For $G$ the centroid in $triangle ABC$, if $AB+GC=AC+GB$, then $triangle ABC$ is isosceles. (Likewise, for...












2












$begingroup$



Let $G$ be the centroid of $triangle ABC$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles!




Of course, the equality is true, when we have isosceles triangle, but the other way is not trivial for me. I have tried using vectors, even the length of the medians.






geometry euclidean-geometry triangle







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $G$ be the centroid of $triangle ABC$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles!




    Of course, the equality is true, when we have isosceles triangle, but the other way is not trivial for me. I have tried using vectors, even the length of the medians.






    geometry euclidean-geometry triangle







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      3



      $begingroup$



      Let $G$ be the centroid of $triangle ABC$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles!




      Of course, the equality is true, when we have isosceles triangle, but the other way is not trivial for me. I have tried using vectors, even the length of the medians.






      geometry euclidean-geometry triangle







      share|cite|improve this question











      $endgroup$





      Let $G$ be the centroid of $triangle ABC$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles!




      Of course, the equality is true, when we have isosceles triangle, but the other way is not trivial for me. I have tried using vectors, even the length of the medians.






      geometry euclidean-geometry triangle




      geometry euclidean-geometry triangle






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 25 at 5:13









      Michael Rozenberg

      107k1894199




      107k1894199










      asked Jan 24 at 15:44









      EmathkeEmathke

      1226




      1226






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          In the standard notation we obtain:
          $$c+frac{1}{3}sqrt{2a^2+2b^2-c^2}=b+frac{1}{3}sqrt{2a^2+2c^2-b^2}$$ or
          $$3(b-c)=frac{3(b^2-c^2)}{sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
          $$sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}=b+c$$ or
          $$sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is impossible for $bc-2a^2leq0.$



          But for $bc>2a^2$ we obtain
          $$(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)=(bc-2a^2)^2$$ or
          $$(b-c)^2(a+b+c)(b+c-a)=0,$$ which gives $b=c$ again.



          The second problem we can solve by the similar way.






          share|cite|improve this answer











          $endgroup$





















            5












            $begingroup$

            Diagram



            Let $D$ be the midpoint of $BC$. Let $G'$ be the reflection of $G$ over $D$. As $BD=DC$ and $GD=DG'$, $GBG'C$ is a parallelogram. Therefore, $GB=G'C$ and $GC=G'B$, whence, $AB+BG'=AC+CG'$. Thus, $B$ and $C$ lie on an ellipse $e$ with foci $A$ and $G'$. Let $e'$ be the reflection of $e$ over $D$. As $BD=DC$, $B$ and $C$ also lie on $e'$, and thus on $ecap e'$. As $e$ and $e'$ are distinct ellipses symmetric about $AD$, $BCperp AD$, whence $AB=AC$.
            $blacksquare$



            (Note that we didn't use the fact that $G$ is the centroid of $ABC$ in the proof, we just used the fact that $G$ lies on $AD$. So, the same proof proves the following more general statement: If $G$ is a point on $AD$ such that $AB+GC=AC+GB$, then, $triangle ABC$ is isosceles.)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection.
              $endgroup$
              – Anubhab Ghosal
              Jan 25 at 15:34











            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%2f3086013%2ffor-g-the-centroid-in-triangle-abc-if-abgc-acgb-then-triangle-abc%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            In the standard notation we obtain:
            $$c+frac{1}{3}sqrt{2a^2+2b^2-c^2}=b+frac{1}{3}sqrt{2a^2+2c^2-b^2}$$ or
            $$3(b-c)=frac{3(b^2-c^2)}{sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
            $$sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}=b+c$$ or
            $$sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is impossible for $bc-2a^2leq0.$



            But for $bc>2a^2$ we obtain
            $$(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)=(bc-2a^2)^2$$ or
            $$(b-c)^2(a+b+c)(b+c-a)=0,$$ which gives $b=c$ again.



            The second problem we can solve by the similar way.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              In the standard notation we obtain:
              $$c+frac{1}{3}sqrt{2a^2+2b^2-c^2}=b+frac{1}{3}sqrt{2a^2+2c^2-b^2}$$ or
              $$3(b-c)=frac{3(b^2-c^2)}{sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
              $$sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}=b+c$$ or
              $$sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is impossible for $bc-2a^2leq0.$



              But for $bc>2a^2$ we obtain
              $$(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)=(bc-2a^2)^2$$ or
              $$(b-c)^2(a+b+c)(b+c-a)=0,$$ which gives $b=c$ again.



              The second problem we can solve by the similar way.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                In the standard notation we obtain:
                $$c+frac{1}{3}sqrt{2a^2+2b^2-c^2}=b+frac{1}{3}sqrt{2a^2+2c^2-b^2}$$ or
                $$3(b-c)=frac{3(b^2-c^2)}{sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
                $$sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}=b+c$$ or
                $$sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is impossible for $bc-2a^2leq0.$



                But for $bc>2a^2$ we obtain
                $$(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)=(bc-2a^2)^2$$ or
                $$(b-c)^2(a+b+c)(b+c-a)=0,$$ which gives $b=c$ again.



                The second problem we can solve by the similar way.






                share|cite|improve this answer











                $endgroup$



                In the standard notation we obtain:
                $$c+frac{1}{3}sqrt{2a^2+2b^2-c^2}=b+frac{1}{3}sqrt{2a^2+2c^2-b^2}$$ or
                $$3(b-c)=frac{3(b^2-c^2)}{sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
                $$sqrt{2a^2+2b^2-c^2}+sqrt{2a^2+2c^2-b^2}=b+c$$ or
                $$sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is impossible for $bc-2a^2leq0.$



                But for $bc>2a^2$ we obtain
                $$(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)=(bc-2a^2)^2$$ or
                $$(b-c)^2(a+b+c)(b+c-a)=0,$$ which gives $b=c$ again.



