Difference between revisions of "Barycentric coordinates"

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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

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

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

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

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

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

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

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

Two pare isogonal points and isogonals

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 $R = PQ \cap P'Q', E = \Omega \cap \odot RPQ', K = EQ' \cap \Omega, AD || PQ.$

Prove that $AK$ and $AD$ are the isogonals with respect $\angle BAC.$

Proof

As was shown before, the line $PQ$ is $f(x, y, z) = (y_P z_Q - z_P y_Q) x + (x_Q z_P - x_P z_Q) y + (x_P y_Q - x_Q y_P)z = 0.$ The point $K$ is $K = \left( \frac {a^2}{x_P - x_Q} : \frac {b^2}{y_P - y_Q} :\frac {c^2}{z_P - z_Q}\right)$ (normalized barycentric coordinates NBC).

Point $K'$ is the isogonal conjugate of a point $K$ with respect to $\triangle ABC.$ $K' = \left( {x_P - x_Q} : {y_P - y_Q} : {z_P - z_Q}\right).$ $f(K') = (y_P z_Q - z_P y_Q) (x_P - x_Q) + (x_Q z_P - x_P z_Q) (y_P - y_Q) + (x_P y_Q - x_Q y_P)(z_P - z_Q) = 0.$ Therefore point in infinity $K'$ lies on $PQ$ and on $AD||PQ.$

Point $K'$ is isogonal to $K$ with respect to $\triangle ABC \implies AK' = AD$ and AK are isogonals with respect $\angle BAC. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 149: / Line 149: @@
 <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>\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>

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Difference between revisions of "Barycentric coordinates"

Revision as of 05:20, 1 October 2023

Contents

Useful formulas

Product of isogonal segments

Ratio of isogonal segments

Point on incircle

Crossing point

Fixed point on circumcircle

Two pare isogonal points

Two pare isogonal points and isogonals