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How do you calculate the weighting of a point inside of an equilateral triangle compared to its vertices?

How to find the inverse position inside a triangleHow to find the other vertices of an equilateral triangle given one vertex and centroidIn an equilateral triangle what is sum of distance from vertices to a point inside the triangle?Is it possible to find the vertices of an equilateral triangle given its center point?Maximize the distance to vertices in an equilateral triangleFinding the length of a side of an equilateral trianglePoint inside a triangle that is the same distance from each vertexhow to distribute the weight of a point among the vertices of a square in which it lies?Finding the largest equilateral triangle inside a given trianglePyramid Triangles

2

$begingroup$

In an equilateral triangle that contains a point, how do you calculate 3 weights that sum to 100% and indicate how much influence each vertex has on the point.

When the point is in the center all the weights are 33%:

And if it's on one edge they should be split between the vertices that share that edge:

This is similar to how an HSL color wheel works:

geometry triangle

share|cite|improve this question

asked Mar 6 at 2:44

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Barycentric_coordinate_system
  $endgroup$
  – Rahul
  Mar 6 at 3:52
  

  
  
  
  

add a comment | 

2

$begingroup$

In an equilateral triangle that contains a point, how do you calculate 3 weights that sum to 100% and indicate how much influence each vertex has on the point.

When the point is in the center all the weights are 33%:

And if it's on one edge they should be split between the vertices that share that edge:

This is similar to how an HSL color wheel works:

geometry triangle

share|cite|improve this question

asked Mar 6 at 2:44

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Barycentric_coordinate_system
  $endgroup$
  – Rahul
  Mar 6 at 3:52
  

  
  
  
  

add a comment | 

$begingroup$

In an equilateral triangle that contains a point, how do you calculate 3 weights that sum to 100% and indicate how much influence each vertex has on the point.

When the point is in the center all the weights are 33%:

And if it's on one edge they should be split between the vertices that share that edge:

This is similar to how an HSL color wheel works:

geometry triangle

share|cite|improve this question

asked Mar 6 at 2:44

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 6 at 2:44

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 6 at 2:44

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 6 at 2:44

DShookDShook

1114

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Check out our Code of Conduct.

asked Mar 6 at 2:44

DShookDShook

1114

2 Answers
2

active

oldest

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1

$begingroup$

Let's think of three different points of the plane, $P$, $Q$ and $R$, not all co-linear, each with some coordinate vector in $mathbb{R}^2$. If we only deal with two at first, $P$ and $Q$, we can write a one parameter interpolation as:
$$
X = P(1-t)+Qt,
$$
for $tin[0,1]$. If you like, $100t$ gives you the percentage of $Q$'s weight, and $100(1-t)$ is the percentage of $P$'s weight. Also, if you pick any point $Xinmathbb{R}^2$ which lies in the segment between $P$ and $Q$, there is only one value of $t$ such that $X = P(1-t)+Qt$, because the equation is linear.

Now, for the third point $R$. Since $P(1-t)+Qt$ already describes all the points in the $PQ$ segment, we interpolate this expression again with the point $R$, obtaining
$$
X=[P(1-t)+Qt](1-s)+Rs.
$$
for $sin[0,1]$. Expanding, we get
$$
X=P(1-t)(1-s)+Qt(1-s)+Rs.
$$
It looks worse than before, but it is the same trick. Any point contained in the triangle $PQR$ can be uniquely identified with two values $tin[0,1]$ and $sin[0,1]$. And again, $100(1-t)(1-s)$ is the percentage weight of $P$, $100t(1-s)$ that of $Q$, and $100s$ that of $R$.

