Barycentric coordinates

This can be used in mass points. http://mathworld.wolfram.com/BarycentricCoordinates.html This article is a stub. Help us out by expanding it.

Barycentric coordinates are triples of numbers $(t_1,t_2,t_3)$ corresponding to masses placed at the vertices of a reference triangle $\Delta{A_1}{A_2}{A_3}$ . These masses then determine a point $P$ , which is the geometric centroid of the three masses and is identified with coordinates $(t_1,t_2,t_3)$ . The vertices of the triangle are given by $(1,0,0)$ , $(0,1,0)$ , and $(0,0,1)$ . Barycentric coordinates were discovered by Möbius in 1827 (Coxeter 1969, p. 217; Fauvel et al. 1993).

The Central NC Math Group published a lecture concerning this topic at https://www.youtube.com/watch?v=KQim7-wrwL0 if you would like to view it.

Useful formulas

Let the triangle $\triangle ABC$ be a given triangle, $a, b, c$ be the lengths of $BC, AC, AB, \angle A = \alpha, \angle B = \beta, \angle C = \gamma.$

We use the following Conway symbols:

$s = \frac {a+b+c}{2}$ is semiperimeter, $2S$ is twice the area of $\triangle ABC,$

$r^2 = \frac {(s-a)(s-b)(s-c)}{s},$ where $r$ is the inradius, $R = \frac {abc}{2 \cdot 2S}$ is the circumradius,

$\cos \omega = \frac {a^2 + b^2 +c^2}{2 \cdot 2S}$ is the cosine of the Brocard angle.

$S_A = bc \cos \alpha = \frac{b^2+c^2-a^2}{2}, S_B = ac \cos \beta =\frac{a^2 +c^2-b^2}{2}, S_C = ab \cos \gamma = \frac {a^2+b^2-c^2}{2}.$ For any point in the plane $ABC$ there are barycentric coordinates: $x \cdot \vec {XA} + y \cdot \vec {YB} + z \cdot \vec {XC} = \vec {0},$ $\vec X = \frac {x \cdot \vec {A} + y \cdot \vec {B} + z \cdot \vec {C}}{x+y+z}.$ The normalized (absolute) barycentric coordinates NBC satisfy the condition $x + y + z = 1,$ they are uniquely determined: $x = \frac{[\vec {XB},\vec {XC}]}{\sigma}, y = \frac{[\vec {XC},\vec {XA}]}{\sigma}, z = \frac{[\vec {XA},\vec {XB}]}{\sigma}, \sigma = [\vec {XB},\vec {XC}] + [\vec {XC},\vec {XA}] + [\vec {XA},\vec {XB}] .$ Triangle vertices $A = (1:0:0), B = (0:1:0), C = (0:0:1).$

The barycentric coordinates of a point do not change under an affine transformation.

The straight line in barycentric coordinates (BC) is given by the equation $kx + ly + mz = 0.$

The lines given in the BC by the equations $k_1x + l_1y + m_1z = 0$ and $k_2x + l_2y + m_2z = 0$ intersect at the point $(l_1m_2 – m_1l_2 : m_1k_2-k_1m_2 : k_1l_2-l_1k_2).$

These lines are parallel iff $l_1m_2 – m_1l_2 + m_1k_2-k_1m_2 + k_1l_2-l_1k_2 = 0.$

The sideline $BC$ contains the points $B = (0:1:0), C = (0:0:1),$ its equation is $x = 0.$

The line $AX, X = (k_1 : l_1 : m_1)$ has equation $l_1z = m_1 y,$ it intersects the sideline $BC$ at the point $A_1 = (0 : l_1 : m_1), \frac {BA_1}{A_1C} = \frac {m_1}{l_1}.$

Iff $A_1 = (0 : l_1 : m_1), B_1 = (k_1 : 0 : m_1 ), C_1 = (k_1 : l_1 : 0),$ then $AA_1 \cap BB_1 \cap CC_1 = (k_1 : l_1 : m_1 ).$

Let NBC of points $P$ and $Q$ be $P = (x_1 : y_1 : z_1), Q = (x_2 : y_2 : z_2).$

Then the square of distance $|PQ|^2 = S_A \cdot (x_1 - x_2)^2 + S_B(y_1 - y_2)^2 + S_C(z_1 - z_2)^2.$ $|PQ|^2 = - a^2 (y_1 - y_2)(z_1 - z_2) - b^2 (x_1 - x_2)(z_1 - z_2) - c^2 (x_1 - x_2)(y_1 - y_2).$

Page

Toolbox

Search

Barycentric coordinates

Useful formulas