Prove triangle area formula for barycentric coordinates












4














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







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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
















4














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







share|cite|improve this question

















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














4












4








4


3





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited 2 days ago









Alex Ravsky

39.4k32181




39.4k32181










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


















  • @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










2 Answers
2






active

oldest

votes


















2














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.






share|cite|improve this answer





























    1














    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.






    share|cite|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      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.






      share|cite|improve this answer


























        2














        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.






        share|cite|improve this answer
























          2












          2








          2






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          MaximMaxim

          4,5781219




          4,5781219























              1














              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.






              share|cite|improve this answer




























                1














                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.






                share|cite|improve this answer


























                  1












                  1








                  1






                  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.






                  share|cite|improve this answer














                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  SomosSomos

                  13.1k11034




                  13.1k11034






























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