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Difference between revisions of "Barycentric coordinates"

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line <math>AP'</math> is <math>y_P \cdot y – z_P \cdot z = 0,</math>
 
line <math>AP'</math> is <math>y_P \cdot y – z_P \cdot z = 0,</math>
  
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line <math>DX</math> is <math>\frac {y_P y_D c^2– z_P z_D b^2}{x_X y_P z_P} \cdot x + z_D \cdot y - y_D \cdot z = 0,</math>
+
line <math>DX</math> is <math>\frac {c^2 y_P y_D – b^2 z_P z_D}{x_X y_P z_P} \cdot x + z_D \cdot y - y_D \cdot z = 0,</math>
  
 
line <math>DP</math> is <math>(z_D y_P - y_D z_P ) \cdot x - z_D x_P \cdot y + y_D x_P \cdot z = 0,</math>
 
line <math>DP</math> is <math>(z_D y_P - y_D z_P ) \cdot x - z_D x_P \cdot y + y_D x_P \cdot z = 0,</math>
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We get the equations for some points:
 
We get the equations for some points:
  
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point <math>Q</math> is <math>(x_Q : y_Q : z_Q) = \left( \frac {a^2 y_P z_P (y_P z_D – z_P y_D)}{x_P (z_P z_D b^2 – y_P y_D c^2)} : y_P : z_P \right),</math>
+
point <math>Q</math> is <math>(x_Q : y_Q : z_Q) = \left( \frac {a^2 y_P z_P (z_P y_D - y_P z_D)}{x_P (c^2 y_P y_D - b^2 z_P z_D)} : y_P : z_P \right),</math>
  
 
point <math>Q'</math> is <math>\left( \frac {a^2}{x_Q}  : \frac{b^2}{y_Q} : \frac{c^2}{z_Q} \right),</math>
 
point <math>Q'</math> is <math>\left( \frac {a^2}{x_Q}  : \frac{b^2}{y_Q} : \frac{c^2}{z_Q} \right),</math>

Revision as of 03:13, 9 September 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

Notation

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,

\[S_A = bc \cos \alpha = \frac{b^2+c^2-a^2}{2}, S_B = ac \cos \beta =\frac{a^2 +c^2-b^2}{2}, S_C = ab \cos \gamma = \frac {a^2+b^2-c^2}{2}.\]

Main

For any point in the plane $ABC$ there are barycentric coordinates(BC): $\vec X = (x : y: z)$: \[x \cdot \vec {XA} + y \cdot \vec {YB} + z \cdot \vec {XC} = \vec {0},\] \[\vec X = \frac {x \cdot \vec {A} + y \cdot \vec {B} + z \cdot \vec {C}}{x+y+z}.\] 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.

Lines

Barycentric.png

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 $AX, X = (k : l  : m)$ has equation $l z = m y,$ it intersects the sideline $BC$ at the point $A_X = (0 : l : m), \frac {BA_X}{A_XC} = \frac {m}{l}, \frac {AX}{XA_X} = \frac {m + l}{k}.$

Iff $A_X = (0 : l : m), B_X = (k  : 0 : m ), C_X = (k  : l  : 0),$ then $AA_X \cap BB_X \cap CC_X = (k  : l : m ).$

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: \[x(c^2(y_2-y_1) + b^2(z_2-z_1)) + y(a^2(z_2-z_1) + c^2(x_2-x_1)) + z(a^2(y_2-y_1) + b^2(x_2-x_1)) + a^2(y_1z_1 - y_2z_2) + b^2(x_1z_1-x_2z_2) + c^2 (x_1y_1 – x_2y_2).\] Nagel line : $(b-c) x + (c-a) y + (a-b) z = 0.$

Circles

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 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 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: \[k_1xa + l_1yb + m_1zc = k_2xa + l_2yb + m_2zc.\] Conjugate

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 $P_2 =\left( \frac {a^2}{x} : \frac {b^2}{y} : \frac {c^2}{z}\right).$

The point $P = (x : y : z)$ is isocircular conjugate with respect to $\triangle ABC$ with the point $P_3 = \left(\frac {x}{a} : \frac {y}{b} : \frac {z}{c}\right).$

Triangle centers

The median $AA_G, A_G \in BC \implies \frac {BA_G}{CA_G} = 1 \implies$ centroid is $G = (1 : 1 : 1).$

The simmedian point $K$ is isogonally conjugate with respect to $\triangle ABC$ with the point $G \implies K = \left( a^2 : b^2 : c^2\right).$

The bisector $AA_I, A_I \in BC \implies \frac {BA_I}{CA_I} = \frac {c}{b} \implies$ the incenter is $I = (a : b: c).$

