Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Barycentric coordinates"

(→‎Product of isogonal segments)
(→‎Set of parallel lines)
 
(61 intermediate revisions by 2 users not shown)
Line 12: Line 12:
 
<i><b>Notation</b></i>
 
<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>
+
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:
 
We use the following Conway symbols:
Line 26: Line 26:
 
<i><b>Main</b></i>
 
<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>:
+
For any point in the plane  <math>ABC</math> there are <i><b>barycentric coordinates(BC):</b></i> <cmath>\vec X = (x_X : y_X : z_X)</cmath>
−
<cmath>x \cdot \vec {XA} + y \cdot \vec {YB} + z \cdot \vec {XC} = \vec {0},</cmath>
+
<cmath>x_X \cdot \vec {XA} + y_X \cdot \vec {XB} + z_X \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>  
+
<cmath>\vec X = \frac {x_X \cdot \vec {A} + y_X \cdot \vec {B} + z_X \cdot \vec {C}}{x_X + y_X + z_X}.</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:
+
The <i><b>normalized (absolute)</b></i> barycentric coordinates <i><b>NBC</b></i> satisfy the condition <math>x_X + y_X + z_X = 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>
+
<cmath>x_X = \frac{[\vec {XB},\vec {XC}]}{\sigma}, y_X = \frac{[\vec {XC},\vec {XA}]}{\sigma}, z_X = \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>
 
Triangle vertices <math>A = (1:0:0), B = (0:1:0), C = (0:0:1).</math>
  
Line 37: Line 37:
 
<i><b>Lines</b></i>
 
<i><b>Lines</b></i>
 
[[File:Barycentric.png|400px|right]]
 
[[File:Barycentric.png|400px|right]]
−
The straight line in barycentric coordinates (BC) is given by the equation <math>kx + ly + mz = 0.</math>
+
The straight line in barycentric coordinates (BC) is given by the equation <cmath>kx + ly + mz = 0.</cmath>
  
 
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  
 
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  
Line 52: Line 52:
 
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>
 
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>
+
Then the square of distance  <cmath>|PQ|^2 = S_A \cdot (x_1 - x_2)^2 + S_B \cdot (y_1 - y_2)^2 + S_C \cdot (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>
 
<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:
 
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>
+
<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) =0.</cmath>
 
Nagel line : <math> (b-c) x + (c-a) y + (a-b) z = 0.</math>
 
Nagel line : <math> (b-c) x + (c-a) y + (a-b) z = 0.</math>
  
Line 63: Line 63:
  
 
Circumcircle contains the points <math>A = (1:0:0), B = (0:1:0), C = (0:0:1) \implies</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 equation of this circle:  <cmath>xyc^2 + xzb^2 + yza^2 = 0.</cmath>
  
 
The incircle contains the tangent points of the incircle with the sides:
 
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>
+
<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} : 0\right).</cmath>
  
 
The equation of the incircle is  
 
The equation of the incircle is  
Line 74: Line 74:
  
 
The radical axis of two circles given by equations of this form is:
 
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>
+
<cmath>(k_1 - k_2) a \cdot x + (l_1 - l_2) b \cdot y + (m_1 - m_2) c \cdot z = 0.</cmath>
 
<i><b>Conjugate</b></i>
 
<i><b>Conjugate</b></i>
  
Line 94: Line 94:
 
The excenters are <math>I_A = (-a : b : c), I_B =(a : -b: c), I_C = (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 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>
 
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>
Line 103: Line 103:
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
== Product of isogonal segments ==
 
== Product of isogonal segments ==
 
[[File:Barycentric M.png|350px|right]]
 
[[File:Barycentric M.png|350px|right]]
Line 111: Line 112:
  
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
 +
 
We fixed <math>\triangle ABC</math> and the point <math>F.</math> So isogonal <math>AG</math> is fixed.
 
We fixed <math>\triangle ABC</math> and the point <math>F.</math> So isogonal <math>AG</math> is fixed.
  
Line 119: Line 121:
 
Line <math>AP</math> has the equation <math>y_P z = z_P y \implies \frac {BD}{DC}  = \frac {z_P}{y_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 system this equation and equation for circumcircle: <math>xyc^2 + xzb^2 + yza^2 = 0.</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) = (-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)).</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  
 
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>
+
<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>
 
and then <math>\frac {BE}{EC}  = \frac {c^2 y_P}{b^2 z_P}.</math>
  
−
<cmath>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).</cmath>
+
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:
 
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.
 
<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|.</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'''
 +
 
 +
== Ratio of isogonal segments ==
 +
[[File:Isogonals division.png|350px|right]]
 +
Let triangle <math>\triangle ABC</math> and point <math>P</math> be given.
 +
Denote  <math>P'</math> the isogonal conjugate of a point <math>P</math> with respect to <math>\triangle ABC, \Omega = \odot ABC,</math>
 +
<cmath>D = AP' \cap BC, E = AP \cap BC, L = AP \cap  \Omega.</cmath>
 +
Prove that  <math>\frac {AP'}{P'D} \cdot \frac {AP}{PE} = \frac {AL}{LE}.</math>
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
We use the formula for isogonal conjugate point and get <cmath>P = (x_P : y_P : z_P), 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>
 +
<cmath>\frac {AP'}{P'D} = \frac {y_{P'} + z_{P'}}{x_{P'}},</cmath>
 +
<cmath> \frac {AP}{PE} = \frac {y_{P} + z_{P}}{x_{P}}.</cmath>
 +
<cmath>L \in \Omega \implies  x_{L'} + y_{L'} + z_{L'} = 0, L \in AP \implies y_L = y_P, z_L = z_P \implies \frac {a^2}{x_L} + y_{P'} + z_{P'} = 0 \implies x_L = \frac{-a^2}{y_{P'} + z_{P'}}.</cmath>
 +
<cmath>\frac {AL}{LE} = \frac {y_{P} + z_{P}}{-x_{L}} =  \frac {(y_{P} + z_{P})(y_{P'} + z_{P'})}{a^2}.</cmath>
 +
<cmath>x_P \cdot x_{P'} = a^2 \implies  \frac {AP'}{P'D} \cdot \frac {AP}{PE} = \frac {AL}{LE}.</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 incircle <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>
 +
 
 +
This configuration can be used as a straight-line mechanism since it allows to create a mechanism that converts the rotational motion of a point Z  to perfect straight-line motion of the X point or vice versa. Of course, we need to use the prismatic joint at the points <math>E</math> and <math>F.</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>c^2 y_P \cdot y - b^2 z_P \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>\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 (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>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>(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>
 +
Let point <math>Z'</math> be the isogonal conjugate of a point <math>Z</math>  with respect to a triangle <math>\triangle ABC.</math>
 +
<cmath>Z' = \left(a^2 (y_D + z_D) : x_X (y_P z_D - z_P y_D) - a^2 y_D : x_X (z_P y_D - y_P z_D) - a^2 z_D \right).</cmath>
 +
The sum of coordinates is equal zero, so <math>Z'</math> is in infinity, therefore the point <math>Z</math> lies on <math>\Omega.\blacksquare</math>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Fixed point on circumcircle==
 +
[[File:Fixed point 2.png|350px|right]]
 +
Let triangle <math>\triangle ABC,</math> point <math>G \ne A</math> on circumcircle <math>\Omega = \odot ABC,</math> and point <math>D \in BC</math> be given.
 +
Point <math>P</math> lies on <math>AG,</math> point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC, Q = DP' \cap AP, F = \odot DPQ \cap \Omega.</math>
 +
 
