Difference between revisions of "Barycentric coordinates"
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<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: | ||
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<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> < | + | 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> | + | <cmath>x_X \cdot \vec {XA} + y_X \cdot \vec {XB} + z_X \cdot \vec {XC} = \vec {0},</cmath> |
− | <cmath>\vec X = \frac { | + | <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> | + | 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> | + | <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> | ||
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<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 < | + | 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 | ||
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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> | ||
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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 | ||
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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> | + | <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> | ||
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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 | + | 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> | ||
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'''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]] | ||
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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 | + | 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(- | + | <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' = ( | + | <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(- | + | 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. | ||
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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''' | '''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== | ==Point on incircle== | ||
− | Let triangle <math>\triangle ABC</math> be given. Denote the | + | [[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> | 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> | Symilarly denote <math>E_3 \in AC.</math> | ||
− | Prove that point <math>AD_3 \cap BE_3</math> lies on <math>\omega.</math> | + | Prove that point <math>F = AD_3 \cap BE_3</math> lies on <math>\omega.</math> |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
− | <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> | + | <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: | We calculate distances (using NBC) and solve the system of equations: | ||
<math>ID_1^2 = ID^2, SD_1^2 = SD^2.</math> | <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: | 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_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> | <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> | + | 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> | 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> | 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. | 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''' | ||
+ | |||
+ | ==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 15:12, 6 July 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 corresponding to masses placed at the vertices of a reference triangle
. These masses then determine a point
, which is the geometric centroid of the three masses and is identified with coordinates
. The vertices of the triangle are given by
,
, and
. 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.
Contents
Useful formulas
Notation
Let the triangle be a given triangle,
be the lengths of
We use the following Conway symbols:
is semiperimeter,
is twice the area of
where
is the inradius,
is the circumradius,
is the cosine of the Brocard angle,
Main
For any point in the plane there are barycentric coordinates(BC):
The normalized (absolute) barycentric coordinates NBC satisfy the condition
they are uniquely determined:
Triangle vertices
The barycentric coordinates of a point do not change under an affine transformation.
Lines
The straight line in barycentric coordinates (BC) is given by the equation
The lines given in the BC by the equations and
intersect at the point
These lines are parallel iff
The sideline contains the points
its equation is
The line has equation
it intersects the sideline
at the point
Iff then
Let NBC of points and
be
Then the square of distance
The equation of bisector of
is:
Nagel line :
Circles
Any circle is given by an equation of the form
Circumcircle contains the points
the equation of this circle:
The incircle contains the tangent points of the incircle with the sides:
The equation of the incircle is
where
The radical axis of two circles given by equations of this form is:
Conjugate
The point is isotomically conjugate with respect to
with the point
The point is isogonally conjugate with respect to
with the point
The point is isocircular conjugate with respect to
with the point
Triangle centers
The median centroid is
The simmedian point is isogonally conjugate with respect to
with the point
The bisector the incenter is
The excenters are
The circumcenter lies at the intersection of the bisectors
and
its BC coordinates
The orthocenter is isogonally conjugate with respect to
with the point
Let Nagel point lies at line
The Gergonne point is the isotomic conjugate of the Nagel point, so
vladimir.shelomovskii@gmail.com, vvsss
Product of isogonal segments
Let triangle the circumcircle
and isogonals
and
of the
be given.
Let point
and
be the isogonal conjugate of a point
and
with respect to
Prove that
Proof
We fixed and the point
So isogonal
is fixed.
Denote
We need to prove that do not depends from
Line has the equation
To find the point we solve the equation:
We use the formula for isogonal cobnjugate point and get
and then
To find the point we solve the equation:
We calculate distances (using NBC) and get:
where
has sufficiently big formula.
Therefore
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Ratio of isogonal segments
Let triangle and point
be given.
Denote
the isogonal conjugate of a point
with respect to
Prove that
Proof
We use the formula for isogonal conjugate point and get
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Point on incircle
Let triangle be given. Denote the incircle
the incenter
, the Spieker center
Let be the point corresponding to the condition
is symmetric
with respect midpoint
Symilarly denote
Prove that point lies on
Proof
We calculate distances (using NBC) and solve the system of equations:
We know one solution of this system (point D), so we get linear equation and get:
Similarly
Therefore
We calculate the length of the segment
and get
The author learned about the existence of such a point from Leonid Shatunov in August 2023.
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Crossing point
Let triangle and points
and
be given. Let point
be the isogonal conjugate of a point
with respect to a triangle
Let
be an arbitrary point at
Prove that
lies on
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 and
Proof
We use the barycentric coordinates:
We get the equations for some lines:
Line is
line is
line is
line is
line is
We get the equations for some points:
point is
point is
point is
Any circle is given by an equation of the form
We find the coefficients for the circles (these formulas are big), but can be used for calculations of the crossing points:
We get the equations for some lines
and
:
We get the equation for the point
Let point
be the isogonal conjugate of a point
with respect to a triangle
The sum of coordinates is equal zero, so
is in infinity, therefore the point
lies on
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Fixed point on circumcircle
Let triangle point
on circumcircle
and point
be given.
Point
lies on
point
be the isogonal conjugate of a point
with respect to a triangle
Prove that is fixed point and not depends from position of
Proof
Denote the coordinates of the points
The line
is
The line is
We find the circle
and get the point
depends only from points
and
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Two pare isogonal points
Let triangle and points
and
(points do not lie on sidelines) be given.
Let point and
be the isogonal conjugate of a point
and
with respect to a triangle
Denote
Prove that and
lies on
Proof
The line is
The line
is
Denote
is the isogonal conjugate of a point
with respect to
If we use NBC, we get
If we use NBC, we get
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Collinearity for two pares of isogonal points
Let triangle and points
and
be given. Let point
and
be the isogonal conjugate of the points
and
with respect to a triangle
Denote is the point isogonal conjugate to line
with respect
Isogonal_bijection_lines_and_points
Prove that points and
are collinear.
Proof
After the simple calculations one can get:
We use the normalized barycentric coordinates NBC and get line
in the form of:
We check the condition of collinearity for points
and
and finishing the proof.
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Points on bisectors
Let a triangle be given.
Let segments and
be the bisectors of
The lines and
meet circumcircle
) at points
respectively.
is the midpoint
Denote
We will find barycentric coordinates of the points and length of the segments.
Line
is
line
is
line
is
Circle is
Line is
Point
Line is
Point
Point
Some simple formulas:
Circumcenter
Tangent is
Line is
is the midpoint
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Small Pascal's theorem
Let and point
be given. Let
be the circumcircle of
Let the tangent line to
at point
cross line
at point
Similarly denote points
and
Prove that the points and
are collinear.
Proof
1. Simplest case, is the Lemoine point,
The equation of is
Line is
The line
is
Similarly,
The line is
2. Simple case, is one of the external Lemoine point,
This point is the crosspoint of the tangent lines to in points
and
so
The line
is
Similarly,
The line is
Similarly, if then the line
is
If then the line
is
These three lines intersect in pairs at points and
of the line of case 1.
3. Common case. Denote the coordinates of the point The equation of
is
Line is
Similarly,
The tangent line to
at
is
The line is
Similarly,
The line
is
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