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 triangle barycentric-coordinates
edited 2 days ago
Alex Ravsky
39.4k32181
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
This question has an open bounty worth +50
reputation from Anubhab Ghosal ending in 4 days.
This question has not received enough attention.
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 triangle barycentric-coordinates
edited 2 days ago
Alex Ravsky
39.4k32181
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
This question has an open bounty worth +50
reputation from Anubhab Ghosal ending in 4 days.
This question has not received enough attention.
@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
Jan 3 at 15:32
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
Jan 3 at 17:27
@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
Jan 3 at 17:58
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
Jan 3 at 18:54
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
Jan 3 at 19:01
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 triangle barycentric-coordinates
edited 2 days ago
Alex Ravsky
39.4k32181
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
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 triangle barycentric-coordinates
geometry triangle barycentric-coordinates
edited 2 days ago
Alex Ravsky
39.4k32181
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
edited 2 days ago
Alex Ravsky
39.4k32181
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
edited 2 days ago
Alex Ravsky
39.4k32181
edited 2 days ago
Alex Ravsky
39.4k32181
edited 2 days ago
Alex Ravsky
39.4k32181
39.4k32181
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
asked Jan 3 at 14:00
Anubhab GhosalAnubhab Ghosal
81617
81617
This question has an open bounty worth +50
reputation from Anubhab Ghosal ending in 4 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from Anubhab Ghosal ending in 4 days.
This question has not received enough attention.
@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
Jan 3 at 15:32
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
Jan 3 at 17:27
@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
Jan 3 at 17:58
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
Jan 3 at 18:54
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
Jan 3 at 19:01
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
Jan 3 at 15:32
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
Jan 3 at 17:27
@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
Jan 3 at 17:58
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
Jan 3 at 18:54
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
Jan 3 at 19:01
@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
Jan 3 at 15:32
@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
Jan 3 at 15:32
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
Jan 3 at 17:27
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
Jan 3 at 17:27
@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
Jan 3 at 17:58
@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
Jan 3 at 17:58
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
Jan 3 at 18:54
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
Jan 3 at 18:54
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
Jan 3 at 19:01
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
Jan 3 at 19:01
add a comment |
2 Answers
2
active
oldest
votes
The area of a triangle whose vertices have cartesian coordinates $(x_i, y_i)$ is
$$frac 1 2 begin{vmatrix}
x_2 - x_1 & y_2 - y_1 \
x_3 - x_1 & y_3 - y_1
end{vmatrix} =
frac 1 2 begin{vmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{vmatrix}.$$
If the three points $(x_i, y_i)$ have normalized barycentric coordinates $(u_i, v_i, w_i)$, then
$$frac 1 2 begin{pmatrix}
u_1 & v_1 & w_1 \
u_2 & v_2 & w_2 \
u_3 & v_3 & w_3
end{pmatrix}
begin{pmatrix}
x_A & y_A & 1 \
x_B & y_B & 1 \
x_C & y_C & 1
end{pmatrix} =
frac 1 2 begin{pmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{pmatrix}.$$
The determinant of a matrix product is the product of the determinants.
answered 2 days ago
MaximMaxim
4,5781219
add a comment |
Without loss of generality, let $,ABC,$ be the three basis vectors in a cartesian coordinate system and let $,O,$ be the origin. The triangle
$,triangle ABC,$ is the convex hull of $,{A,B,C},$ and is the base of a tetrahedron with vertex at $,O,$. Any three points
$,{P_1,P_2,P_3},$ in the plane of $,triangle ABC,$ also form the base of a tetrahedron with vertex at $,O.,$ It is well-known that the volume of such a tetrahedon is $,1/6,$ the area of the base times the altitude to that base, and also that the volume is $,1/6,$ the determinant of the matrix given by the coordinates of the three points. The requested result follows. The key fact needed is that relative length, area, or volume is an affine invariant.
answered 2 days ago
SomosSomos
13.1k11034
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The area of a triangle whose vertices have cartesian coordinates $(x_i, y_i)$ is
$$frac 1 2 begin{vmatrix}
x_2 - x_1 & y_2 - y_1 \
x_3 - x_1 & y_3 - y_1
end{vmatrix} =
frac 1 2 begin{vmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{vmatrix}.$$
If the three points $(x_i, y_i)$ have normalized barycentric coordinates $(u_i, v_i, w_i)$, then
$$frac 1 2 begin{pmatrix}
u_1 & v_1 & w_1 \
u_2 & v_2 & w_2 \
u_3 & v_3 & w_3
end{pmatrix}
begin{pmatrix}
x_A & y_A & 1 \
x_B & y_B & 1 \
x_C & y_C & 1
end{pmatrix} =
frac 1 2 begin{pmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{pmatrix}.$$
The determinant of a matrix product is the product of the determinants.
answered 2 days ago
MaximMaxim
4,5781219
add a comment |
The area of a triangle whose vertices have cartesian coordinates $(x_i, y_i)$ is
$$frac 1 2 begin{vmatrix}
x_2 - x_1 & y_2 - y_1 \
x_3 - x_1 & y_3 - y_1
end{vmatrix} =
frac 1 2 begin{vmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{vmatrix}.$$
If the three points $(x_i, y_i)$ have normalized barycentric coordinates $(u_i, v_i, w_i)$, then
$$frac 1 2 begin{pmatrix}
u_1 & v_1 & w_1 \
u_2 & v_2 & w_2 \
u_3 & v_3 & w_3
end{pmatrix}
begin{pmatrix}
x_A & y_A & 1 \
x_B & y_B & 1 \
x_C & y_C & 1
end{pmatrix} =
frac 1 2 begin{pmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{pmatrix}.$$
The determinant of a matrix product is the product of the determinants.