 +
Prove that <math>F</math> is fixed point and not depends from position of <math>P.</math>
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
Denote the coordinates of the points <math>D = (0 : y_D : z_D), G = (x_G : y_G : z_G).</math>
 +
<cmath>G \in \Omega \implies a^2 y_G z_G + b^2 x_G z_G + c^2 x_G y_G = 0 \implies</cmath>
 +
<cmath>G = \left( \frac {- a^2 y_G z_G}{b^2 z_G + c^2 y_G} : y_G : z_G\right).</cmath>
 +
<cmath>P = (x_P : y_G : z_G) \implies P' = \left( \frac {a^2}{x_P} : \frac {b^2}{y_G} : \frac {c^2}{z_G}\right).</cmath>
 +
The line <math>AG</math> is <math>z_G y = y_G z.</math>
 +
 
 +
The line <math>DP'</math> is <math>(y_{P'} z_D - z_{P'} y_D) x  - x_{P'} z_D y + x_{P'} y_D z = 0 \implies</math>
 +
<cmath>Q =\left( \frac {a^2 (y_D z_G - y_G z_D) \cdot y_G z_G}{x_P (c^2 y_D y_G - b^2 z_D z_P)} : y_G : z_G \right).</cmath>
 +
We find the circle <math>\odot PQD</math> and get the point
 +
<cmath>F =\left( \frac {a^2}{\frac {c^2}{z_D \cdot z_G} - \frac{b^2}{y_D \cdot y_G}} : y_G \cdot y_D : - z_G \cdot z_D \right).</cmath>
 +
<math>F</math> depends only from points <math>G</math> and <math>D.</math>
 +
 
 +
[[Isogonal conjugate | Fixed point on circumcircle]]
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Two pare isogonal points==
 +
[[File:2 pare Miquel.png|400px|right]]
 +
Let triangle <math>\triangle ABC,</math> and points <math>P</math> and <math>Q</math> (points do not lie on sidelines) 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 a triangle <math>\triangle ABC, \Omega = \odot ABC.</math>
 +
 
 +
Denote <math>R = PQ \cap P'Q', E = \Omega \cap RPQ', F = \Omega \cap RQP'.</math>
 +
 
 +
Prove that <math>L= EP \cap FQ</math> and <math>K = EQ' \cap FP'</math> lies on <math>\Omega.</math>
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
The line <math>PQ</math> is
 +
<cmath>(y_P z_Q - z_P y_Q) x + (x_Q z_P – x_P z_Q) y + (x_P y_Q - x_Q y_P)z = 0.</cmath>
 +
The line <math>P'Q'</math> is
 +
<cmath>(y_P z_Q - z_P y_Q) \frac {x_P x_Q}{a^2} x + (x_Q z_P – x_P z_Q) \frac {y_P y_Q}{b^2}y + (x_P y_Q - x_Q y_P) \frac {z_P z_Q}{c^2}z = 0.</cmath>
 +
<cmath>R = PQ \cap P'Q' = \left ( a^2 \frac {(b^2 z_P z_Q – c^2 y_P y_Q)}{y_P z_Q – z_P y_Q} : b^2 \frac {(a^2 z_P z_Q – c^2 x_P x_Q)}{x_P z_Q – z_P x_Q}  : c^2 \frac {(a^2 y_P y_Q – b^2 x_P x_Q)}{x_P y_Q – y_P x_Q} \right)</cmath>
 +
<cmath>E = \Omega \cap RPQ' = \left( \frac {a^2}{x_Q (z_P y_Q – y_P z_Q)} : \frac {b^2}{y_Q (x_P z_Q – z_P y_Q)} :  \frac {c^2}{z_Q (y_P x_Q – x_P y_Q)} \right).</cmath>
 +
<cmath>F = \Omega \cap RP'Q = \left( \frac {a^2}{x_P (z_P y_Q – y_P z_Q)} : \frac {b^2}{y_P (x_P z_Q – z_P y_Q)} :  \frac {c^2}{z_P (y_P x_Q – x_P y_Q)} \right).</cmath>
 +
<cmath>K = EQ' \cap FP' =  \left( \frac {a^2}{x_Q (y_P + z_P) - x_P (y_Q + z_Q)} : \frac {b^2}{y_Q (x_P + z_P) - y_P (z_Q + x_Q)} :  \frac {c^2}{z_Q (x_P + y_P) - z_P (x_Q + y_Q)} \right).</cmath>
 +
Denote  <math>K'</math> is the isogonal conjugate of a point <math>K</math> with respect to  <math>\triangle ABC.</math>
 +
<cmath>K' =  \left( x_Q (y_P + z_P) - x_P (y_Q + z_Q) : y_Q (x_P + z_P) – y_P (z_Q + x_Q) : z_Q (x_P + y_P) – z_P (x_Q + y_Q) \right).</cmath>
 +
<cmath>x_{K'} + y_{K'} + z_{K'} = 0 \implies K \in \Omega.</cmath>
 +
If we use NBC, we get <cmath>K = \left( \frac {a^2}{x_Q - x_P} : \frac {b^2}{y_Q - y_P } :  \frac {c^2}{z_Q - z_P } \right) \implies K' = (x_Q - x_P : y_Q - y_P : z_Q - z_P).</cmath>
 +
<cmath>L = EP \cap FQ =  \left( \frac {1}{\frac {b^2} {x_Q y_P} - \frac {b^2} {x_P y_Q}+ \frac {c^2} {x_Q z_P}- \frac {c^2} {x_Q z_P}} : \frac {1}{\frac {a^2} {x_Q y_P} - \frac {a^2} {x_P y_Q}+ \frac {c^2} {z_Q y_P}- \frac {c^2} {y_Q z_P}}  : \frac {1}{\frac {a^2} {x_Q z_P} - \frac {a^2} {x_P z_Q}+ \frac {b^2} {y_Q z_P}- \frac {b^2} {y_P z_Q}} \right).</cmath>
 +
<cmath>x_{L'} + y_{L'} + z_{L'} = 0 \implies L \in \Omega. </cmath>
 +
If we use NBC, we get <cmath>L = \left( \frac {a^2}{x_{Q'} - x_{P'}} : \frac {b^2}{y_{Q'} - y_{P'} } :  \frac {c^2}{z_{Q'} - z_{P'} } \right) \implies L' = (x_{Q'} - x_{P'} : y_{Q'} - y_{P'} : z_{Q'} - z_{P'}).\blacksquare</cmath>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Collinearity for two pares of isogonal points==
 +
[[File:D P' F collinearity.png|370px|right]]
 +
Let triangle <math>\triangle ABC,</math> and points <math>P</math> and <math>Q</math> be given. Let point <math>P'</math> and  <math>Q'</math>  be the isogonal conjugate of the points <math>P</math> and <math>Q</math> with respect to a triangle <math>\triangle ABC, \Omega = \odot ABC.</math>
 +
 
 +
Denote <math>R = PQ \cap P'Q', \theta = \odot P'QR, F = \Omega \cap \theta \notin \odot PQ'R, D \in \Omega</math> is the point isogonal conjugate to line <math>PQ</math> with respect <math>\triangle ABC.</math>
 +
[[Isogonal_conjugate | Isogonal_bijection_lines_and_points]]
 +
 
 +
Prove that points <math>D, P',</math> and <math>F</math> are collinear.
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
<cmath>P = (x_P :  y_P : z_P), Q = (x_Q : y_Q : z_Q), P' = \left (\frac {a^2}{x_P} : \frac {b^2}{y_P} : \frac {c^2}{z_P} \right ).</cmath>
 +
After the simple calculations one can get:
 +
 