I hope this helps!

share|cite|improve this answer

answered Mar 6 at 3:03

R_BR_B

563110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not following here unfortunately. Did you rename A,B,C in the diagrams to P,Q,R? What is X? Where do the coordinates of the point fit into the equation?
  $endgroup$
  – DShook
  Mar 6 at 3:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed,$P$, $Q$ and $R$ are your $A$, $B$ and $C$. $X$ is the point in the triangle which you want to express as a combination of the vertex points. The coordinates of the points come into it because all the equations I wrote are vector equations. Writing $X=P(1-t)+Qt$ is shorthand for $X_1=P_1(1-t)+Q_1 t$ and $X_2=P_2(1-t)+Q_2 t$, where $X$ is the vector with coordinates $(X_1, X_2)$, and likewise for $P$, $Q$ and $R$.
  $endgroup$
  – R_B
  Mar 6 at 3:29
  

  
  
  
  

add a comment | 

0

$begingroup$

I ended up using another method to solve this. In the diagram below to calculate the weight for a point, find the distance from the control point to the line opposite it and then divide by the triangle's height.

For the weight of A:

weightOfA = lengthOfx / triangleHeight

share|cite|improve this answer

answered yesterday

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

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1

$begingroup$

Let's think of three different points of the plane, $P$, $Q$ and $R$, not all co-linear, each with some coordinate vector in $mathbb{R}^2$. If we only deal with two at first, $P$ and $Q$, we can write a one parameter interpolation as:
$$
X = P(1-t)+Qt,
$$
for $tin[0,1]$. If you like, $100t$ gives you the percentage of $Q$'s weight, and $100(1-t)$ is the percentage of $P$'s weight. Also, if you pick any point $Xinmathbb{R}^2$ which lies in the segment between $P$ and $Q$, there is only one value of $t$ such that $X = P(1-t)+Qt$, because the equation is linear.

Now, for the third point $R$. Since $P(1-t)+Qt$ already describes all the points in the $PQ$ segment, we interpolate this expression again with the point $R$, obtaining
$$
X=[P(1-t)+Qt](1-s)+Rs.
$$
for $sin[0,1]$. Expanding, we get
$$
X=P(1-t)(1-s)+Qt(1-s)+Rs.
$$
It looks worse than before, but it is the same trick. Any point contained in the triangle $PQR$ can be uniquely identified with two values $tin[0,1]$ and $sin[0,1]$. And again, $100(1-t)(1-s)$ is the percentage weight of $P$, $100t(1-s)$ that of $Q$, and $100s$ that of $R$.

I hope this helps!

share|cite|improve this answer

answered Mar 6 at 3:03

R_BR_B

563110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not following here unfortunately. Did you rename A,B,C in the diagrams to P,Q,R? What is X? Where do the coordinates of the point fit into the equation?
  $endgroup$
  – DShook
  Mar 6 at 3:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed,$P$, $Q$ and $R$ are your $A$, $B$ and $C$. $X$ is the point in the triangle which you want to express as a combination of the vertex points. The coordinates of the points come into it because all the equations I wrote are vector equations. Writing $X=P(1-t)+Qt$ is shorthand for $X_1=P_1(1-t)+Q_1 t$ and $X_2=P_2(1-t)+Q_2 t$, where $X$ is the vector with coordinates $(X_1, X_2)$, and likewise for $P$, $Q$ and $R$.
  $endgroup$
  – R_B
  Mar 6 at 3:29
  

  
  
  
  

add a comment | 

1

$begingroup$

Let's think of three different points of the plane, $P$, $Q$ and $R$, not all co-linear, each with some coordinate vector in $mathbb{R}^2$. If we only deal with two at first, $P$ and $Q$, we can write a one parameter interpolation as:
$$
X = P(1-t)+Qt,
$$
for $tin[0,1]$. If you like, $100t$ gives you the percentage of $Q$'s weight, and $100(1-t)$ is the percentage of $P$'s weight. Also, if you pick any point $Xinmathbb{R}^2$ which lies in the segment between $P$ and $Q$, there is only one value of $t$ such that $X = P(1-t)+Qt$, because the equation is linear.

