Finding the incentre, circumcentre of a triangle.












0












$begingroup$


I am trying to discover another type of geometry. I want to transform a geometry problem to a coordinate geometry problem as a coordinate geometry problem that is easier than a pure geometry (Euclidean geometry) problem. A geometry problem needs deep thinking (I am telling about hard problems i.e. Olympiad level problems), but a coordinate geometry problem doesn't need deep thinking ,it just needs calculations. So, let us go to the original problem.




The vertices of a triangle $ABC$ are $(a_1,b_1),(a_2,b_2),(a_3,b_3)$. Find the following coordinates:



$1)$ Incentre



$2)$ Circumcentre



$3)$ Orthocentre



$4)$ Centre of the nine point circle




Please tell me with good explanation. Thank you.






geometry analytic-geometry







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am trying to discover another type of geometry. I want to transform a geometry problem to a coordinate geometry problem as a coordinate geometry problem that is easier than a pure geometry (Euclidean geometry) problem. A geometry problem needs deep thinking (I am telling about hard problems i.e. Olympiad level problems), but a coordinate geometry problem doesn't need deep thinking ,it just needs calculations. So, let us go to the original problem.




    The vertices of a triangle $ABC$ are $(a_1,b_1),(a_2,b_2),(a_3,b_3)$. Find the following coordinates:



    $1)$ Incentre



    $2)$ Circumcentre



    $3)$ Orthocentre



    $4)$ Centre of the nine point circle




    Please tell me with good explanation. Thank you.






    geometry analytic-geometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to discover another type of geometry. I want to transform a geometry problem to a coordinate geometry problem as a coordinate geometry problem that is easier than a pure geometry (Euclidean geometry) problem. A geometry problem needs deep thinking (I am telling about hard problems i.e. Olympiad level problems), but a coordinate geometry problem doesn't need deep thinking ,it just needs calculations. So, let us go to the original problem.




      The vertices of a triangle $ABC$ are $(a_1,b_1),(a_2,b_2),(a_3,b_3)$. Find the following coordinates:



      $1)$ Incentre



      $2)$ Circumcentre



      $3)$ Orthocentre



      $4)$ Centre of the nine point circle




      Please tell me with good explanation. Thank you.






      geometry analytic-geometry







      share|cite|improve this question











      $endgroup$




      I am trying to discover another type of geometry. I want to transform a geometry problem to a coordinate geometry problem as a coordinate geometry problem that is easier than a pure geometry (Euclidean geometry) problem. A geometry problem needs deep thinking (I am telling about hard problems i.e. Olympiad level problems), but a coordinate geometry problem doesn't need deep thinking ,it just needs calculations. So, let us go to the original problem.




      The vertices of a triangle $ABC$ are $(a_1,b_1),(a_2,b_2),(a_3,b_3)$. Find the following coordinates:



      $1)$ Incentre



      $2)$ Circumcentre



      $3)$ Orthocentre



      $4)$ Centre of the nine point circle




      Please tell me with good explanation. Thank you.






      geometry analytic-geometry




      geometry analytic-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 18 '17 at 3:12







      Sufaid Saleel

















      asked Sep 18 '17 at 2:53









      Sufaid SaleelSufaid Saleel

      1,760829




      1,760829






















          1 Answer
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          3












          $begingroup$

          There are well known formula for the triangle centers in trilinear coordinates $x : y : z$,



          $$begin{array}{rc:c}
          text{name} & & x : y : z \
          hline
          text{incenter} & I & 1 : 1 : 1\
          text{circumcenter} & O & cos A : cos B : cos C\
          text{orthocenter} & H & sec A : sec B :sec C\
          text{nine-point center} & N & cos(B-C) : cos(C-A) : cos(A-B)
          end{array}
          $$
          You can convert them to Cartesian coordinates by the recipe:
          $$P = (x,y,z) quad leftrightarrow quad vec{P} = frac{ax vec{A} + byvec{B} + czvec{C}}{ax + by + cz}$$



