 - λ 2

In[]:=

ToPlanarBarycentricCoordinatesExplicit[cartesian_List,vecs_List]:=Module{x1,x2,x3,y1,y2,y3,x,y,λ1,λ2,λ3},x=cartesian[[1]];y=cartesian[[2]];x1=vecs[[1,1]];y1=vecs[[1,2]];x2=vecs[[2,1]];y2=vecs[[2,2]];x3=vecs[[3,1]];y3=vecs[[3,2]];λ1=

(y2-y3)(x-x3)+(x3-x2)(y-y3)

(y2-y3)(x1-x3)+(x3-x2)(y1-y3)

;λ2=

(y3-y1)(x-x3)+(x1-x3)(y-y3)

(y2-y3)(x1-x3)+(x3-x2)(y1-y3)

;λ3=1-λ1-λ2;{λ1,λ2,λ3}/;Dimensions[vecs]=={3,2}

In[]:=

ToPlanarBarycentricCoordinatesExplicit[{76,21},{{8,1},{8,2},{4,5}}]

Out[]=

{-70,88,-17}

There is also a Resource Function BarycentricCoordinates that includes generalized barycentric coordinates (see below)

In[]:=

PersistResourceFunction["BarycentricCoordinates"]

Out[]=

Success

✓

Message: Successfully stored BarycentricCoordinates as an initialization symbol. PersistenceLocation: PersistenceLocation Type: Local 



In[]:=

Vertex Approach

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

Rλ=r

with

R=(

r

1

|

r

2

|

r

3

)andλ=



(

λ

1

,

λ

2

,

λ

3

)

, i.e.



x 1	x 2	x 3
y 1	y 2	y 3



λ 1

λ 2

λ 3

=

x

y



To get the normalized solution we need to add the condition

λ

1

+

λ

2

+

λ

3

=1

. The barycentric coordinates are thus the solutions of the linear system

1	1	1
x 1	x 2	x 3
y 1	y 2	y 3

λ 1

λ 2

λ 3

=

1

x

y

which is

λ 1

λ 2

λ 3

=

1

2A

x 2 y 3 - x 3 y 2	y 2 - y 3	x 3 - x 1
x 3 y 1 - x 1 y 3	y 3 - y 1	x 1 - x 3
x 1 y 2 - x 2 y 1	y 1 - y 2	x 2 - x 1

1

x

y

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.

In[]:=

Solve

1	1	1
x1	x2	x3
y1	y2	y3

.

λ1

λ2

λ3

==

1

x

y

,{λ1,λ2,λ3}

Out[]=

λ1-

-21x2+21x3+76y2-x3y2-76y3+x2y3

x2y1-x3y1-x1y2+x3y2+x1y3-x2y3

,λ2-

-21x1+21x3+76y1-x3y1-76y3+x1y3

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

,λ3-

21x1-21x2-76y1+x2y1+76y2-x1y2

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3



Conversion between barycentric and trilinear coordinates



Equations in Barycentric Coordinates



Distance between points



Applications

In[]:=

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

λ

1…3

>=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 on the open interval (0,1). If a point lies on the edge of a triangle, but not at a vertex, of the area coordinates

λ

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 r lies inside the triangle if and only if

0<

λ

i

<1∀i

in 1, 2, 3. r lies on the edge or corner of the triangle if

0<=

λ

i

<=1∀i

in 1, 2, 3 and

λ

i

=0

, for some i in 1, 2, 3.

Otherwise r 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.

I will design a function that displays a string indicating where a triangle point is with relation to the triangle.

In[]:=

ContainedInTriangleQ[points_]:=AllTrue[ToPlanarBarycentricCoordinatesExplicit[points,trianglePoints],0<#<1&]

In[]:=

ContainedInTriangleQ[{76,21}]

Out[]=

False

In[]:=

ContainedInTriangleQ[{7.8,1.5}]

Out[]=

True

In[]:=

Graphics[{{Opacity[.5],Green,Triangle[trianglePoints]},{Red,PointSize[Large],Point[{7.8,1.5}]}}]