 +
<cmath>PQ:    (y_P z_Q - z_P y_Q) x + (x_Q z_P - x_P  z_Q) y + (x_P  y_Q  - x_Q  y_P)z = 0.</cmath>
 +
<cmath>R =  \left (\frac {a^2(c^2 y_P y_Q - b^2 z_P z_Q)}{y_Q z_P - z_Q y_P} : \frac {b^2(a^2 z_P z_Q - c^2 x_P x_Q)}{z_Q x_P - x_Q z_P} : \frac {c^2(b^2 x_P x_Q - a^2 y_P y_Q)}{x_Q y_P - y_Q x_P} \right ),</cmath>
 +
<cmath>F = \left (\frac {a^2}{x_P (y_Q z_P - z_Q y_P)} : \frac {b^2}{y_P (z_Q x_P - x_Q z_P)} : \frac {c^2}{z_P (x_Q y_P - y_Q x_P)} \right ).</cmath>
 +
We use the normalized barycentric coordinates NBC and get line <math>PQ</math> in the form of:
 +
<cmath>PQ = (x_P - x_Q :  y_P - y_Q : z_P - z_Q).</cmath>
 +
<cmath>(x_P - x_Q) + (y_P - y_Q) + (z_P - z_Q) = (x_P + y_P + z_P) - (x_Q + y_Q + z_Q) = 1 - 1 = 0 \implies</cmath>
 +
<cmath>D = \left (\frac {a^2}{x_P - x_Q} : \frac {b^2}{y_P - y_Q} : \frac {c^2}{z_P - z_Q} \right ).</cmath>
 +
We check the condition of collinearity for points <math>D, F,</math> and <math>P'</math> and finishing the proof. <math>\blacksquare</math>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
==Points on bisectors==
 +
[[File:Points on bisectors.png|350px|right]]
 +
Let a triangle <math>\triangle ABC, BC = a, AC = b, AB = c</math> be given.
 +
 
 +
Let segments <math>AA', BB',</math> and <math>CC'</math> be the bisectors of <math>\triangle ABC.</math>
 +
 
 +
The lines <math>AA', BB',</math> and <math>CC'</math> meet circumcircle <math>ABC (\Omega</math>) at points <math>D, E, F,</math> respectively. <math>M</math> is the midpoint <math>AB.</math> Denote <math>G = FM \cap AD, H = FM \cap BE, K = BE \cap  A'C', L = BE \cap FD.</math>
 +
 
 +
We will find barycentric coordinates of the points and length of the segments.
 +
<cmath>A= (1:0:0), B= (0:1:0), C= (0:0:1), I=(a:b:c),</cmath>
 +
<cmath>A'= (0:b:c), B'= (a:0:c), C'= (a:b:0), M = (1:1:0).</cmath>
 +
Line <math>AA'</math> is <math>cy=bz,</math> line <math>BB'</math> is <math>cx=az,</math> line <math>CC'</math> is <math>bx=ay.</math>
 +
 
 +
Circle <math>\Omega</math> is  <math>xyc^2 + xzb^2 + yza^2 = 0.</math>
 +
<cmath> D = \Omega \cap AA', D= \left( - \frac {a^2}{b+c}:b:c \right), E = \Omega \cap BB', E = \left(a: - \frac {b^2}{a+c}:c \right),</cmath>
 +
<cmath>F = \Omega \cap CC', F = \left(a:b: - \frac {c^2}{a+b} \right), K = BE \cap A'C', K = (a : 2b : c)</cmath>
 +
 
 +
Line <math>DF</math> is <math>x \frac {b+c}{a} +y + z\frac {a+b}{c} = 0. </math>
 +
 
 +
Point <math>L = BE \cap DF, L = (a : a+2b+c : c).</math>
 +
 
 +
Line <math>FM</math> is <math>x = y + z \frac {b^2 - a^2}{c^2}.</math>
 +
 
 +
Point <math>G = AD \cap FM, G = \left( b+ \frac {b^2-a^2}{c} :b: c \right).</math>
 +
 
 +
Point <math>H = BE \cap FM, H = \left( a: a - \frac {b^2-a^2}{c} : c \right).</math>
 +
[[File:Bisector division A.png|400px|right]]
 +
Some simple formulas:
 +
<cmath>\frac {FM}{GM} = \frac {b+c-a}{a+b-c}; \frac {FM}{FG} = \frac {b+c-a}{2b};</cmath>
 +
<cmath>\frac {FM}{FH} = \frac {a+c-b}{2a}; \frac {GM}{FG} = \frac {a+b-c}{2b};</cmath>
 +
<cmath>\frac {EH}{B'E} = \frac {a(a+c)}{b^2}; \frac {IH}{B'E} = \frac {|a-b|(a+c)}{b^2};</cmath>
 +
<cmath>\frac {B'I}{IE} = 1 - \frac {b}{a+c}; \frac {IB}{2} = IL = BL.</cmath>
 +
Circumcenter <math>O \in FM , O = \left( a^2(b^2+c^2 -a^2) : b^2(a^2 + c^2 – b^2) : c^2(a^2 + b^2 - c^2) \right).</math>
 +
 
 +
Tangent <math>BN</math> is <math>c^2 x + a^2 z = 0.</math>
 +
 
 +
Line <math>NI || AC</math> is <math>\left( \frac {b}{a+c} = \frac{y}{x+z}  \right).</math>
 +
<cmath>N = BN \cap IN, N = (a^2 : b(a - c) : -c^2).</cmath>
 +
<cmath>BN = IN = \frac {abc}{|a-c|(a+b+c)}.</cmath>
 +
<cmath>\frac{HO}{BO} = \frac {|a-c|}{b}; \frac{GO}{BO} = \frac {|b-c|}{a}.</cmath>
 +
<math>Q</math> is the midpoint <math>BB', QP \perp BB', P \in AD \implies P = \left( \frac {a(b-a)}{c} : b : c\right). </math>
 +
<cmath>G = FM \cap AD = \left( \frac {b^2-a^2}{c} : b : c\right).</cmath>
 +
<cmath>\frac {PD}{GP} = \frac {a}{c}; \frac {FM}{GM} = \frac {b+c-a}{a+b-c};</cmath>
 +
<cmath>\frac {IC'}{FC'} = \frac {a+b-c}{c}; \frac {IA'}{DA'} = \frac {b+c-a}{a};</cmath>
 +
<cmath>\frac {ND}{NF} = \frac {a(a+b-c)}{c(b+c-a)}.</cmath>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Crosspoint of median and set of secants==
 +
[[File:Incircle and secants.png|350px|right]]
 +
Triangle <math>ABC</math> and point <math>P \in BC</math> be given. The incircle <math>\omega</math> of <math>\triangle ABC</math> touches side <math>BC</math> at point <math>D.</math>  Point <math>P'</math> is symmetrical to point <math>P</math> with respect midpoint <math>M</math> of <math>BC.</math> The common points of segments <math>AP</math> and <math>AP'</math> with <math>\omega</math> form a convex quadrilateral <math>EFE'F'.</math>
 +
 
 +
Prove that point <math>G = DI \cap AM</math> lies on <math>EF.</math>
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
Denote <math>p_a = \frac{b+c-a}{2}, p_b = \frac{a-b+c}{2}, p_c = \frac{a+b-c}{2}, m = \sqrt{\frac {CP}{BP}}.</math>
 +
<math>D = \left (0 : p_c : p_b \right), P= \left (0: m: \frac{1}{m}\right), P' = \left ( 0: \frac{1}{m}: m\right).</math>
 +
<cmath>\omega:\hspace{10mm}  {p_a}^2x^2 + p_b^2y^2 + p_c^2z^2 - 2p_a p_b  xy - 2p_a p_c xz - 2p_bp_cyz = 0,</cmath>
 +
Line <math>AP:  \frac {y}{m} = z \cdot m,</math> line <math>AP' : y \cdot m =  \frac {z}{m}.</math>
 +
 