          In terms of coordinates, this means



          $$vec{P} = (p_1,p_2) =
          left(
          frac{ax a_1 + by a_2 + cz a_3}{ax + by + cz},
          frac{ax b_1 + by b_2 + cz b_3}{ax + by + cz}
          right)
          $$
          The parameters $a,b,c$ and $A,B,C$ are given by the formula:
          $$begin{cases}
          a &= sqrt{(a_2-a_3)^2 + (b_2 - b_3)^2}\
          b &= sqrt{(a_3-a_1)^2 + (b_3 - b_1)^2}\
          c &= sqrt{(a_1-a_2)^2 + (b_1 - b_2)^2}
          end{cases}
          quadtext{ and }quad
          begin{cases}
          A &= cos^{-1}frac{b^2+c^2-a^2}{2bc}\
          B &= cos^{-1}frac{c^2+a^2-b^2}{2ca}\
          C &= cos^{-1}frac{a^2+b^2-c^2}{2ab}
          end{cases}
          $$
          The parameters $x, y, z$ can be obtained by plugging these expressions of $a,b,c, A,B,C$ into corresponding formula in above table.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @achilille hui I want the coordinate not the trilinear coordinate.I need the Cartesian coordinate .So please tell me the right way
            $endgroup$
            – Sufaid Saleel
            Sep 18 '17 at 12:17










          • $begingroup$
            Thanks for the great write-up. In the equation defining c, I think that $b_2 - b_3$ should be replaced by $b_1 - b_2$.
            $endgroup$
            – Alex338207
            Jan 14 at 7:52












          • $begingroup$
            @Alex338207 yes, that's a typo, thanks for pointing that out.
            $endgroup$
            – achille hui
            Jan 14 at 13:28











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          1 Answer
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          1 Answer
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          3












          $begingroup$

          There are well known formula for the triangle centers in trilinear coordinates $x : y : z$,



          $$begin{array}{rc:c}
          text{name} & & x : y : z \
          hline
          text{incenter} & I & 1 : 1 : 1\
          text{circumcenter} & O & cos A : cos B : cos C\
          text{orthocenter} & H & sec A : sec B :sec C\
          text{nine-point center} & N & cos(B-C) : cos(C-A) : cos(A-B)
          end{array}
          $$
          You can convert them to Cartesian coordinates by the recipe:
          $$P = (x,y,z) quad leftrightarrow quad vec{P} = frac{ax vec{A} + byvec{B} + czvec{C}}{ax + by + cz}$$



          In terms of coordinates, this means



          $$vec{P} = (p_1,p_2) =
          left(
          frac{ax a_1 + by a_2 + cz a_3}{ax + by + cz},
          frac{ax b_1 + by b_2 + cz b_3}{ax + by + cz}
          right)
          $$
          The parameters $a,b,c$ and $A,B,C$ are given by the formula:
          $$begin{cases}
          a &= sqrt{(a_2-a_3)^2 + (b_2 - b_3)^2}\
          b &= sqrt{(a_3-a_1)^2 + (b_3 - b_1)^2}\
          c &= sqrt{(a_1-a_2)^2 + (b_1 - b_2)^2}
          end{cases}
          quadtext{ and }quad
          begin{cases}
          A &= cos^{-1}frac{b^2+c^2-a^2}{2bc}\
          B &= cos^{-1}frac{c^2+a^2-b^2}{2ca}\
          C &= cos^{-1}frac{a^2+b^2-c^2}{2ab}
          end{cases}
          $$
          The parameters $x, y, z$ can be obtained by plugging these expressions of $a,b,c, A,B,C$ into corresponding formula in above table.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @achilille hui I want the coordinate not the trilinear coordinate.I need the Cartesian coordinate .So please tell me the right way
            $endgroup$
            – Sufaid Saleel
            Sep 18 '17 at 12:17










          • $begingroup$
            Thanks for the great write-up. In the equation defining c, I think that $b_2 - b_3$ should be replaced by $b_1 - b_2$.
            $endgroup$
            – Alex338207
            Jan 14 at 7:52












          • $begingroup$
            @Alex338207 yes, that's a typo, thanks for pointing that out.
            $endgroup$
            – achille hui
            Jan 14 at 13:28
















