Editing Barycentric coordinate system (section)

==Barycentric coordinates on triangles==
{{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 [[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>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 [[Determinant#Properties of the determinant|''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 coordinates ===

==== Edge approach ====
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 [[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 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 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 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 [[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 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 approach ====
Another way to solve the conversion from Cartesian to barycentric coordinates is to write the relation in the [[Matrix (mathematics)|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 [[System of linear equations|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 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 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 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 display=block>\overset{}\overrightarrow{P Q}=(p_1-q_1,p_2-q_2,p_3-q_3).</math>

The distance {{mvar|d}} between {{mvar|P}} and {{mvar|Q}}, 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 ''d''<sub>''i''</sub> 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>

=== Applications===
[[File:3_jugs_puzzle_barycentric_plot.svg|thumb|upright=1.25|Two solutions to the 8, 5 and 3&thinsp;L [[water pouring puzzle]] 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&thinsp;L has been measured.]]

====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 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 grid ====
[[File:Piecewise linear function2D.svg|thumb|upright=1.3|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 {{mvar|f}} 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 {{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 <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 tetrahedron ====

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 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 {{mvar|A}} 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>. 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 {{mvar|V}} is the volume of the tetrahedron.

===Examples of special points===
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 [[vertex (geometry)|vertices]] {{mvar|A}}, {{mvar|B}}, and {{mvar|C}} 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 {{mvar|a}}, {{mvar|b}}, {{mvar|c}} are the [[length|edge lengths]] <math>BC</math>, <math>CA</math>, <math>AB</math> respectively, <math>\alpha</math>, <math>\beta</math>, <math>\gamma</math> are the [[angle | angle measures]] <math>\angle CAB</math>, <math>\angle ABC</math>, and <math>\angle BCA</math> respectively, and {{mvar|s}} 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>{{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}}</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 |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 }}</ref><ref name=":0">[http://mathworld.wolfram.com/BarycentricCoordinates.html 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 [[excenter]]s 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|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" />