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Difference between revisions of "Barycentric coordinates"

(→‎Product of isogonal segments)
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Therefore <cmath>\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</cmath>
 
Therefore <cmath>\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</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
==Point on incircle==
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Let triangle <math>\triangle ABC</math> be given. Denote the incicle <math>\omega,</math> the incenter <math>I</math>, the Spieker center <math>S, D = \omega \cap BC, E = \omega \cap AC.</math>
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Let <math>D_1 \in \omega</math> be the point corresponding to the condition <math>SD = SD_1, D_2 = AD_1 \cap BC, D_3</math> is symmetric <math>D_2</math> with respect midpoint <math>BC.</math>
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Symilarly denote <math>E_3 \in AC.</math>
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Prove that point <math>AD_3 \cap BE_3</math> lies on <math>\omega.</math>
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<i><b>Proof</b></i>
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<cmath>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 ).</cmath>
 +
We calculate distances (using NBC) and solve the system of equations:
 +
<math>ID_1^2 = ID^2, SD_1^2 = SD^2.</math>
 +
We know one solution of this system (point D), so we get linear equation and get:
 +
<math>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</math>
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<cmath>D_2 =  \left(0 : (a-b)^2 \cdot(b+c-a) : (b-c)^2 \cdot(b+a-c) \right) \implies </cmath>
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<cmath>D_3 =  \left(0  : (b-c)^2 \cdot(b+a-c): (a-b)^2 \cdot(b+c-a) \right) . </cmath>
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Similarly <cmath>E_3 =  \left((b-c)^2 \cdot(b+a-c) : 0 : (a-b)^2 \cdot(b+c-a) \right) \implies </cmath>.
 +
Therefore <cmath>F = \left(\frac {(b-c)^2}{b+c-a} : \frac{(a-c)^2}{a+c-b} : \frac{(a-b)^2}{a+b-c}) \right).</cmath>
 +
We calculate the  length of the segment <math>FI</math> and get <math>FI^2 = r^2.</math>
 +
The author learned about the existence of such a point from Leonid Shatunov in August 2023.
 +
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 12:52, 30 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

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 system this equation and equation for circumcircle: \[xyc^2 + xzb^2 + yza^2 = 0.\]

\[F = (x_F : y_F : z_F) = \left(-y_P z_P a^2 : y_P(c^2 y_P +b^2 z_P) : z_P(c^2 y_P +b^2 z_P)\right).\] 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 $\frac {BE}{EC} = \frac {c^2 y_P}{b^2 z_P}.$

\[G = (x_G : y_G : z_G) = \left(-y_p z_p a^2 : y_p(c^2 y_p +b^2 z_p) : z_p(c^2 y_p +b^2 z_p)\right).\] 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

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 $AD_3 \cap BE_3$ 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

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