                The second problem we can solve by the similar way.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 24 at 18:51









                Emathke

                1226




                1226










                answered Jan 24 at 16:03









                Michael RozenbergMichael Rozenberg

                107k1894199




                107k1894199























                    5












                    $begingroup$

                    Diagram



                    Let $D$ be the midpoint of $BC$. Let $G'$ be the reflection of $G$ over $D$. As $BD=DC$ and $GD=DG'$, $GBG'C$ is a parallelogram. Therefore, $GB=G'C$ and $GC=G'B$, whence, $AB+BG'=AC+CG'$. Thus, $B$ and $C$ lie on an ellipse $e$ with foci $A$ and $G'$. Let $e'$ be the reflection of $e$ over $D$. As $BD=DC$, $B$ and $C$ also lie on $e'$, and thus on $ecap e'$. As $e$ and $e'$ are distinct ellipses symmetric about $AD$, $BCperp AD$, whence $AB=AC$.
                    $blacksquare$



                    (Note that we didn't use the fact that $G$ is the centroid of $ABC$ in the proof, we just used the fact that $G$ lies on $AD$. So, the same proof proves the following more general statement: If $G$ is a point on $AD$ such that $AB+GC=AC+GB$, then, $triangle ABC$ is isosceles.)






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection.
                      $endgroup$
                      – Anubhab Ghosal
                      Jan 25 at 15:34
















                    5












                    $begingroup$

                    Diagram



                    Let $D$ be the midpoint of $BC$. Let $G'$ be the reflection of $G$ over $D$. As $BD=DC$ and $GD=DG'$, $GBG'C$ is a parallelogram. Therefore, $GB=G'C$ and $GC=G'B$, whence, $AB+BG'=AC+CG'$. Thus, $B$ and $C$ lie on an ellipse $e$ with foci $A$ and $G'$. Let $e'$ be the reflection of $e$ over $D$. As $BD=DC$, $B$ and $C$ also lie on $e'$, and thus on $ecap e'$. As $e$ and $e'$ are distinct ellipses symmetric about $AD$, $BCperp AD$, whence $AB=AC$.
                    $blacksquare$



                    (Note that we didn't use the fact that $G$ is the centroid of $ABC$ in the proof, we just used the fact that $G$ lies on $AD$. So, the same proof proves the following more general statement: If $G$ is a point on $AD$ such that $AB+GC=AC+GB$, then, $triangle ABC$ is isosceles.)






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection.
                      $endgroup$
                      – Anubhab Ghosal
                      Jan 25 at 15:34














                    5












                    5








                    5





                    $begingroup$

                    Diagram



                    Let $D$ be the midpoint of $BC$. Let $G'$ be the reflection of $G$ over $D$. As $BD=DC$ and $GD=DG'$, $GBG'C$ is a parallelogram. Therefore, $GB=G'C$ and $GC=G'B$, whence, $AB+BG'=AC+CG'$. Thus, $B$ and $C$ lie on an ellipse $e$ with foci $A$ and $G'$. Let $e'$ be the reflection of $e$ over $D$. As $BD=DC$, $B$ and $C$ also lie on $e'$, and thus on $ecap e'$. As $e$ and $e'$ are distinct ellipses symmetric about $AD$, $BCperp AD$, whence $AB=AC$.
                    $blacksquare$



                    (Note that we didn't use the fact that $G$ is the centroid of $ABC$ in the proof, we just used the fact that $G$ lies on $AD$. So, the same proof proves the following more general statement: If $G$ is a point on $AD$ such that $AB+GC=AC+GB$, then, $triangle ABC$ is isosceles.)






                    share|cite|improve this answer











                    $endgroup$



                    Diagram



                    Let $D$ be the midpoint of $BC$. Let $G'$ be the reflection of $G$ over $D$. As $BD=DC$ and $GD=DG'$, $GBG'C$ is a parallelogram. Therefore, $GB=G'C$ and $GC=G'B$, whence, $AB+BG'=AC+CG'$. Thus, $B$ and $C$ lie on an ellipse $e$ with foci $A$ and $G'$. Let $e'$ be the reflection of $e$ over $D$. As $BD=DC$, $B$ and $C$ also lie on $e'$, and thus on $ecap e'$. As $e$ and $e'$ are distinct ellipses symmetric about $AD$, $BCperp AD$, whence $AB=AC$.
                    $blacksquare$



                    (Note that we didn't use the fact that $G$ is the centroid of $ABC$ in the proof, we just used the fact that $G$ lies on $AD$. So, the same proof proves the following more general statement: If $G$ is a point on $AD$ such that $AB+GC=AC+GB$, then, $triangle ABC$ is isosceles.)







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 25 at 14:50

























                    answered Jan 25 at 14:24









                    Anubhab GhosalAnubhab Ghosal

                    1,22319




                    1,22319












                    • $begingroup$
                      @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection.
                      $endgroup$
                      – Anubhab Ghosal
                      Jan 25 at 15:34


















                    • $begingroup$
                      @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection.
                      $endgroup$
                      – Anubhab Ghosal
                      Jan 25 at 15:34
















                    $begingroup$
                    @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection.
                    $endgroup$
                    – Anubhab Ghosal
                    Jan 25 at 15:34




                    $begingroup$
                    @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection.
                    $endgroup$
                    – Anubhab Ghosal
                    Jan 25 at 15:34


















                    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%2f3086013%2ffor-g-the-centroid-in-triangle-abc-if-abgc-acgb-then-triangle-abc%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

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                    Read more

                    Antonio Litta Visconti Arese

                    Image
                    Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                    Read more