The excenters are $I_A = (-a : b : c), I_B =(a : -b: c), I_C = (a : b : -c).$

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 orthocenter $H$ is isogonally conjugate with respect to $\triangle ABC$ with the point $O \implies H =\left( \frac {1}{S_A} : \frac {1}{S_B} : \frac {1}{S_C}\right).$

Let Nagel point $N$ lies at line $AA_N, A_N \in BC \implies \frac {BA_N}{A_NC} = \frac {a+c-b}{a+b-c} \implies N=(b+c-a: a+c-b:a+b-c).$

The Gergonne point is the isotomic conjugate of the Nagel point, so $Ge=\left( \frac {1}{b+c-a} : \frac{1}{a+c-b} : \frac{1}{a+b-c}\right ).$

vladimir.shelomovskii@gmail.com, vvsss

Product of isogonal segments

Barycentric M.png
Isogonal formulas.png

Let triangle $\triangle ABC,$ the circumcircle $\Omega$ and isogonals $AF$ and $AG (F,G \in \Omega)$ of the $\angle BAC$ be given. Let point $P'$ and $Q'$ be the isogonal conjugate of a point $P$ and $Q$ with respect to $\triangle ABC.$ Prove that $PF \cdot P'G = Q'F \cdot QG.$

Proof

We fixed $\triangle ABC$ and the point $F.$ So isogonal $AG$ is fixed.

Denote $D = BC \cap AF, E = BC \cap AG.$

We need to prove that $PF \cdot P'G$ do not depends from $P.$

Line $AP$ has the equation $y_P z = z_P y \implies \frac {BD}{DC} = \frac {z_P}{y_P}.$

To find the point $F$ we solve the equation: \[x_F y_P c^2 + x_F z_P b^2 + y_P z_P a^2 = 0.\]

\[F = (x_F : y_F : z_F) = \left(\frac{- a^2 y_P z_P}{c^2 y_P +b^2 z_P} : y_P : z_P \right).\] We use the formula for isogonal cobnjugate point and get \[P' = (x_p : y_p : z_p) = \left(\frac {a^2}{x_P} : \frac {b^2}{y_P} : \frac {c^2}{z_P}\right)\] and then $\frac {BE}{EC} = \frac {c^2 y_P}{b^2 z_P}.$

To find the point $G$ we solve the equation: \[x_G \cdot \frac {b^2}{y_P} \cdot c^2 + x_G \cdot \frac {c^2}{z_P} \cdot b^2 + \frac {b^2 c^2}{y_P \cdot z_P} \cdot a^2 = 0.\] \[G = (x_G : y_G : z_G) = \left(\frac{- a^2}{y_p + z_p} : \frac {b^2}{y_p} : \frac {c^2}{z_p}\right).\] We calculate distances (using NBC) and get: \[PF \cdot P'G = \frac {a^2 bc y_P z_P}{\psi},\] \[FG = \frac {a|b^2 z_P^2 - c^2 y_P^2|}{\psi},\] where $\psi$ has sufficiently big formula.

Therefore \[\frac {FG\cdot a\cdot c}{b\cdot PF \cdot P'G} = \left|\frac {z_P}{y_P} - \frac {c^2 y_P}{b^2 z_P}\right| = \left|\frac {BD}{DC}- \frac {BE}{EC}\right|. \blacksquare\] vladimir.shelomovskii@gmail.com, vvsss

Point on incircle

Point on incicle.png

Let triangle $\triangle ABC$ be given. Denote the incicle $\omega,$ the incenter $I$, the Spieker center $S, D = \omega \cap BC, E = \omega \cap AC.$

Let $D_1 \in \omega$ be the point corresponding to the condition $SD = SD_1, D_2 = AD_1 \cap BC, D_3$ is symmetric $D_2$ with respect midpoint $BC.$

Symilarly denote $E_3 \in AC.$

Prove that point $F = AD_3 \cap BE_3$ lies on $\omega.$

Proof \[I = (a : b : c), S = (b+c : a +c : a+b),\] \[D = \left(0 : a+b-c : a-b+c \right), D_1 = (x : y : z ).\] We calculate distances (using NBC) and solve the system of equations: $ID_1^2 = ID^2, SD_1^2 = SD^2.$

