# Barycentric coordinates

This can be used in mass points.
http://mathworld.wolfram.com/BarycentricCoordinates.html
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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

*barycentric coordinates*

**normalized (absolute)***satisfy the condition they are uniquely determined: Triangle vertices*

**NBC**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

**vladimir.shelomovskii@gmail.com, vvsss**

## 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**

## 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.

**vladimir.shelomovskii@gmail.com, vvsss**

## 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

**vladimir.shelomovskii@gmail.com, vvsss**

## 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

**vladimir.shelomovskii@gmail.com, vvsss**