 +
We solve the system of these equations and get:
 +
<cmath>E =  \left( \left(\sqrt{\frac {p_c}{m}} + \sqrt {p_b \cdot m} \right)^2 : mp_a : \frac {p_a}{m} \right),</cmath>
 +
<cmath>F' = \left( \left(\sqrt{\frac {p_c}{m}} - \sqrt {p_b \cdot m} \right)^2 : mp_a : \frac {p_a}{m} \right),</cmath>
 +
<cmath>F =  \left( \left(\sqrt{p_c \cdot m} + \sqrt {\frac {p_b}{m}} \right)^2 : \frac {p_a}{m} : mp_a \right),</cmath>
 +
<cmath>E' = \left( \left(\sqrt{p_c \cdot m} - \sqrt {\frac {p_b}{m}} \right)^2 : \frac {p_a}{m} : mp_a \right).</cmath>
 +
We find the lines <math>EE'</math> and <math>FF',</math> we solve the system of equations for this lines and get:
 +
<cmath>G =  \left (a: pa : pa \right ).</cmath> This point lies at the line <math>AM, \frac {AG}{GM} =\frac {2 p_a}{a}.</math>
 +
Point <math>G</math> lies at line <math>DI</math> and <math>\frac {DI}{GI} = \frac {b+c}{a}.</math>
 +
 
 +
<i><b>Corollary</b></i>
 +
 
 +
Denote <math>D' = AB \cap \omega, D'' = AC \cap \omega.</math> Then <math>\frac {D'G}{D''G} = \frac {b}{c}.</math>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Set of lines in triangle==
 +
[[File:Set of lines 30 34 .png|395px|right]]
 +
Let triangle <math>\triangle ABC</math> and points <math>D</math> at the line <math>BC, E \in AC, F \in AB</math> be given.
 +
 
 +
Denote <math>D'</math> point in <math>AB</math> such that <math>DD'||AC.</math> Similarly,  <math>E' \in BC, EE'||AB, F' \in AC, FF'||BC.</math>
 +
 
 +
<cmath>K = AD \cap CD', K' = CE \cap AE', L = BE \cap AE',</cmath>  <cmath>L' = AD \cap BF', N = CF \cap BF', N' = BE \cap CD'.</cmath>
 +
 
 +
Prove that lines <math>KK', LL',</math> and <math>NN'</math> are concurrent.
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
Let <math>d = \frac {CD}{BD}, e = \frac{BF}{AF}, f = \frac{BF}{AF}.</math>
 +
 
 +
Then <math>D = (0: d: 1), E = (1: 0: e), F = (f: 1: 0).</math>
 +
<cmath>DD' || AC \implies D' = (1 :d : 0), EE' || AB \implies E' = (0 :1 : e), FF' || BC \implies F' = (f : 0 : 1).</cmath>
 +
<cmath>K = (1 : d : 1), L = (1 : 1 : e), N = (f : 1 : 1), K' = (f : 1 : e), L' = (f : d : 1), N' = (1 : d : e).</cmath>
 +
 
 +
Point <math>G = (1 +f : 1 + d : 1 + e)</math> lies at lines <math>KK', LL',</math> and <math>NN'.</math>
 +
 
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Set of parallel lines==
 +
[[File:Set of lines small.png|400px|right]]
 +
 
 +
Let triangle <math>\triangle ABC</math> and points <math>D</math> at the line <math>BC, E \in AC, F \in AB</math> be given.
 +
 
 +
Denote <math>d = \frac {CD}{BD}, e = \frac{BF}{AF}, f = \frac{BF}{AF}.</math>
 +
 
 +
Let <math>A'</math> be the point such that <math>A'E||AB, A'D||AC.</math>
 +
 
 +
Similarly,  <math>B'E||AB, B'F||BC, C'D||AC, C'F||BC.</math>
 +
 
 +
Prove that lines <math>AA', BB',</math> and <math>CC'</math> are concurrent.
 +
 
 +
Find the condition that <math>\triangle ABC = \triangle A'B'C'.</math>
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
One can get <math>A' = \left ( \frac {1}{d \cdot e}-1 : 1 + \frac {1}{e} : 1 + \frac {1}{d} \right ),</math>
 +
<cmath>B' = \left (1 + \frac {1}{e} : \frac {1}{e \cdot f}-1 : 1 + \frac {1}{f} \right ),
 +
C' = \left (1 + \frac {1}{d} : 1 + \frac {1}{f} : \frac {1}{d \cdot f}-1 \right ),</cmath>
 +
<cmath>O = \left (\frac {f}{1+f} :  \frac {d}{1+d} : \frac {e}{1 + e} \right ),
 +
\frac {A'O}{AO} = \frac{B'O}{BO} = \frac {C'O}{CO} = \frac{2def + df + de + ef - 1}{(1+d)(1+ e)(1+f)}.</cmath>
 +
If <math>\frac {A'O}{AO} = 1 </math> then <math>def = d+e+f+1.</math>
 +
[[File:Set of lines small С.png|350px|right]]
 +
<i><b>Corollary</b></i>
 +
 
 +
Let points <math>D, E,</math> and <math>F</math> lie at the lines <math>BC, AC,</math> and <math>AB.</math>
 +
 
 +
Denote circle <math>\omega = \odot DEF, D' = \omega \cap BC, E' = \omega \cap AC, F' = \omega \cap AB.</math>
 +
 
 +
Let <math>ED' || DE', D'F' || E'F, DF' || FE,</math>
 +
<cmath>A' = DF' \cap D'E, B' = D'E \cap E'F, C' = DF' \cap E'F.</cmath>
 +
 
 +
Then lines <math>AA', BB',</math> and <math>CC'</math> are concurrent.
 +
 +
WLOG, situation is shown on diagram.
 +
 
 +
The proof contain calculations started from <math>\triangle A'B'C'</math> and finished at <math>\triangle ABC.</math>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Small Pascal's theorem==
 +
[[File:PascalS Lemoine.png|390px|right]]
 +
[[File:Pascal S Lemoine E.png|390px|right]]
 +
 
 +
Let <math>\triangle ABC</math> and point <math>P</math> be given. Let <math>\Omega</math> be the circumcircle of <math>\triangle ABC,</math>
 +
<cmath>A' = AP \cap \Omega, B' = BP \cap \Omega, C' = CP \cap \Omega.</cmath>
 +
Let the tangent line to <math>\Omega</math> at point <math>A</math> cross line <math> B'C'</math> at point <math>D.</math> Similarly denote points <math>E</math> and <math>F.</math>
 +
 
 +
Prove that the points <math>D, E</math> and <math>F</math> are collinear.
 +
 +
<i><b>Proof</b></i>
 +
 
 +
1. Simplest case, <math>P</math> is the Lemoine point, <math>P  = L = (a^2 : b^2 : c^2).</math>
 +
 
 +
The equation of <math>\Omega</math> is <math> xyc^2 + xzb^2 + yza^2 = 0.</math>
 +
 
 +
Line <math>AP</math> is <math>k = 0, l y_P + m z_P = 0 \implies y c^2 - z b^2 = 0 \implies </math>
 +
<cmath>A' = \left(-\frac{a^2}{2} : b^2 : c^2 \right), B' = \left(a^2 :  -\frac{b^2}{2} : c^2 \right), C' = \left(a^2 :  b^2 : -\frac{c^2}{2} \right).</cmath>
 +
The line <math>B'C'</math> is <math>\frac{x}{2a^2} - \frac{y}{b^2} - \frac{z}{c^2} = 0 \implies D = \left( 0 : b^2 : - c^2 \right).</math>
 +
 
 +
Similarly, <math>E = ( a^2 : 0 : - c^2), F = (a^2 : - b^2 : 0).</math>
 +
 
 +
The line <math>DEF</math> is <math>\frac {x}{a^2} + \frac {y}{b^2} + \frac {z}{c^2} = 0.</math>
 +
 
 +
2. Simple case, <math>P</math> is one of the external Lemoine point, <math>P = L' = (a^2 : - b^2 : c^2).</math>
 +
 
