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







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    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
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          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











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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

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            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


















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