Prove triangle area formula for barycentric coordinates
Let $P_1, P_2, P_3$ be points with barycentric
coordinates (with reference triangle $ABC$) $P_i = (x_i, y_i, z_i )$ for $i = 1, 2, 3$. Then the signed area of $Delta P_1P_2P_3$ is given by the determinant $$frac{[P_1P_2P_3]}{[ABC]}=begin{vmatrix} x_1& y_1& z_1 \ x_2& y_2& z_2\x_3& y_3& z_3 end{vmatrix}$$
I came across this theorem in Evan Chen's "Euclidean Geometry in Mathematical Olympiads" where the proof is skipped. I failed to prove this myself and cannot find the proof online. Any help will be appreciated.
geometry barycentric-coordinates
add a comment |
Let $P_1, P_2, P_3$ be points with barycentric
coordinates (with reference triangle $ABC$) $P_i = (x_i, y_i, z_i )$ for $i = 1, 2, 3$. Then the signed area of $Delta P_1P_2P_3$ is given by the determinant $$frac{[P_1P_2P_3]}{[ABC]}=begin{vmatrix} x_1& y_1& z_1 \ x_2& y_2& z_2\x_3& y_3& z_3 end{vmatrix}$$
I came across this theorem in Evan Chen's "Euclidean Geometry in Mathematical Olympiads" where the proof is skipped. I failed to prove this myself and cannot find the proof online. Any help will be appreciated.
geometry barycentric-coordinates
@ja72, I fail to understand how notation plays a role when I have specifically mentioned that the coordinates are barycentric. Moreover, the conversion between barycentric and cartesian is not that simple. See en.wikipedia.org/wiki/….
– Anubhab Ghosal
yesterday
You are right. I was just trying to understand the question, and in the process, I confused myself and possibly others. I am deleting my original comment.
– ja72
yesterday
@ja72, will my changing z to w perhaps lessen the probability of people being confused? If it will, I shall edit my question.(You know, I am unaware of the popular notation)
– Anubhab Ghosal
yesterday
Barycentric coordinates are ratios of distances (weights) and using $x$, $y$ and $z$ is confusing. I prefer $w_A$, $w_B$ and $w_C$ for the coordinates such that the a homogeneous point is defined by $$boldsymbol{P}(w_A,w_B,w_C) = w_A boldsymbol{A}+w_B boldsymbol{B} + w_C boldsymbol{C}$$
– ja72
yesterday
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
yesterday
add a comment |
Let $P_1, P_2, P_3$ be points with barycentric
coordinates (with reference triangle $ABC$) $P_i = (x_i, y_i, z_i )$ for $i = 1, 2, 3$. Then the signed area of $Delta P_1P_2P_3$ is given by the determinant $$frac{[P_1P_2P_3]}{[ABC]}=begin{vmatrix} x_1& y_1& z_1 \ x_2& y_2& z_2\x_3& y_3& z_3 end{vmatrix}$$
I came across this theorem in Evan Chen's "Euclidean Geometry in Mathematical Olympiads" where the proof is skipped. I failed to prove this myself and cannot find the proof online. Any help will be appreciated.
geometry barycentric-coordinates
Let $P_1, P_2, P_3$ be points with barycentric
coordinates (with reference triangle $ABC$) $P_i = (x_i, y_i, z_i )$ for $i = 1, 2, 3$. Then the signed area of $Delta P_1P_2P_3$ is given by the determinant $$frac{[P_1P_2P_3]}{[ABC]}=begin{vmatrix} x_1& y_1& z_1 \ x_2& y_2& z_2\x_3& y_3& z_3 end{vmatrix}$$
I came across this theorem in Evan Chen's "Euclidean Geometry in Mathematical Olympiads" where the proof is skipped. I failed to prove this myself and cannot find the proof online. Any help will be appreciated.
geometry barycentric-coordinates
geometry barycentric-coordinates
edited yesterday
asked yesterday
Anubhab Ghosal
84915
84915
@ja72, I fail to understand how notation plays a role when I have specifically mentioned that the coordinates are barycentric. Moreover, the conversion between barycentric and cartesian is not that simple. See en.wikipedia.org/wiki/….
– Anubhab Ghosal
yesterday
You are right. I was just trying to understand the question, and in the process, I confused myself and possibly others. I am deleting my original comment.