 +
This point is the crosspoint of the tangent lines to <math>\Omega</math> in points <math>A</math> and <math>C,</math> so
 +
<cmath>A' = A, C' = C, B' = \left (a^2 :  -\frac{b^2}{2} : c^2 \right ).</cmath>
 +
The line <math>B'C'</math> is <math>b^2 x + 2y a^2 = 0 \implies D = \left( 2a^2 : -b^2 : c^2 \right).</math>
 +
 
 +
Similarly, <math>E = ( a^2 : 0 : - c^2), F = (a^2 : - b^2 : 2c^2).</math>
 +
 
 +
The line <math>DEF</math> is <math>\frac {x}{a^2} + \frac {3y}{b^2} + \frac {z}{c^2} = 0.</math>
 +
 
 +
Similarly, if <math>P  = (-a^2 : b^2 : c^2),</math> then the line <math>DEF</math> is <math>\frac {3x}{a^2} + \frac {y}{b^2} + \frac {z}{c^2} = 0.</math>
 +
 
 +
If <math>P  = (a^2 : b^2 : -c^2),</math> then the line <math>DEF</math> is <math>\frac {x}{a^2} + \frac {y}{b^2} + \frac {3z}{c^2} = 0.</math>
 +
 
 +
These three lines intersect in pairs at points <math>D, E,</math> and <math>F</math> of the line of case 1.
 +
 
 +
3. Common case. Denote the coordinates of the point <math>P = (x_P : y_P : z_P).</math> The equation of <math>\Omega</math> is <math> xyc^2 + xzb^2 + yza^2 = 0.</math>
 +
 
 +
Line <math>AP</math> is <math>l z_P + m y_P = 0 \implies A' = \left( \frac{-y_P \cdot z_P a^2}{y_P c^2 + z_P b^2} : y_P :z_P \right ).</math>
 +
 
 +
Similarly, <math>B' = \left (x_P :  \frac{-x_P \cdot z_P b^2}{x_P c^2 + z_P a^2} : z_P \right ), C' = \left (x_P : y_P : \frac{-x_P \cdot y_P c^2}{x_P b^2 + y_P a^2} \right ).</math>
 +
 
 +
The tangent line <math>l_A</math> to <math>\Omega</math> at <math>A</math> is <math>yC^2 +zb^2=0.</math>
 +
 
 +
The line <math>B'C'</math> is <math>\frac{y_P \cdot z_P a^2}{x_P} x - (x_P \cdot c^2 + z_P a^2)y - (x_P b^2 +y_Pa^2)z = 0.</math>
 +
 
 +
<math>D = l_A \cap B'C' = \left( x_P (y_P c^2 - z_P b^2) : -y_P \cdot z_P b^2 :  y_P \cdot z_P c^2 \right).</math>
 +
 
 +
Similarly, <math>E = l_B \cap A'C' = \left( x_P z_P a^2 : -y_P (x_P c^2 - z_P a^2) : -x_P \cdot z_P c^2 \right).</math>
 +
<cmath>F = l_C \cap A'B' = \left( x_P y_P a^2 : -x_P y_P b^2 : z_P (y_P a^2 - x_P b^2) \right).</cmath>
 +
The line <math>DEF</math> is <cmath> \frac {x}{x_P} (-\frac {a^2}{x_P} + \frac {b^2}{y_P} + \frac {c^2}{z_P})  + \frac {y}{y_P} (\frac {a^2}{x_P} - \frac {b^2}{y_P} + \frac {c^2}{z_P})  +\frac {z}{z_P} (\frac {a^2}{x_P} + \frac {b^2}{y_P} - \frac {c^2}{z_P}) = 0.</cmath>
 +
 
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Latest revision as of 16:24, 22 November 2024

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): \[\vec X = (x_X : y_X : z_X)\] \[x_X \cdot \vec {XA} + y_X \cdot \vec {XB} + z_X \cdot \vec {XC} = \vec {0},\] \[\vec X = \frac {x_X \cdot \vec {A} + y_X \cdot \vec {B} + z_X \cdot \vec {C}}{x_X + y_X + z_X}.\] The normalized (absolute) barycentric coordinates NBC satisfy the condition $x_X + y_X + z_X = 1,$ they are uniquely determined: \[x_X = \frac{[\vec {XB},\vec {XC}]}{\sigma}, y_X = \frac{[\vec {XC},\vec {XA}]}{\sigma}, z_X = \frac{[\vec {XA},\vec {XB}]}{\sigma},\] \[\sigma = [\vec {XB},\vec {XC}] + [\vec {XC},\vec {XA}] + [\vec {XA},\vec {XB}] .\] 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 \[|PQ|^2 = S_A \cdot (x_1 - x_2)^2 + S_B \cdot (y_1 - y_2)^2 + S_C \cdot (z_1 - z_2)^2.\] \[|PQ|^2 = - a^2 (y_1 - y_2)(z_1 - z_2) - b^2 (x_1 - x_2)(z_1 - z_2) - c^2 (x_1 - x_2)(y_1 - y_2).\] The equation of bisector of $PQ$ is: \[x(c^2(y_2-y_1) + b^2(z_2-z_1)) + y(a^2(z_2-z_1) + c^2(x_2-x_1)) + z(a^2(y_2-y_1) + b^2(x_2-x_1)) + a^2(y_1z_1 - y_2z_2) + b^2(x_1z_1-x_2z_2) + c^2 (x_1y_1 – x_2y_2) =0.\] 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: \[xyc^2 + xzb^2 + yza^2 = 0.\]

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

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: \[(k_1 - k_2) a \cdot x + (l_1 - l_2) b \cdot y + (m_1 - m_2) c \cdot z = 0.\] 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

Ratio of isogonal segments

Isogonals division.png

Let triangle $\triangle ABC$ and point $P$ be given. Denote $P'$ the isogonal conjugate of a point $P$ with respect to $\triangle ABC, \Omega = \odot ABC,$ \[D = AP' \cap BC, E = AP \cap BC, L = AP \cap \Omega.\] Prove that $\frac {AP'}{P'D} \cdot \frac {AP}{PE} = \frac {AL}{LE}.$

Proof

We use the formula for isogonal conjugate point and get \[P = (x_P : y_P : z_P), P' = (x_{P'} : y_{P'} : z_{P'}) = \left( \frac {a^2}{x_P} : \frac {b^2}{y_P} : \frac {c^2}{z_P} \right).\] \[\frac {AP'}{P'D} = \frac {y_{P'} + z_{P'}}{x_{P'}},\] \[\frac {AP}{PE} = \frac {y_{P} + z_{P}}{x_{P}}.\] \[L \in \Omega \implies x_{L'} + y_{L'} + z_{L'} = 0, L \in AP \implies y_L = y_P, z_L = z_P \implies \frac {a^2}{x_L} + y_{P'} + z_{P'} = 0 \implies x_L = \frac{-a^2}{y_{P'} + z_{P'}}.\] \[\frac {AL}{LE} = \frac {y_{P} + z_{P}}{-x_{L}} = \frac {(y_{P} + z_{P})(y_{P'} + z_{P'})}{a^2}.\] \[x_P \cdot x_{P'} = a^2 \implies \frac {AP'}{P'D} \cdot \frac {AP}{PE} = \frac {AL}{LE}.\]

vladimir.shelomovskii@gmail.com, vvsss

Point on incircle

Point on incicle.png

Let triangle $\triangle ABC$ be given. Denote the incircle $\omega,$ 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 $\Omega.$

This configuration can be used as a straight-line mechanism since it allows to create a mechanism that converts the rotational motion of a point Z to perfect straight-line motion of the X point or vice versa. Of course, we need to use the prismatic joint at the points $E$ and $F.$

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 $c^2 y_P \cdot y - b^2 z_P \cdot z = 0,$