Out[]=

In[]:=

Interpolation on a triangular unstructured grid

If

f(

r

1

),f(

r

2

),f(

r

3

)

are known quantities, but the values of f inside the triangle defined by

r

1

,

r

2

,

r

3

is unkown, they can be approximated using linear interpolation. Barycentric coordinates provide a convenient way to compute this interpolation. If r is a point inside the triangle with barycentric coordinates

λ

1

,

λ

2

,

λ

3

, then

f(r)≈

λ

1

f(

r

1

)+

λ

2

f(

r

2

)+

λ

3

f(

r

3

)

In general, given any unstructured grid or polygon mesh, this kind of technique can be used to approximate the value of f at all points, as long as the function's value is known to all vertices of the mesh. In this case, we have many triangles, each corresponding to a different part of the space. To incorporate a function f at a point r, first a triangle must be found that contains r. To do so, r is transformed into the barycentric coordinates of each triangle. If some triangle is found such that the coordinates satisfy

0<=

λ

i

<=1∀i

in 1, 2, 3, then the point lies in that triangle on its edge (explained in previous section). Then the value of

f(r)

can be interpolated as described above.

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

In[]:=

Integration over a triangle or tetrahedron

The integral of a function over the domain of a triangle can be annoying to compute 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.

λ

1

,

λ

2

. Under this change of variables,

∫

T

f(r)dr=2A

1

∫

0

1-

λ

2

∫

0

f(

λ

1

r

1

+

λ

2

r

2

+(1-

λ

1

-

λ

2

)

r

3

)

λ

1



λ

2

where 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.

∫

∫

T

f(r)dr=6V

1

∫

0

1-

λ

3

∫

0

1-

λ

2

-

λ

3

∫

0

f(

λ

1

r

1

+

λ

2

r

2

+

λ

3

r

3

+(1-

λ

1

-

λ

2

-

λ

3

)

r

4

)

λ

1



λ

2



λ

3

where

V

is the volume of the tetrahedron.

Examples of special points

The three vertices of a triangle have barycentric coordinates 1:0:0, 0:1:0, and 0:0:1.

The centroid has barycentrics 1/3:1/3:1/3.

The circumcenter of a triangle ABC has barycentric coordinates

2

a

(-

2

a

+

2

b

+

2

c

):

2

b

(

2

a

-

2

b

+

2

c

):

2

c

(

2

a

+

2

b

-

2

c

)=sin2A:sin2B:sin2C=(1-cosBcosC):(1-cosCcosA):(1-cosAcosB).

where

a,b,c

are edge lengths BC, CA, AB respectively of the triangle.

The orthocenter has barycentric coordinates

(

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

a

+

2

b

+

2

c

)=tanA:tanB:tanC=acosBcosC:bcosCcosA:ccosAcosB

The incenter has barycentric coordinates

a:b:c=sinA:sinB:sinC

The excenter's barycentrics are

a:b:c=sinA:sinB:sinC

The nine-point center has barycentric coordinates

acos(B-C):bcos(C-A):ccos(A-B)=(1+cosBcosC):(1+cosCcosA):(1+cosAcosB)=

2

a

(

2

b

+

2

c

)-

2

(

2

b

-

2

c

)

:

2

b

(

2

c

+

2

a

)-

2

(

2

c

-

2

a

)

:

2

c

(

2

a

+

2

b

)-

2

(

2

a

-

2

b

)



I wonder if there is a generalization of the centroid, circumcenter, orthocenter, incenter, excenter, nine-point center in higher dimensions. This could be connected to generalized barycentric points. This could be based on simplexes in higher dimensions n and

n



such as the tetrahedron. The cyclical sums could be used to find these.

Generalized Centroid

If two dimensional barycentric coordinates for a simplex's centroid (a triangle in a two dimensional plane) is 1/3:1/3:1/3 then a higher dimensional centroid of a simplex could be 1/n+1:1/n+1:…:1/n+1:1/n+1 n+1 times.

