Barycentric coordinates

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. \[A= (1:0:0), B= (0:1:0), C= (0:0:1), I=(a:b:c),\] \[A'= (0:b:c), B'= (a:0:c), C'= (a:b:0), M = (1:1:0).\] Lines $AA': cy=bz, BB': cx=az, CC': bx=ay.$ \[\Omega:  xyc^2 + xzb^2 + yza^2 = 0.\] \[D = \Omega \cap AA' : \left( - \frac {a^2}{b+c}:b:c \right), E = \Omega \cap BB' : \left(a: - \frac {b^2}{a+c}:c \right),\] \[F = \Omega \cap CC' : \left(a:b: - \frac {c^2}{a+b} \right).\] \[K = BE \cap A'C' : (a : 2b : c)\]

Line $DF: x \frac {b+c}{a} +y + z\frac {a+b}{c} = 0.$ Point $L = BE \cap DF : (a : a+2b+c : c).$

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

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

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 $O \in FM : \left( a^2(b^2+c^2 -a^2) : b^2(a^2 + c^2 – b^2) : c^2(a^2 + b^2 - c^2) \right).$

Tangent $BN : c^2 x + a^2 z = 0.$

Line $NI || AC : \left( \frac {b}{a+c} = \frac{y}{x+y}  \right).$ \[N = BN \cap IN : (a^2 : b(a – c) : -c^2).\] \[BN = IN = \frac {abc}{|a-c|(a+b+c)}.\] \[\frac{HO}{BO} = \frac {|a-c|}{b}; \frac{GO}{BO} = \frac {|b-c|}{a}.\] $Q$ 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