Difference between revisions of "Barycentric coordinates"
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− | 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 |
Revision as of 03:38, 1 October 2023
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
[hide]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 vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
Two pare isogonal points and isogonals
Let triangle and points and be given. Let point and be the isogonal conjugate of the points and with respect to a triangle
Denote
Prove that and are the isogonals with respect
Proof
As was shown before, the line is The point is (normalized barycentric coordinates NBC).
Point is the isogonal conjugate of a point with respect to Therefore point in infinity lies on and on
Point is isogonal to with respect to and AK are isogonals with respect
vladimir.shelomovskii@gmail.com, vvsss