          3












          $begingroup$

          There are well known formula for the triangle centers in trilinear coordinates $x : y : z$,



          $$begin{array}{rc:c}
          text{name} & & x : y : z \
          hline
          text{incenter} & I & 1 : 1 : 1\
          text{circumcenter} & O & cos A : cos B : cos C\
          text{orthocenter} & H & sec A : sec B :sec C\
          text{nine-point center} & N & cos(B-C) : cos(C-A) : cos(A-B)
          end{array}
          $$
          You can convert them to Cartesian coordinates by the recipe:
          $$P = (x,y,z) quad leftrightarrow quad vec{P} = frac{ax vec{A} + byvec{B} + czvec{C}}{ax + by + cz}$$



          In terms of coordinates, this means



          $$vec{P} = (p_1,p_2) =
          left(
          frac{ax a_1 + by a_2 + cz a_3}{ax + by + cz},
          frac{ax b_1 + by b_2 + cz b_3}{ax + by + cz}
          right)
          $$
          The parameters $a,b,c$ and $A,B,C$ are given by the formula:
          $$begin{cases}
          a &= sqrt{(a_2-a_3)^2 + (b_2 - b_3)^2}\
          b &= sqrt{(a_3-a_1)^2 + (b_3 - b_1)^2}\
          c &= sqrt{(a_1-a_2)^2 + (b_1 - b_2)^2}
          end{cases}
          quadtext{ and }quad
          begin{cases}
          A &= cos^{-1}frac{b^2+c^2-a^2}{2bc}\
          B &= cos^{-1}frac{c^2+a^2-b^2}{2ca}\
          C &= cos^{-1}frac{a^2+b^2-c^2}{2ab}
          end{cases}
          $$
          The parameters $x, y, z$ can be obtained by plugging these expressions of $a,b,c, A,B,C$ into corresponding formula in above table.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @achilille hui I want the coordinate not the trilinear coordinate.I need the Cartesian coordinate .So please tell me the right way
            $endgroup$
            – Sufaid Saleel
            Sep 18 '17 at 12:17










          • $begingroup$
            Thanks for the great write-up. In the equation defining c, I think that $b_2 - b_3$ should be replaced by $b_1 - b_2$.
            $endgroup$
            – Alex338207
            Jan 14 at 7:52












          • $begingroup$
            @Alex338207 yes, that's a typo, thanks for pointing that out.
            $endgroup$
            – achille hui
            Jan 14 at 13:28














          3












          3








          3





          $begingroup$

          There are well known formula for the triangle centers in trilinear coordinates $x : y : z$,



          $$begin{array}{rc:c}
          text{name} & & x : y : z \
          hline
          text{incenter} & I & 1 : 1 : 1\
          text{circumcenter} & O & cos A : cos B : cos C\
          text{orthocenter} & H & sec A : sec B :sec C\
          text{nine-point center} & N & cos(B-C) : cos(C-A) : cos(A-B)
          end{array}
          $$
          You can convert them to Cartesian coordinates by the recipe:
          $$P = (x,y,z) quad leftrightarrow quad vec{P} = frac{ax vec{A} + byvec{B} + czvec{C}}{ax + by + cz}$$



          In terms of coordinates, this means



          $$vec{P} = (p_1,p_2) =
          left(
          frac{ax a_1 + by a_2 + cz a_3}{ax + by + cz},
          frac{ax b_1 + by b_2 + cz b_3}{ax + by + cz}
          right)
          $$
          The parameters $a,b,c$ and $A,B,C$ are given by the formula:
          $$begin{cases}
          a &= sqrt{(a_2-a_3)^2 + (b_2 - b_3)^2}\
          b &= sqrt{(a_3-a_1)^2 + (b_3 - b_1)^2}\
          c &= sqrt{(a_1-a_2)^2 + (b_1 - b_2)^2}
          end{cases}
          quadtext{ and }quad
          begin{cases}
          A &= cos^{-1}frac{b^2+c^2-a^2}{2bc}\
          B &= cos^{-1}frac{c^2+a^2-b^2}{2ca}\
          C &= cos^{-1}frac{a^2+b^2-c^2}{2ab}
          end{cases}
          $$
          The parameters $x, y, z$ can be obtained by plugging these expressions of $a,b,c, A,B,C$ into corresponding formula in above table.






