View source for Barycentric coordinates

This can be used in [[mass points]].
http://mathworld.wolfram.com/BarycentricCoordinates.html
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Barycentric coordinates are triples of numbers <math> (t_1,t_2,t_3) </math> corresponding to masses placed at the vertices of a reference triangle <math> \Delta{A_1}{A_2}{A_3} </math>. These masses then determine a point <math> P </math>, which is the geometric centroid of the three masses and is identified with coordinates <math> (t_1,t_2,t_3) </math>. The vertices of the triangle are given by <math> (1,0,0) </math>, <math> (0,1,0) </math>, and <math> (0,0,1) </math>. 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==
<i><b>Notation</b></i>

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

We use the following Conway symbols:

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

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

<math>\cos \omega = \frac {a^2 + b^2 +c^2}{2 \cdot 2S}</math> is the cosine of the Brocard angle,

<cmath>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}.</cmath>

<i><b>Main</b></i>

For any point in the plane  <math>ABC</math> there are <i><b>barycentric coordinates(BC):</b></i> <math>\vec X = (x : y: z)</math>:
<cmath>x \cdot \vec {XA} + y \cdot \vec {YB} + z \cdot \vec {XC} = \vec {0},</cmath>
<cmath>\vec X = \frac {x \cdot \vec {A} + y \cdot \vec {B} + z \cdot \vec {C}}{x+y+z}.</cmath> 
The <i><b>normalized (absolute)</b></i> barycentric coordinates <i><b>NBC</b></i> satisfy the condition <math>x + y + z = 1,</math> they are uniquely determined:
<cmath>x = \frac{[\vec {XB},\vec {XC}]}{\sigma}, y = \frac{[\vec {XC},\vec {XA}]}{\sigma}, z = \frac{[\vec {XA},\vec {XB}]}{\sigma},</cmath>  <cmath>\sigma = [\vec {XB},\vec {XC}] + [\vec {XC},\vec {XA}] + [\vec {XA},\vec {XB}] .</cmath>
Triangle vertices <math>A = (1:0:0), B = (0:1:0), C = (0:0:1).</math>

The barycentric coordinates of a point <i><b> do not change </b></i> under an affine transformation.

<i><b>Lines</b></i>
[[File:Barycentric.png|400px|right]]
The straight line in barycentric coordinates (BC) is given by the equation <math>kx + ly + mz = 0.</math>

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

These lines are parallel iff <math>l_1m_2 – m_1l_2 + m_1k_2-k_1m_2 + k_1l_2-l_1k_2 = 0.</math>

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

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

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

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

Then the square of distance  <cmath>|PQ|^2 = S_A \cdot (x_1 - x_2)^2 + S_B(y_1 - y_2)^2 + S_C(z_1 - z_2)^2.</cmath>
<cmath>|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).</cmath>
The equation of bisector of <math>PQ</math> is:
<cmath>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).</cmath>
Nagel line : <math> (b-c) x + (c-a) y + (a-b) z = 0.</math>

<i><b>Circles</b></i>

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

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

The incircle contains the tangent points of the incircle with the sides:
<cmath>\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).</cmath>

The equation of the incircle is 
<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>
<i><b>Conjugate</b></i>

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>

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

<i><b>Triangle centers</b></i>

The median <math>AA_G, A_G \in BC \implies \frac {BA_G}{CA_G} = 1 \implies</math> <i><b>centroid</b></i> is  <math>G = (1 : 1 : 1).</math> 

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

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

The excenters are <math>I_A = (-a : b : c), I_B =(a : -b: c), I_C = (a : b : -c).</math> 

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

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

Let Nagel point <math>N</math> lies at line <math>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).</math>
 
The Gergonne point is the isotomic conjugate of the Nagel point, so <math>Ge=\left( \frac {1}{b+c-a} : \frac{1}{a+c-b} : \frac{1}{a+b-c}\right ).</math>

'''vladimir.shelomovskii@gmail.com, vvsss'''
== Product of isogonal segments ==
[[File:Barycentric M.png|350px|right]]
[[File:Isogonal formulas.png|350px|right]]
Let triangle <math>\triangle ABC,</math> the circumcircle  <math>\Omega</math> and isogonals <math>AF</math> and <math>AG (F,G \in \Omega)</math> of the <math>\angle BAC</math> be given.
Let point <math>P'</math> and <math>Q'</math> be the isogonal conjugate of a point <math>P</math> and <math>Q</math> with respect to <math>\triangle ABC.</math>
Prove that  <math>PF \cdot P'G = Q'F \cdot QG.</math>

<i><b>Proof</b></i>

We fixed <math>\triangle ABC</math> and the point <math>F.</math> So isogonal <math>AG</math> is fixed.