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 $Z$ \[\left(\frac {1}{y_D + z_D} : \frac {b^2}{ x_X (y_P z_D - z_P y_D) - a^2 y_D} :\frac {c^2}{x_X (z_P y_D - y_P z_D) - a^2 z_D} \right).\] Let point $Z'$ be the isogonal conjugate of a point $Z$ with respect to a triangle $\triangle ABC.$ \[Z' = \left(a^2 (y_D + z_D) : x_X (y_P z_D - z_P y_D) - a^2 y_D : x_X (z_P y_D - y_P z_D) - a^2 z_D \right).\] The sum of coordinates is equal zero, so $Z'$ is in infinity, therefore the point $Z$ lies on $\Omega.\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Fixed point on circumcircle

Fixed point 2.png

Let triangle $\triangle ABC,$ point $G \ne A$ on circumcircle $\Omega = \odot ABC,$ and point $D \in BC$ be given. Point $P$ lies on $AG,$ point $P'$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC, Q = DP' \cap AP, F = \odot DPQ \cap \Omega.$

Prove that $F$ is fixed point and not depends from position of $P.$

Proof

Denote the coordinates of the points $D = (0 : y_D : z_D), G = (x_G : y_G : z_G).$ \[G \in \Omega \implies a^2 y_G z_G + b^2 x_G z_G + c^2 x_G y_G = 0 \implies\] \[G = \left( \frac {- a^2 y_G z_G}{b^2 z_G + c^2 y_G} : y_G : z_G\right).\] \[P = (x_P : y_G : z_G) \implies P' = \left( \frac {a^2}{x_P} : \frac {b^2}{y_G} : \frac {c^2}{z_G}\right).\] The line $AG$ is $z_G y = y_G z.$

The line $DP'$ is $(y_{P'} z_D - z_{P'} y_D) x - x_{P'} z_D y + x_{P'} y_D z = 0 \implies$ \[Q =\left( \frac {a^2 (y_D z_G - y_G z_D) \cdot y_G z_G}{x_P (c^2 y_D y_G - b^2 z_D z_P)} : y_G : z_G \right).\] We find the circle $\odot PQD$ and get the point \[F =\left( \frac {a^2}{\frac {c^2}{z_D \cdot z_G} - \frac{b^2}{y_D \cdot y_G}} : y_G \cdot y_D : - z_G \cdot z_D \right).\] $F$ depends only from points $G$ and $D.$

Fixed point on circumcircle

vladimir.shelomovskii@gmail.com, vvsss

Two pare isogonal points

2 pare Miquel.png

Let triangle $\triangle ABC,$ and points $P$ and $Q$ (points do not lie on sidelines) be given.

Let point $P'$ and $Q'$ be the isogonal conjugate of a point $P$ and $Q$ with respect to a triangle $\triangle ABC, \Omega = \odot ABC.$

Denote $R = PQ \cap P'Q', E = \Omega \cap RPQ', F = \Omega \cap RQP'.$

Prove that $L= EP \cap FQ$ and $K = EQ' \cap FP'$ lies on $\Omega.$

Proof

The line $PQ$ is \[(y_P z_Q - z_P y_Q) x + (x_Q z_P – x_P z_Q) y + (x_P y_Q - x_Q y_P)z = 0.\] The line $P'Q'$ is \[(y_P z_Q - z_P y_Q) \frac {x_P x_Q}{a^2} x + (x_Q z_P – x_P z_Q) \frac {y_P y_Q}{b^2}y + (x_P y_Q - x_Q y_P) \frac {z_P z_Q}{c^2}z = 0.\] \[R = PQ \cap P'Q' = \left ( a^2 \frac {(b^2 z_P z_Q – c^2 y_P y_Q)}{y_P z_Q – z_P y_Q} : b^2 \frac {(a^2 z_P z_Q – c^2 x_P x_Q)}{x_P z_Q – z_P x_Q}  : c^2 \frac {(a^2 y_P y_Q – b^2 x_P x_Q)}{x_P y_Q – y_P x_Q} \right)\] \[E = \Omega \cap RPQ' = \left( \frac {a^2}{x_Q (z_P y_Q – y_P z_Q)} : \frac {b^2}{y_Q (x_P z_Q – z_P y_Q)} : \frac {c^2}{z_Q (y_P x_Q – x_P y_Q)} \right).\] \[F = \Omega \cap RP'Q = \left( \frac {a^2}{x_P (z_P y_Q – y_P z_Q)} : \frac {b^2}{y_P (x_P z_Q – z_P y_Q)} : \frac {c^2}{z_P (y_P x_Q – x_P y_Q)} \right).\] \[K = EQ' \cap FP' = \left( \frac {a^2}{x_Q (y_P + z_P) - x_P (y_Q + z_Q)} : \frac {b^2}{y_Q (x_P + z_P) - y_P (z_Q + x_Q)} : \frac {c^2}{z_Q (x_P + y_P) - z_P (x_Q + y_Q)} \right).\] Denote $K'$ is the isogonal conjugate of a point $K$ with respect to $\triangle ABC.$ \[K' = \left( x_Q (y_P + z_P) - x_P (y_Q + z_Q) : y_Q (x_P + z_P) – y_P (z_Q + x_Q) : z_Q (x_P + y_P) – z_P (x_Q + y_Q) \right).\] \[x_{K'} + y_{K'} + z_{K'} = 0 \implies K \in \Omega.\] If we use NBC, we get \[K = \left( \frac {a^2}{x_Q - x_P} : \frac {b^2}{y_Q - y_P } : \frac {c^2}{z_Q - z_P } \right) \implies K' = (x_Q - x_P : y_Q - y_P : z_Q - z_P).\] \[L = EP \cap FQ = \left( \frac {1}{\frac {b^2} {x_Q y_P} - \frac {b^2} {x_P y_Q}+ \frac {c^2} {x_Q z_P}- \frac {c^2} {x_Q z_P}} : \frac {1}{\frac {a^2} {x_Q y_P} - \frac {a^2} {x_P y_Q}+ \frac {c^2} {z_Q y_P}- \frac {c^2} {y_Q z_P}}  : \frac {1}{\frac {a^2} {x_Q z_P} - \frac {a^2} {x_P z_Q}+ \frac {b^2} {y_Q z_P}- \frac {b^2} {y_P z_Q}} \right).\] \[x_{L'} + y_{L'} + z_{L'} = 0 \implies L \in \Omega.\] If we use NBC, we get \[L = \left( \frac {a^2}{x_{Q'} - x_{P'}} : \frac {b^2}{y_{Q'} - y_{P'} } : \frac {c^2}{z_{Q'} - z_{P'} } \right) \implies L' = (x_{Q'} - x_{P'} : y_{Q'} - y_{P'} : z_{Q'} - z_{P'}).\blacksquare\]

vladimir.shelomovskii@gmail.com, vvsss

Collinearity for two pares of isogonal points

D P' F collinearity.png

Let triangle $\triangle ABC,$ and points $P$ and $Q$ be given. Let point $P'$ and $Q'$ be the isogonal conjugate of the points $P$ and $Q$ with respect to a triangle $\triangle ABC, \Omega = \odot ABC.$

Denote $R = PQ \cap P'Q', \theta = \odot P'QR, F = \Omega \cap \theta \notin \odot PQ'R, D \in \Omega$ is the point isogonal conjugate to line $PQ$ with respect $\triangle ABC.$ Isogonal_bijection_lines_and_points

Prove that points $D, P',$ and $F$ are collinear.