We know one solution of this system (point D), so we get linear equation and get: \[D_1 = \left((b-c)^2 \cdot (3a-b-c)^2 : (a-b)^2 \cdot(b+c-a)\cdot(-b+a+c) : (a-c)^2\cdot(b+c-a) \cdot(b+a-c) \right) \implies\] \[D_2 = \left(0 : (a-b)^2 \cdot(b+c-a) : (b-c)^2 \cdot(b+a-c) \right) \implies\] \[D_3 = \left(0  : (b-c)^2 \cdot(b+a-c): (a-b)^2 \cdot(b+c-a) \right) .\] Similarly \[E_3 = \left((b-c)^2 \cdot(b+a-c) : 0 : (a-b)^2 \cdot(b+c-a) \right) \implies\] Therefore \[F = \left(\frac {(b-c)^2}{b+c-a} : \frac{(a-c)^2}{a+c-b} : \frac{(a-b)^2}{a+b-c}) \right).\] We calculate the length of the segment $FI$ and get $FI^2 = r^2.$

The author learned about the existence of such a point from Leonid Shatunov in August 2023.

vladimir.shelomovskii@gmail.com, vvsss

Crossing point

Point on circumcircle.png

Let triangle $\triangle ABC,$ and points $P$ and $D \in BC$ be given. Let point $P'$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC, \Omega = \odot ABC.$ Let $X$ be an arbitrary point at $AP', Y = DX \cap AP, Q = DP' \cap AP,$ \[Q' = DP \cap AP', E = \odot DP'Q' \cap \Omega, F = \odot DPQ \cap \Omega.\] Prove that $EX \cap FY$ lies on $\Omega.$

Proof

We use the barycentric coordinates: \[P = (x_P : y_P : z_P), D = (0 : y_D : z_D),\] \[X = \left ( x_X : \frac {b^2}{y_P} : \frac {c^2}{z_P} \right), P' = \left (\frac {a^2}{x_P} : \frac {b^2}{y_P} : \frac {c^2}{z_P} \right).\] We get the equations for some lines:

Line $AP$ is $z_P \cdot y – y_P \cdot z = 0,$

line $AP'$ is $y_P \cdot y – z_P \cdot z = 0,$

line $DX$ is $\frac {c^2 y_P y_D – b^2 z_P z_D}{x_X y_P z_P} \cdot x + z_D \cdot y - y_D \cdot z = 0,$

line $DP$ is $(z_D y_P - y_D z_P ) \cdot x - z_D x_P \cdot y + y_D x_P \cdot z = 0,$

line $DP'$ is $\left( \frac {b^2 z_D}{y_P}- \frac{c^2 y_D}{z_P}\right) \cdot x - \frac {a^2 z_D}{x_P}\cdot y + \frac{a^2 y_D}{x_P} \cdot z = 0.$

We get the equations for some points:

point $Q$ is $(x_Q : y_Q : z_Q) = \left( \frac {a^2 y_P z_P (z_P y_D - y_P z_D)}{x_P (c^2 y_P y_D - b^2 z_P z_D)} : y_P : z_P \right),$

point $Q'$ is $\left( \frac {a^2}{x_Q}  : \frac{b^2}{y_Q} : \frac{c^2}{z_Q} \right),$

point $Y$ is $\left( \frac {x_X y_P z_P ( y_D z_P - y_P z_D)}{c^2 y_D y_P – b^2 z_P z_D}  : y_P : z_P \right).$

Any circle is given by an equation of the form $(kx + ly + mz)(x +y + z) = xyc^2 + xzb^2 + yza^2.$ We find the coefficients for the circles (these formulas are big), but can be used for calculations of the crossing points: \[E = \left( a^2 : b^2 (\frac {z_P y_D}{y_P z_D} - 1) : c^2 (\frac {y_P z_D}{z_P y_D} - 1 \right),\] \[F = \left( a^2 : -b^2 + c^2 \frac {y_P y_D}{z_P z_D} : b^2 \frac {z_P z_D}{y_P y_D} - c^2 \right),\] We get the equations for some lines $EX$ and $FY$: \[(y_D + z_D) \cdot (y_D z_P – y_P z_D)\cdot x + \frac{y_P z_D ((y_P z_D – y_D z_P) x_X + y_D a^2}{b^2}\cdot y + \frac {(y_P z_D - y_D z_P) x_X + a^2 z_D ) z_P y_D}{c^2} \cdot z = 0,\]

\[(y_D + z_D)(c^2 y_D y_P - b^2 z_D z_P) x + ((y_P z_D - y_D z_P) x_X - y_D a^2) z_D z_P y + y_D y_P(y_P z_D - y_D z_P) x_X + a^2 z_D) z = 0.\] We get the equation for the point $Z$ \[\left(\frac {1}{y_D + z_D} : \frac {b^2}{ x_X (y_P z_D - z_P y_D) - a^2 y_D} :\frac {c^2}{x_X (z_P y_D - y_P z_D) - a^2 z_D} \right).\] This point satisfies the equation of a circle $\Omega.\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

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