Now, for the third point $R$. Since $P(1-t)+Qt$ already describes all the points in the $PQ$ segment, we interpolate this expression again with the point $R$, obtaining
$$
X=[P(1-t)+Qt](1-s)+Rs.
$$
for $sin[0,1]$. Expanding, we get
$$
X=P(1-t)(1-s)+Qt(1-s)+Rs.
$$
It looks worse than before, but it is the same trick. Any point contained in the triangle $PQR$ can be uniquely identified with two values $tin[0,1]$ and $sin[0,1]$. And again, $100(1-t)(1-s)$ is the percentage weight of $P$, $100t(1-s)$ that of $Q$, and $100s$ that of $R$.

I hope this helps!

share|cite|improve this answer

answered Mar 6 at 3:03

R_BR_B

563110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not following here unfortunately. Did you rename A,B,C in the diagrams to P,Q,R? What is X? Where do the coordinates of the point fit into the equation?
  $endgroup$
  – DShook
  Mar 6 at 3:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed,$P$, $Q$ and $R$ are your $A$, $B$ and $C$. $X$ is the point in the triangle which you want to express as a combination of the vertex points. The coordinates of the points come into it because all the equations I wrote are vector equations. Writing $X=P(1-t)+Qt$ is shorthand for $X_1=P_1(1-t)+Q_1 t$ and $X_2=P_2(1-t)+Q_2 t$, where $X$ is the vector with coordinates $(X_1, X_2)$, and likewise for $P$, $Q$ and $R$.
  $endgroup$
  – R_B
  Mar 6 at 3:29
  

  
  
  
  

add a comment | 

$begingroup$

Let's think of three different points of the plane, $P$, $Q$ and $R$, not all co-linear, each with some coordinate vector in $mathbb{R}^2$. If we only deal with two at first, $P$ and $Q$, we can write a one parameter interpolation as:
$$
X = P(1-t)+Qt,
$$
for $tin[0,1]$. If you like, $100t$ gives you the percentage of $Q$'s weight, and $100(1-t)$ is the percentage of $P$'s weight. Also, if you pick any point $Xinmathbb{R}^2$ which lies in the segment between $P$ and $Q$, there is only one value of $t$ such that $X = P(1-t)+Qt$, because the equation is linear.

Now, for the third point $R$. Since $P(1-t)+Qt$ already describes all the points in the $PQ$ segment, we interpolate this expression again with the point $R$, obtaining
$$
X=[P(1-t)+Qt](1-s)+Rs.
$$
for $sin[0,1]$. Expanding, we get
$$
X=P(1-t)(1-s)+Qt(1-s)+Rs.
$$
It looks worse than before, but it is the same trick. Any point contained in the triangle $PQR$ can be uniquely identified with two values $tin[0,1]$ and $sin[0,1]$. And again, $100(1-t)(1-s)$ is the percentage weight of $P$, $100t(1-s)$ that of $Q$, and $100s$ that of $R$.

I hope this helps!

share|cite|improve this answer

answered Mar 6 at 3:03

R_BR_B

563110

$endgroup$

share|cite|improve this answer

answered Mar 6 at 3:03

R_BR_B

563110

answered Mar 6 at 3:03

R_BR_B

563110

answered Mar 6 at 3:03

R_BR_B

563110

0

$begingroup$

I ended up using another method to solve this. In the diagram below to calculate the weight for a point, find the distance from the control point to the line opposite it and then divide by the triangle's height.

For the weight of A:

weightOfA = lengthOfx / triangleHeight

share|cite|improve this answer

answered yesterday

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

I ended up using another method to solve this. In the diagram below to calculate the weight for a point, find the distance from the control point to the line opposite it and then divide by the triangle's height.

For the weight of A:

weightOfA = lengthOfx / triangleHeight

share|cite|improve this answer

answered yesterday

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I ended up using another method to solve this. In the diagram below to calculate the weight for a point, find the distance from the control point to the line opposite it and then divide by the triangle's height.

For the weight of A:

weightOfA = lengthOfx / triangleHeight

share|cite|improve this answer

answered yesterday

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

weightOfA = lengthOfx / triangleHeight

share|cite|improve this answer

answered yesterday

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

DShookDShook

1114

New contributor

DShook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

DShookDShook

1114