Proof

\[P = (x_P : y_P : z_P), Q = (x_Q : y_Q : z_Q), P' = \left (\frac {a^2}{x_P} : \frac {b^2}{y_P} : \frac {c^2}{z_P} \right ).\] After the simple calculations one can get:

\[PQ: (y_P z_Q - z_P y_Q) x + (x_Q z_P - x_P z_Q) y + (x_P y_Q - x_Q y_P)z = 0.\] \[R = \left (\frac {a^2(c^2 y_P y_Q - b^2 z_P z_Q)}{y_Q z_P - z_Q y_P} : \frac {b^2(a^2 z_P z_Q - c^2 x_P x_Q)}{z_Q x_P - x_Q z_P} : \frac {c^2(b^2 x_P x_Q - a^2 y_P y_Q)}{x_Q y_P - y_Q x_P} \right ),\] \[F = \left (\frac {a^2}{x_P (y_Q z_P - z_Q y_P)} : \frac {b^2}{y_P (z_Q x_P - x_Q z_P)} : \frac {c^2}{z_P (x_Q y_P - y_Q x_P)} \right ).\] We use the normalized barycentric coordinates NBC and get line $PQ$ in the form of: \[PQ = (x_P - x_Q : y_P - y_Q : z_P - z_Q).\] \[(x_P - x_Q) + (y_P - y_Q) + (z_P - z_Q) = (x_P + y_P + z_P) - (x_Q + y_Q + z_Q) = 1 - 1 = 0 \implies\] \[D = \left (\frac {a^2}{x_P - x_Q} : \frac {b^2}{y_P - y_Q} : \frac {c^2}{z_P - z_Q} \right ).\] We check the condition of collinearity for points $D, F,$ and $P'$ and finishing the proof. $\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Points on bisectors

Points on bisectors.png

Let a triangle $\triangle ABC, BC = a, AC = b, AB = c$ be given.

Let segments $AA', BB',$ and $CC'$ be the bisectors of $\triangle ABC.$

The lines $AA', BB',$ and $CC'$ meet circumcircle $ABC (\Omega$) at points $D, E, F,$ respectively. $M$ is the midpoint $AB.$ Denote $G = FM \cap AD, H = FM \cap BE, K = BE \cap A'C', L = BE \cap FD.$

We will find barycentric coordinates of the points and length of the segments. \[A= (1:0:0), B= (0:1:0), C= (0:0:1), I=(a:b:c),\] \[A'= (0:b:c), B'= (a:0:c), C'= (a:b:0), M = (1:1:0).\] Line $AA'$ is $cy=bz,$ line $BB'$ is $cx=az,$ line $CC'$ is $bx=ay.$

Circle $\Omega$ is $xyc^2 + xzb^2 + yza^2 = 0.$ \[D = \Omega \cap AA', D= \left( - \frac {a^2}{b+c}:b:c \right), E = \Omega \cap BB', E = \left(a: - \frac {b^2}{a+c}:c \right),\] \[F = \Omega \cap CC', F = \left(a:b: - \frac {c^2}{a+b} \right), K = BE \cap A'C', K = (a : 2b : c)\]

Line $DF$ is $x \frac {b+c}{a} +y + z\frac {a+b}{c} = 0.$

Point $L = BE \cap DF, L = (a : a+2b+c : c).$

Line $FM$ is $x = y + z \frac {b^2 - a^2}{c^2}.$

Point $G = AD \cap FM, G = \left( b+ \frac {b^2-a^2}{c} :b: c \right).$

Point $H = BE \cap FM, H = \left( a: a - \frac {b^2-a^2}{c} : c \right).$

Bisector division A.png

Some simple formulas: \[\frac {FM}{GM} = \frac {b+c-a}{a+b-c}; \frac {FM}{FG} = \frac {b+c-a}{2b};\] \[\frac {FM}{FH} = \frac {a+c-b}{2a}; \frac {GM}{FG} = \frac {a+b-c}{2b};\] \[\frac {EH}{B'E} = \frac {a(a+c)}{b^2}; \frac {IH}{B'E} = \frac {|a-b|(a+c)}{b^2};\] \[\frac {B'I}{IE} = 1 - \frac {b}{a+c}; \frac {IB}{2} = IL = BL.\] Circumcenter $O \in FM , O = \left( a^2(b^2+c^2 -a^2) : b^2(a^2 + c^2 – b^2) : c^2(a^2 + b^2 - c^2) \right).$

Tangent $BN$ is $c^2 x + a^2 z = 0.$

Line $NI || AC$ is $\left( \frac {b}{a+c} = \frac{y}{x+z} \right).$ \[N = BN \cap IN, N = (a^2 : b(a - c) : -c^2).\] \[BN = IN = \frac {abc}{|a-c|(a+b+c)}.\] \[\frac{HO}{BO} = \frac {|a-c|}{b}; \frac{GO}{BO} = \frac {|b-c|}{a}.\] $Q$ is the midpoint $BB', QP \perp BB', P \in AD \implies P = \left( \frac {a(b-a)}{c} : b : c\right).$ \[G = FM \cap AD = \left( \frac {b^2-a^2}{c} : b : c\right).\] \[\frac {PD}{GP} = \frac {a}{c}; \frac {FM}{GM} = \frac {b+c-a}{a+b-c};\] \[\frac {IC'}{FC'} = \frac {a+b-c}{c}; \frac {IA'}{DA'} = \frac {b+c-a}{a};\] \[\frac {ND}{NF} = \frac {a(a+b-c)}{c(b+c-a)}.\]

vladimir.shelomovskii@gmail.com, vvsss

Crosspoint of median and set of secants

Incircle and secants.png

Triangle $ABC$ and point $P \in BC$ be given. The incircle $\omega$ of $\triangle ABC$ touches side $BC$ at point $D.$ Point $P'$ is symmetrical to point $P$ with respect midpoint $M$ of $BC.$ The common points of segments $AP$ and $AP'$ with $\omega$ form a convex quadrilateral $EFE'F'.$

Prove that point $G = DI \cap AM$ lies on $EF.$

Proof

Denote $p_a = \frac{b+c-a}{2}, p_b = \frac{a-b+c}{2}, p_c = \frac{a+b-c}{2}, m = \sqrt{\frac {CP}{BP}}.$ $D = \left (0 : p_c : p_b \right), P= \left (0: m: \frac{1}{m}\right), P' = \left ( 0: \frac{1}{m}: m\right).$ \[\omega:\hspace{10mm} {p_a}^2x^2 + p_b^2y^2 + p_c^2z^2 - 2p_a p_b xy - 2p_a p_c xz - 2p_bp_cyz = 0,\] Line $AP: \frac {y}{m} = z \cdot m,$ line $AP' : y \cdot m = \frac {z}{m}.$

We solve the system of these equations and get: \[E = \left( \left(\sqrt{\frac {p_c}{m}} + \sqrt {p_b \cdot m} \right)^2 : mp_a : \frac {p_a}{m} \right),\] \[F' = \left( \left(\sqrt{\frac {p_c}{m}} - \sqrt {p_b \cdot m} \right)^2 : mp_a : \frac {p_a}{m} \right),\] \[F = \left( \left(\sqrt{p_c \cdot m} + \sqrt {\frac {p_b}{m}} \right)^2 : \frac {p_a}{m} : mp_a \right),\] \[E' = \left( \left(\sqrt{p_c \cdot m} - \sqrt {\frac {p_b}{m}} \right)^2 : \frac {p_a}{m} : mp_a \right).\] We find the lines $EE'$ and $FF',$ we solve the system of equations for this lines and get: \[G = \left (a: pa : pa \right ).\] This point lies at the line $AM, \frac {AG}{GM} =\frac {2 p_a}{a}.$ Point $G$ lies at line $DI$ and $\frac {DI}{GI} = \frac {b+c}{a}.$

Corollary

Denote $D' = AB \cap \omega, D'' = AC \cap \omega.$ Then $\frac {D'G}{D''G} = \frac {b}{c}.$

vladimir.shelomovskii@gmail.com, vvsss

Set of lines in triangle

Set of lines 30 34 .png

Let triangle $\triangle ABC$ and points $D$ at the line $BC, E \in AC, F \in AB$ be given.