          share|cite|improve this answer











          $endgroup$



          There are well known formula for the triangle centers in trilinear coordinates $x : y : z$,



          $$begin{array}{rc:c}
          text{name} & & x : y : z \
          hline
          text{incenter} & I & 1 : 1 : 1\
          text{circumcenter} & O & cos A : cos B : cos C\
          text{orthocenter} & H & sec A : sec B :sec C\
          text{nine-point center} & N & cos(B-C) : cos(C-A) : cos(A-B)
          end{array}
          $$
          You can convert them to Cartesian coordinates by the recipe:
          $$P = (x,y,z) quad leftrightarrow quad vec{P} = frac{ax vec{A} + byvec{B} + czvec{C}}{ax + by + cz}$$



          In terms of coordinates, this means



          $$vec{P} = (p_1,p_2) =
          left(
          frac{ax a_1 + by a_2 + cz a_3}{ax + by + cz},
          frac{ax b_1 + by b_2 + cz b_3}{ax + by + cz}
          right)
          $$
          The parameters $a,b,c$ and $A,B,C$ are given by the formula:
          $$begin{cases}
          a &= sqrt{(a_2-a_3)^2 + (b_2 - b_3)^2}\
          b &= sqrt{(a_3-a_1)^2 + (b_3 - b_1)^2}\
          c &= sqrt{(a_1-a_2)^2 + (b_1 - b_2)^2}
          end{cases}
          quadtext{ and }quad
          begin{cases}
          A &= cos^{-1}frac{b^2+c^2-a^2}{2bc}\
          B &= cos^{-1}frac{c^2+a^2-b^2}{2ca}\
          C &= cos^{-1}frac{a^2+b^2-c^2}{2ab}
          end{cases}
          $$
          The parameters $x, y, z$ can be obtained by plugging these expressions of $a,b,c, A,B,C$ into corresponding formula in above table.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 13:27

























          answered Sep 18 '17 at 8:46









          achille huiachille hui

          96.3k5132261




          96.3k5132261












          • $begingroup$
            @achilille hui I want the coordinate not the trilinear coordinate.I need the Cartesian coordinate .So please tell me the right way
            $endgroup$
            – Sufaid Saleel
            Sep 18 '17 at 12:17










          • $begingroup$
            Thanks for the great write-up. In the equation defining c, I think that $b_2 - b_3$ should be replaced by $b_1 - b_2$.
            $endgroup$
            – Alex338207
            Jan 14 at 7:52












          • $begingroup$
            @Alex338207 yes, that's a typo, thanks for pointing that out.
            $endgroup$
            – achille hui
            Jan 14 at 13:28


















          • $begingroup$
            @achilille hui I want the coordinate not the trilinear coordinate.I need the Cartesian coordinate .So please tell me the right way
            $endgroup$
            – Sufaid Saleel
            Sep 18 '17 at 12:17










          • $begingroup$
            Thanks for the great write-up. In the equation defining c, I think that $b_2 - b_3$ should be replaced by $b_1 - b_2$.
            $endgroup$
            – Alex338207
            Jan 14 at 7:52












          • $begingroup$
            @Alex338207 yes, that's a typo, thanks for pointing that out.
            $endgroup$
            – achille hui
            Jan 14 at 13:28
















          $begingroup$
          @achilille hui I want the coordinate not the trilinear coordinate.I need the Cartesian coordinate .So please tell me the right way
          $endgroup$
          – Sufaid Saleel
          Sep 18 '17 at 12:17




          $begingroup$
          @achilille hui I want the coordinate not the trilinear coordinate.I need the Cartesian coordinate .So please tell me the right way
          $endgroup$
          – Sufaid Saleel
          Sep 18 '17 at 12:17












          $begingroup$
          Thanks for the great write-up. In the equation defining c, I think that $b_2 - b_3$ should be replaced by $b_1 - b_2$.
          $endgroup$
          – Alex338207
          Jan 14 at 7:52






          $begingroup$
          Thanks for the great write-up. In the equation defining c, I think that $b_2 - b_3$ should be replaced by $b_1 - b_2$.
          $endgroup$
          – Alex338207
          Jan 14 at 7:52














          $begingroup$
          @Alex338207 yes, that's a typo, thanks for pointing that out.
          $endgroup$
          – achille hui
          Jan 14 at 13:28




          $begingroup$
          @Alex338207 yes, that's a typo, thanks for pointing that out.
          $endgroup$
          – achille hui
          Jan 14 at 13:28


















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