Difference between revisions of "Barycentric coordinates"
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Tangent <math>BN</math> is <math>c^2 x + a^2 z = 0.</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+ | + | 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>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>BN = IN = \frac {abc}{|a-c|(a+b+c)}.</cmath> | ||
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<cmath>\frac {IC'}{FC'} = \frac {a+b-c}{c}; \frac {IA'}{DA'} = \frac {b+c-a}{a};</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> | <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
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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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