Difference between revisions of "Barycentric coordinates"

Revision as of 14:59, 25 August 2023

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 $|PQ|^2 = S_A \cdot (x_1 - x_2)^2 + S_B(y_1 - y_2)^2 + S_C(z_1 - z_2)^2.$ $|PQ|^2 = - a^2 (y_1 - y_2)(z_1 - z_2) - b^2 (x_1 - x_2)(z_1 - z_2) - c^2 (x_1 - x_2)(y_1 - y_2).$ The equation of bisector of $PQ$ is: $x(c^2(y_2-y_1) + b^2(z_2-z_1)) + y(a^2(z_2-z_1) + c^2(x_2-x_1)) + z(a^2(y_2-y_1) + b^2(x_2-x_1)) + a^2(y_1z_1 - y_2z_2) + b^2(x_1z_1-x_2z_2) + c^2 (x_1y_1 – x_2y_2).$

Any circle is given by an equation of the form $(kx + ly + mz)(x +y + z) = xyc^2 + xzb^2 + yza^2.$

Circumcircle contains the points $A = (1:0:0), B = (0:1:0), C = (0:0:1) \implies$ the equation of this circle: $xyc^2 + xzb^2 + yza^2.$

The circumcenter $O$ lies at the intersection of the bisectors $AB (c^2(x - y) + z(a^2 – b^2) =0)$ and $AC (b^2(x - z) + y(a^2 – c^2) =0) \implies$ its BC coordinates $O = (a^2S_A : b^2S_B : c^2S_C).$

The incircle contains the tangent points of the incircle with the sides: $\left(0 : \frac {a+b-c}{2a} : \frac {a-b+c}{2a}\right), \left(\frac {a+b-c}{2b} : 0 : \frac {-a+b+c}{2b}\right), \left(\frac {a-b+c}{2c} : \frac {-a+b+c}{2c}\right).$

The incenter is $I = (a : b: c).$ The equation of the incircle is ${k_a}^2x^2 + k_b^2y^2 + k_c^2z^2 - 2k_a k_b xy - 2k_a k_c xz - 2k_bk_cyz = 0,$ where $k_a = b+c - a, k_b = a+c - b, k_c = a + b - c.$

The radical axis of two circles given by equations of this form is: $k_1xa + l_1yb + m_1zc = k_2xa + l_2yb + m_2zc.$ The point $P = (x : y : z)$ is isotomically conjugate with respect to $\triangle ABC$ with the point $P_1 =\left( \frac {1}{x} : \frac {1}{y} : \frac {1}{z}\right).$

The point $P = (x : y : z)$ is isogonally conjugate with respect to $\triangle ABC$ with the point $P_2 =\left( \frac {a^2}{x} : \frac {b^2}{y} : \frac {c^2}{z}\right).$

@@ Line 64: / Line 64: @@
 <cmath>{k_a}^2x^2 + k_b^2y^2 + k_c^2z^2 - 2k_a k_b  xy - 2k_a k_c xz - 2k_bk_cyz = 0,</cmath> where <math>k_a = b+c - a, k_b = a+c - b, k_c = a + b - c.</math>
+The radical axis of two circles given by equations of this form is:
+<cmath>k_1xa + l_1yb + m_1zc = k_2xa + l_2yb + m_2zc.</cmath>
 The point <math>P = (x : y : z)</math> is isotomically conjugate with respect to <math>\triangle ABC</math> with the point <math>P_1 =\left( \frac {1}{x} :  \frac {1}{y} :  \frac {1}{z}\right).</math>
 The point <math>P = (x : y : z)</math> is  isogonally conjugate with respect to <math>\triangle ABC</math> with the point <math>P_2
   =\left( \frac {a^2}{x} :  \frac {b^2}{y} :  \frac {c^2}{z}\right).</math>

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Difference between revisions of "Barycentric coordinates"

Revision as of 14:59, 25 August 2023

Useful formulas