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
          1






          active

          oldest

          votes


















          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2433914%2ffinding-the-incentre-circumcentre-of-a-triangle%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2433914%2ffinding-the-incentre-circumcentre-of-a-triangle%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Human spaceflight

          Image
          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
          Read more

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          Image
          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
          Read more

          File:DeusFollowingSea.jpg

          Image
          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
          Read more