Denote $D'$ point in $AB$ such that $DD'||AC.$ Similarly, $E' \in BC, EE'||AB, F' \in AC, FF'||BC.$

\[K = AD \cap CD', K' = CE \cap AE', L = BE \cap AE',\] \[L' = AD \cap BF', N = CF \cap BF', N' = BE \cap CD'.\]

Prove that lines $KK', LL',$ and $NN'$ are concurrent.

Proof

Let $d = \frac {CD}{BD}, e = \frac{BF}{AF}, f = \frac{BF}{AF}.$

Then $D = (0: d: 1), E = (1: 0: e), F = (f: 1: 0).$ \[DD' || AC \implies D' = (1 :d : 0), EE' || AB \implies E' = (0 :1 : e), FF' || BC \implies F' = (f : 0 : 1).\] \[K = (1 : d : 1), L = (1 : 1 : e), N = (f : 1 : 1), K' = (f : 1 : e), L' = (f : d : 1), N' = (1 : d : e).\]

Point $G = (1 +f : 1 + d : 1 + e)$ lies at lines $KK', LL',$ and $NN'.$


vladimir.shelomovskii@gmail.com, vvsss

Set of parallel lines

Set of lines small.png

Let triangle $\triangle ABC$ and points $D$ at the line $BC, E \in AC, F \in AB$ be given.

Denote $d = \frac {CD}{BD}, e = \frac{BF}{AF}, f = \frac{BF}{AF}.$

Let $A'$ be the point such that $A'E||AB, A'D||AC.$

Similarly, $B'E||AB, B'F||BC, C'D||AC, C'F||BC.$

Prove that lines $AA', BB',$ and $CC'$ are concurrent.

Find the condition that $\triangle ABC = \triangle A'B'C'.$

Proof

One can get $A' = \left ( \frac {1}{d \cdot e}-1 : 1 + \frac {1}{e} : 1 + \frac {1}{d} \right ),$ \[B' = \left (1 + \frac {1}{e} : \frac {1}{e \cdot f}-1 : 1 + \frac {1}{f} \right ), C' = \left (1 + \frac {1}{d} : 1 + \frac {1}{f} : \frac {1}{d \cdot f}-1 \right ),\] \[O = \left (\frac {f}{1+f} : \frac {d}{1+d} : \frac {e}{1 + e} \right ), \frac {A'O}{AO} = \frac{B'O}{BO} = \frac {C'O}{CO} = \frac{2def + df + de + ef - 1}{(1+d)(1+ e)(1+f)}.\] If $\frac {A'O}{AO} = 1$ then $def = d+e+f+1.$

Set of lines small С.png

Corollary

Let points $D, E,$ and $F$ lie at the lines $BC, AC,$ and $AB.$

Denote circle $\omega = \odot DEF, D' = \omega \cap BC, E' = \omega \cap AC, F' = \omega \cap AB.$

Let $ED' || DE', D'F' || E'F, DF' || FE,$ \[A' = DF' \cap D'E, B' = D'E \cap E'F, C' = DF' \cap E'F.\]

Then lines $AA', BB',$ and $CC'$ are concurrent.

WLOG, situation is shown on diagram.

The proof contain calculations started from $\triangle A'B'C'$ and finished at $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Small Pascal's theorem

PascalS Lemoine.png
Pascal S Lemoine E.png

Let $\triangle ABC$ and point $P$ be given. Let $\Omega$ be the circumcircle of $\triangle ABC,$ \[A' = AP \cap \Omega, B' = BP \cap \Omega, C' = CP \cap \Omega.\] Let the tangent line to $\Omega$ at point $A$ cross line $B'C'$ at point $D.$ Similarly denote points $E$ and $F.$

Prove that the points $D, E$ and $F$ are collinear.

Proof

1. Simplest case, $P$ is the Lemoine point, $P = L = (a^2 : b^2 : c^2).$

The equation of $\Omega$ is $xyc^2 + xzb^2 + yza^2 = 0.$

Line $AP$ is $k = 0, l y_P + m z_P = 0 \implies y c^2 - z b^2 = 0 \implies$ \[A' = \left(-\frac{a^2}{2} : b^2 : c^2 \right), B' = \left(a^2 : -\frac{b^2}{2} : c^2 \right), C' = \left(a^2 : b^2 : -\frac{c^2}{2} \right).\] The line $B'C'$ is $\frac{x}{2a^2} - \frac{y}{b^2} - \frac{z}{c^2} = 0 \implies D = \left( 0 : b^2 : - c^2 \right).$

Similarly, $E = ( a^2 : 0 : - c^2), F = (a^2 : - b^2 : 0).$

The line $DEF$ is $\frac {x}{a^2} + \frac {y}{b^2} + \frac {z}{c^2} = 0.$

2. Simple case, $P$ is one of the external Lemoine point, $P = L' = (a^2 : - b^2 : c^2).$

This point is the crosspoint of the tangent lines to $\Omega$ in points $A$ and $C,$ so \[A' = A, C' = C, B' = \left (a^2 : -\frac{b^2}{2} : c^2 \right ).\] The line $B'C'$ is $b^2 x + 2y a^2 = 0 \implies D = \left( 2a^2 : -b^2 : c^2 \right).$

Similarly, $E = ( a^2 : 0 : - c^2), F = (a^2 : - b^2 : 2c^2).$

The line $DEF$ is $\frac {x}{a^2} + \frac {3y}{b^2} + \frac {z}{c^2} = 0.$

Similarly, if $P = (-a^2 : b^2 : c^2),$ then the line $DEF$ is $\frac {3x}{a^2} + \frac {y}{b^2} + \frac {z}{c^2} = 0.$

If $P = (a^2 : b^2 : -c^2),$ then the line $DEF$ is $\frac {x}{a^2} + \frac {y}{b^2} + \frac {3z}{c^2} = 0.$

These three lines intersect in pairs at points $D, E,$ and $F$ of the line of case 1.

3. Common case. Denote the coordinates of the point $P = (x_P : y_P : z_P).$ The equation of $\Omega$ is $xyc^2 + xzb^2 + yza^2 = 0.$

Line $AP$ is $l z_P + m y_P = 0 \implies A' = \left( \frac{-y_P \cdot z_P a^2}{y_P c^2 + z_P b^2} : y_P :z_P \right ).$

Similarly, $B' = \left (x_P : \frac{-x_P \cdot z_P b^2}{x_P c^2 + z_P a^2} : z_P \right ), C' = \left (x_P : y_P : \frac{-x_P \cdot y_P c^2}{x_P b^2 + y_P a^2} \right ).$

The tangent line $l_A$ to $\Omega$ at $A$ is $yC^2 +zb^2=0.$

The line $B'C'$ is $\frac{y_P \cdot z_P a^2}{x_P} x - (x_P \cdot c^2 + z_P a^2)y - (x_P b^2 +y_Pa^2)z = 0.$

$D = l_A \cap B'C' = \left( x_P (y_P c^2 - z_P b^2) : -y_P \cdot z_P b^2 : y_P \cdot z_P c^2 \right).$

Similarly, $E = l_B \cap A'C' = \left( x_P z_P a^2 : -y_P (x_P c^2 - z_P a^2) : -x_P \cdot z_P c^2 \right).$ \[F = l_C \cap A'B' = \left( x_P y_P a^2 : -x_P y_P b^2 : z_P (y_P a^2 - x_P b^2) \right).\] The line $DEF$ is \[\frac {x}{x_P} (-\frac {a^2}{x_P} + \frac {b^2}{y_P} + \frac {c^2}{z_P}) + \frac {y}{y_P} (\frac {a^2}{x_P} - \frac {b^2}{y_P} + \frac {c^2}{z_P}) +\frac {z}{z_P} (\frac {a^2}{x_P} + \frac {b^2}{y_P} - \frac {c^2}{z_P}) = 0.\]

vladimir.shelomovskii@gmail.com, vvsss

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Barycentric_coordinates&oldid=235305"