– ja72
yesterday
@ja72, will my changing z to w perhaps lessen the probability of people being confused? If it will, I shall edit my question.(You know, I am unaware of the popular notation)
– Anubhab Ghosal
yesterday
Barycentric coordinates are ratios of distances (weights) and using $x$, $y$ and $z$ is confusing. I prefer $w_A$, $w_B$ and $w_C$ for the coordinates such that the a homogeneous point is defined by $$boldsymbol{P}(w_A,w_B,w_C) = w_A boldsymbol{A}+w_B boldsymbol{B} + w_C boldsymbol{C}$$
– ja72
yesterday
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
yesterday
add a comment |
@ja72, I fail to understand how notation plays a role when I have specifically mentioned that the coordinates are barycentric. Moreover, the conversion between barycentric and cartesian is not that simple. See en.wikipedia.org/wiki/….
– Anubhab Ghosal
yesterday
You are right. I was just trying to understand the question, and in the process, I confused myself and possibly others. I am deleting my original comment.
– ja72
yesterday
@ja72, will my changing z to w perhaps lessen the probability of people being confused? If it will, I shall edit my question.(You know, I am unaware of the popular notation)
– Anubhab Ghosal
yesterday
Barycentric coordinates are ratios of distances (weights) and using $x$, $y$ and $z$ is confusing. I prefer $w_A$, $w_B$ and $w_C$ for the coordinates such that the a homogeneous point is defined by $$boldsymbol{P}(w_A,w_B,w_C) = w_A boldsymbol{A}+w_B boldsymbol{B} + w_C boldsymbol{C}$$
– ja72
yesterday
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
yesterday
@ja72, I fail to understand how notation plays a role when I have specifically mentioned that the coordinates are barycentric. Moreover, the conversion between barycentric and cartesian is not that simple. See en.wikipedia.org/wiki/….
– Anubhab Ghosal
yesterday
@ja72, I fail to understand how notation plays a role when I have specifically mentioned that the coordinates are barycentric. Moreover, the conversion between barycentric and cartesian is not that simple. See en.wikipedia.org/wiki/….
– Anubhab Ghosal
yesterday
You are right. I was just trying to understand the question, and in the process, I confused myself and possibly others. I am deleting my original comment.
– ja72
yesterday
You are right. I was just trying to understand the question, and in the process, I confused myself and possibly others. I am deleting my original comment.
– ja72
yesterday
@ja72, will my changing z to w perhaps lessen the probability of people being confused? If it will, I shall edit my question.(You know, I am unaware of the popular notation)
– Anubhab Ghosal
yesterday
@ja72, will my changing z to w perhaps lessen the probability of people being confused? If it will, I shall edit my question.(You know, I am unaware of the popular notation)
– Anubhab Ghosal
yesterday
Barycentric coordinates are ratios of distances (weights) and using $x$, $y$ and $z$ is confusing. I prefer $w_A$, $w_B$ and $w_C$ for the coordinates such that the a homogeneous point is defined by $$boldsymbol{P}(w_A,w_B,w_C) = w_A boldsymbol{A}+w_B boldsymbol{B} + w_C boldsymbol{C}$$
– ja72
yesterday
Barycentric coordinates are ratios of distances (weights) and using $x$, $y$ and $z$ is confusing. I prefer $w_A$, $w_B$ and $w_C$ for the coordinates such that the a homogeneous point is defined by $$boldsymbol{P}(w_A,w_B,w_C) = w_A boldsymbol{A}+w_B boldsymbol{B} + w_C boldsymbol{C}$$
– ja72
yesterday
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
yesterday
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
yesterday
add a comment |
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@ja72, I fail to understand how notation plays a role when I have specifically mentioned that the coordinates are barycentric. Moreover, the conversion between barycentric and cartesian is not that simple. See en.wikipedia.org/wiki/….
– Anubhab Ghosal
yesterday
You are right. I was just trying to understand the question, and in the process, I confused myself and possibly others. I am deleting my original comment.
– ja72
yesterday
@ja72, will my changing z to w perhaps lessen the probability of people being confused? If it will, I shall edit my question.(You know, I am unaware of the popular notation)
– Anubhab Ghosal
yesterday
Barycentric coordinates are ratios of distances (weights) and using $x$, $y$ and $z$ is confusing. I prefer $w_A$, $w_B$ and $w_C$ for the coordinates such that the a homogeneous point is defined by $$boldsymbol{P}(w_A,w_B,w_C) = w_A boldsymbol{A}+w_B boldsymbol{B} + w_C boldsymbol{C}$$
– ja72
yesterday
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
yesterday