In[]:=

NDimensionalSimplexCentroid[n_]:=

1

Table[n+1,n+1]

A 4 dimensional centroid

In[]:=

NDimensionalSimplexCentroid[4]

Out[]=



1

5

,

1

5

,

1

5

,

1

5

,

1

5



Triangle Centers

In[]:=

Centroid

Verifying the centroid's barycentric coordinates are 1/3:1/3:1/3

In[]:=

TriangleCenter[Triangle[trianglePoints],"Centroid"]​​BarycentricCoordinates[trianglePoints,TriangleCenter[Triangle[trianglePoints],"Centroid"]]

Out[]=



20

3

,

8

3



Out[]=



1

3

,

1

3

,

1

3



Verifying the centroid in barycentric coordinates symbolically

In[]:=

symbolicPoints={{x1,y1},{x2,y2},{x3,y3}};​​{a,b,c}=RotateRight[EuclideanDistance@@@Subsets[symbolicPoints,{2}]/.{

2

Abs[x___]

:>

2

x

}](*Ihavetorotateleftbecausex1andy1shouldhavethedistanceof{x2,y2}to{x3,y3}*)

Out[]=



2

(x2-x3)

+

2

(y2-y3)

,

2

(x1-x2)

+

2

(y1-y2)

,

2

(x1-x3)

+

2

(y1-y3)



In[]:=

TriangleCenter[Triangle[symbolicPoints],"Centroid"]​​BarycentricCoordinates[symbolicPoints,TriangleCenter[Triangle[symbolicPoints],"Centroid"]]​​BarycentricCoordinates[symbolicPoints,TriangleCenter[Triangle[symbolicPoints],"Centroid"]]//FullSimplify

Out[]=



1

3

(x1+x2+x3),

1

3

(y1+y2+y3)

Out[]=



-x3+

1

3

(x1+x2+x3)(y2-y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

+

(-x2+x3)-y3+

1

3

(y1+y2+y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

,

-x3+

1

3

(x1+x2+x3)(-y1+y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

+

(x1-x3)-y3+

1

3

(y1+y2+y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

,1-

-x3+

1

3

(x1+x2+x3)(y2-y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

-

-x3+

1

3

(x1+x2+x3)(-y1+y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

-

(x1-x3)-y3+

1

3

(y1+y2+y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3

-

(-x2+x3)-y3+

1

3

(y1+y2+y3)

-x2y1+x3y1+x1y2-x3y2-x1y3+x2y3



Out[]=



1

3

,

1

3

,

1

3



Make a geometric scene for the centroid. The centroid is the intersection of the a line from each vertex to the midpoint of the opposite side named the median.

In[]:=

SimpleCentroidScene=GeometricScene[{a,b,c,g},{Triangle[{a,b,c}],g==TriangleConstruct[{a,b,c},"Centroid"]}];

In[]:=

RandomInstance[SimpleCentroidScene,RandomSeeding4]

Out[]=

In[]:=

Orthocenter

Find the orthocenter of the triangle example.

In[]:=

exampleOrthocenter=TriangleCenter[Triangle[trianglePoints],"Orthocenter"]

Out[]=

{11,5}

The value of

(

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

a

+

2

b

+

2

c

)

is not normalized so the orthocenter value will be scaled by a certain factor.

In[]:=

({(

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

a

+

2

b

+

2

c

)}.symbolicPoints/.{

2

a

:>

2

(x3-x2)

+

2

(y3-y2)

,

2

b

:>

2

(x3-x1)

+

2

(y3-y1)

,

2

c

:>

2

(x2-x1)

+

2

(y2-y1)

})/.{x1:>8,x2:>8,x3:>4,y1:>1,y2:>2,y3:>5}

Out[]=

{4176,10116}

In this case the scale factor without normalizing the barycentric coordinates for the orthocenter is 64.

In[]:=

{704,320}/{11,5}

Out[]=

{64,64}

Calculate the scaling value symbolically to normalize the orthocenter's barycentric coordinates.

Scale the barycentric coordinates of the orthocenter by the sum to obtain normalized barycentric coordinates and convert to the orthocenter's Cartesian coordinates

In[]:=

{(

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

a

+

2

b

+

2

c

)}Total[{(

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

a

+

2

b

+

2

c

)}].symbolicPoints/.{

2

a

:>

2

(x3-x2)

+

2

(y3-y2)

,

2

b

:>

2

(x3-x1)

+

2

(y3-y1)

,

2

c

:>

2

(x2-x1)

+

2

(y2-y1)

}/.{x1:>8,x2:>8,x3:>4,y1:>1,y2:>2,y3:>5}

Out[]=