answered 2 days ago
MaximMaxim
4,5781219
add a comment |
The area of a triangle whose vertices have cartesian coordinates $(x_i, y_i)$ is
$$frac 1 2 begin{vmatrix}
x_2 - x_1 & y_2 - y_1 \
x_3 - x_1 & y_3 - y_1
end{vmatrix} =
frac 1 2 begin{vmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{vmatrix}.$$
If the three points $(x_i, y_i)$ have normalized barycentric coordinates $(u_i, v_i, w_i)$, then
$$frac 1 2 begin{pmatrix}
u_1 & v_1 & w_1 \
u_2 & v_2 & w_2 \
u_3 & v_3 & w_3
end{pmatrix}
begin{pmatrix}
x_A & y_A & 1 \
x_B & y_B & 1 \
x_C & y_C & 1
end{pmatrix} =
frac 1 2 begin{pmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{pmatrix}.$$
The determinant of a matrix product is the product of the determinants.
answered 2 days ago
MaximMaxim
4,5781219
The area of a triangle whose vertices have cartesian coordinates $(x_i, y_i)$ is
$$frac 1 2 begin{vmatrix}
x_2 - x_1 & y_2 - y_1 \
x_3 - x_1 & y_3 - y_1
end{vmatrix} =
frac 1 2 begin{vmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{vmatrix}.$$
If the three points $(x_i, y_i)$ have normalized barycentric coordinates $(u_i, v_i, w_i)$, then
$$frac 1 2 begin{pmatrix}
u_1 & v_1 & w_1 \
u_2 & v_2 & w_2 \
u_3 & v_3 & w_3
end{pmatrix}
begin{pmatrix}
x_A & y_A & 1 \
x_B & y_B & 1 \
x_C & y_C & 1
end{pmatrix} =
frac 1 2 begin{pmatrix}
x_1 & y_1 & 1 \
x_2 & y_2 & 1 \
x_3 & y_3 & 1
end{pmatrix}.$$
The determinant of a matrix product is the product of the determinants.
answered 2 days ago
MaximMaxim
4,5781219
answered 2 days ago
MaximMaxim
4,5781219
answered 2 days ago
MaximMaxim
4,5781219
answered 2 days ago
MaximMaxim
4,5781219
4,5781219
add a comment |
add a comment |
Without loss of generality, let $,ABC,$ be the three basis vectors in a cartesian coordinate system and let $,O,$ be the origin. The triangle
$,triangle ABC,$ is the convex hull of $,{A,B,C},$ and is the base of a tetrahedron with vertex at $,O,$. Any three points
$,{P_1,P_2,P_3},$ in the plane of $,triangle ABC,$ also form the base of a tetrahedron with vertex at $,O.,$ It is well-known that the volume of such a tetrahedon is $,1/6,$ the area of the base times the altitude to that base, and also that the volume is $,1/6,$ the determinant of the matrix given by the coordinates of the three points. The requested result follows. The key fact needed is that relative length, area, or volume is an affine invariant.
answered 2 days ago
SomosSomos
13.1k11034
add a comment |
Without loss of generality, let $,ABC,$ be the three basis vectors in a cartesian coordinate system and let $,O,$ be the origin. The triangle
$,triangle ABC,$ is the convex hull of $,{A,B,C},$ and is the base of a tetrahedron with vertex at $,O,$. Any three points
$,{P_1,P_2,P_3},$ in the plane of $,triangle ABC,$ also form the base of a tetrahedron with vertex at $,O.,$ It is well-known that the volume of such a tetrahedon is $,1/6,$ the area of the base times the altitude to that base, and also that the volume is $,1/6,$ the determinant of the matrix given by the coordinates of the three points. The requested result follows. The key fact needed is that relative length, area, or volume is an affine invariant.
answered 2 days ago
SomosSomos
13.1k11034
add a comment |
Without loss of generality, let $,ABC,$ be the three basis vectors in a cartesian coordinate system and let $,O,$ be the origin. The triangle
$,triangle ABC,$ is the convex hull of $,{A,B,C},$ and is the base of a tetrahedron with vertex at $,O,$. Any three points
$,{P_1,P_2,P_3},$ in the plane of $,triangle ABC,$ also form the base of a tetrahedron with vertex at $,O.,$ It is well-known that the volume of such a tetrahedon is $,1/6,$ the area of the base times the altitude to that base, and also that the volume is $,1/6,$ the determinant of the matrix given by the coordinates of the three points. The requested result follows. The key fact needed is that relative length, area, or volume is an affine invariant.
answered 2 days ago
SomosSomos
13.1k11034
Without loss of generality, let $,ABC,$ be the three basis vectors in a cartesian coordinate system and let $,O,$ be the origin. The triangle
$,triangle ABC,$ is the convex hull of $,{A,B,C},$ and is the base of a tetrahedron with vertex at $,O,$. Any three points
$,{P_1,P_2,P_3},$ in the plane of $,triangle ABC,$ also form the base of a tetrahedron with vertex at $,O.,$ It is well-known that the volume of such a tetrahedon is $,1/6,$ the area of the base times the altitude to that base, and also that the volume is $,1/6,$ the determinant of the matrix given by the coordinates of the three points. The requested result follows. The key fact needed is that relative length, area, or volume is an affine invariant.
answered 2 days ago
SomosSomos
13.1k11034
edited 2 days ago
answered 2 days ago
SomosSomos
13.1k11034
answered 2 days ago
SomosSomos
13.1k11034
answered 2 days ago
SomosSomos
13.1k11034
13.1k11034
add a comment |
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
Jan 3 at 15:32
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
Jan 3 at 17:27
@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
Jan 3 at 17:58
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
Jan 3 at 18:54
@ja72, two subscripts is cumbersome.
– Anubhab Ghosal
Jan 3 at 19:01