Denote <math>D = BC \cap AF, E = BC \cap AG.</math>

We need to prove that  <math>PF \cdot P'G</math> do not depends from <math>P.</math>

Line <math>AP</math> has the equation <math>y_P z = z_P y \implies \frac {BD}{DC}  = \frac {z_P}{y_P}.</math>

To find the point <math>F</math> we solve the equation: <cmath>x_F y_P c^2 + x_F z_P b^2 + y_P z_P a^2 = 0.</cmath>

<cmath>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).</cmath>
We use the formula for isogonal cobnjugate point and get 
<cmath>P' = (x_p : y_p : z_p) = \left(\frac {a^2}{x_P} : \frac {b^2}{y_P} :  \frac {c^2}{z_P}\right)</cmath>
and then <math>\frac {BE}{EC}  = \frac {c^2 y_P}{b^2 z_P}.</math>

To find the point <math>G</math> we solve the equation: <cmath>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.</cmath>
<cmath>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).</cmath>
We calculate distances (using NBC) and get:
<cmath>PF \cdot P'G = \frac {a^2 bc y_P z_P}{\psi},</cmath>  <cmath> FG =  \frac {a|b^2 z_P^2 - c^2 y_P^2|}{\psi},</cmath> where <math>\psi</math> has sufficiently big  formula.
  
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==
[[File:Point on incicle.png|350px|right]]
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>

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>

Symilarly denote <math>E_3 \in AC.</math> 

Prove that point <math>F = AD_3 \cap BE_3</math> lies on <math>\omega.</math>

<i><b>Proof</b></i>
<cmath>I = (a : b : c), S = (b+c : a +c : a+b),</cmath> <cmath>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:
<cmath>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</cmath>
<cmath>D_2 =  \left(0 : (a-b)^2 \cdot(b+c-a) : (b-c)^2 \cdot(b+a-c) \right) \implies </cmath>
<cmath>D_3 =  \left(0  : (b-c)^2 \cdot(b+a-c): (a-b)^2 \cdot(b+c-a) \right) . </cmath>
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'''
==Crossing point==
[[File:Point on circumcircle.png|400px|right]]
Let triangle <math>\triangle ABC,</math> and points <math>P</math> and <math>D \in BC</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math>  with respect to a triangle <math>\triangle ABC, \Omega = \odot ABC.</math>
Let <math>X</math> be an arbitrary point at <math>AP', Y = DX \cap AP, Q = DP' \cap AP,</math>
<cmath>Q' = DP \cap AP', E = \odot DP'Q' \cap \Omega, F = \odot DPQ \cap \Omega.</cmath>
Prove that <math>EX \cap FY</math> lies on <math>\Omega.</math>

<i><b>Proof</b></i>

We use the barycentric coordinates:  <cmath>P = (x_P : y_P : z_P), D = (0 : y_D : z_D),</cmath>
<cmath>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). </cmath>
We get the equations for some lines:

Line <math>AP</math> is <math>z_P \cdot y – y_P \cdot z = 0,</math>

line <math>AP'</math> is <math>y_P \cdot y – z_P \cdot z = 0,</math>

line <math>DX</math> is <math>\frac {y_P y_D c^2– z_P z_D b^2}{x_X} \cdot x + b^2 z_P \cdot y + c^2 y_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>

line <math>DP'</math> is <math>\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.</math>

We get the equations for some points:

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>\left( \frac {a^2}{x_Q}  : \frac{b^2}{y_Q} : \frac{c^2}{z_Q} \right),</math>

point <math>Y</math> is <math>\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).</math>

Any circle is given by an equation of the form <math>(kx + ly + mz)(x +y + z) = xyc^2 + xzb^2 + yza^2.</math>
We find the coefficients for the circles (these formulas are big), but can be used for calculations of the crossing points:
<cmath>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),</cmath> 
<cmath>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),</cmath> 
We get the equations for some lines <math>EX</math> and <math>FY</math>:
<cmath>(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,</cmath>

<cmath>kk = (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.</cmath>
We get the equation for the point <math>Z</math>
<cmath>\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).</cmath>
This point satisfies the equation of a circle <math>\Omega.\blacksquare</math>

'''vladimir.shelomovskii@gmail.com, vvsss'''