Is there a solution to this geometry problem?












3












$begingroup$


Triangle



The point $G$ is centroid in the triangle $ABC$.



$m(BGC)=90^o$



$|AB|=8 ;cm$



$|CG|=? , cm$



I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?






geometry euclidean-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
    $endgroup$
    – user10354138
    Dec 4 '18 at 12:20










  • $begingroup$
    BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
    $endgroup$
    – Andrei
    Dec 4 '18 at 15:35
















3












$begingroup$


Triangle



The point $G$ is centroid in the triangle $ABC$.



$m(BGC)=90^o$



$|AB|=8 ;cm$



$|CG|=? , cm$



I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?






geometry euclidean-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
    $endgroup$
    – user10354138
    Dec 4 '18 at 12:20










  • $begingroup$
    BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
    $endgroup$
    – Andrei
    Dec 4 '18 at 15:35














3












3








3


1



$begingroup$


Triangle



The point $G$ is centroid in the triangle $ABC$.



$m(BGC)=90^o$



$|AB|=8 ;cm$



$|CG|=? , cm$



I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?






geometry euclidean-geometry







share|cite|improve this question











$endgroup$




Triangle



The point $G$ is centroid in the triangle $ABC$.



$m(BGC)=90^o$



$|AB|=8 ;cm$



$|CG|=? , cm$



I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?






geometry euclidean-geometry




geometry euclidean-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 18:50









greedoid

40.7k1149100




40.7k1149100










asked Dec 4 '18 at 11:51









Eldar RahimliEldar Rahimli

1269




1269












  • $begingroup$
    I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
    $endgroup$
    – user10354138
    Dec 4 '18 at 12:20










  • $begingroup$
    BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
    $endgroup$
    – Andrei
    Dec 4 '18 at 15:35


















  • $begingroup$
    I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
    $endgroup$
    – user10354138
    Dec 4 '18 at 12:20










  • $begingroup$
    BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
    $endgroup$
    – Andrei
    Dec 4 '18 at 15:35
















$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20




$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20












$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35




$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35










3 Answers
3






active

oldest

votes


















0












$begingroup$

The point $D$ is not uniqely determined. Everything is on a picture:



enter image description here



We have $$4b^2+c^2=16$$
but $$ CG = 2c $$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)



    Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.



    Thus, the perpendicular medians tell us this about $D$:
    $$|AD| = 3,|BD|$$



    Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.



    Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.



    enter image description here



    enter image description here



    enter image description here



    However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$



      Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)



      Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.



      Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        The point $D$ is not uniqely determined. Everything is on a picture:



        enter image description here



        We have $$4b^2+c^2=16$$
        but $$ CG = 2c $$






        share|cite|improve this answer











        $endgroup$


















          0












          $begingroup$

          The point $D$ is not uniqely determined. Everything is on a picture:



          enter image description here



          We have $$4b^2+c^2=16$$
          but $$ CG = 2c $$






          share|cite|improve this answer











          $endgroup$
















            0












            0








            0





            $begingroup$

            The point $D$ is not uniqely determined. Everything is on a picture:



            enter image description here



            We have $$4b^2+c^2=16$$
            but $$ CG = 2c $$






            share|cite|improve this answer











            $endgroup$



            The point $D$ is not uniqely determined. Everything is on a picture:



            enter image description here



            We have $$4b^2+c^2=16$$
            but $$ CG = 2c $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 '18 at 18:20

























            answered Dec 4 '18 at 18:07









            greedoidgreedoid

            40.7k1149100




            40.7k1149100























                2












                $begingroup$

                I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)



                Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.



                Thus, the perpendicular medians tell us this about $D$:
                $$|AD| = 3,|BD|$$



                Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.



                Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.



                enter image description here



                enter image description here



                enter image description here



                However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)



                  Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.



                  Thus, the perpendicular medians tell us this about $D$:
                  $$|AD| = 3,|BD|$$



                  Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.



                  Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.



                  enter image description here



                  enter image description here



                  enter image description here



                  However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)



                    Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.



                    Thus, the perpendicular medians tell us this about $D$:
                    $$|AD| = 3,|BD|$$



                    Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.



                    Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.



                    enter image description here



                    enter image description here



                    enter image description here



                    However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$






                    share|cite|improve this answer









                    $endgroup$



                    I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)



                    Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.



                    Thus, the perpendicular medians tell us this about $D$:
                    $$|AD| = 3,|BD|$$



                    Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.



                    Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.



                    enter image description here



                    enter image description here



                    enter image description here



                    However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 4 '18 at 17:40









                    BlueBlue

                    48k870153




                    48k870153























                        0












                        $begingroup$

                        Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$



                        Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)



                        Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.



                        Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$



                          Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)



                          Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.



                          Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$



                            Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)



                            Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.



                            Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.






                            share|cite|improve this answer









                            $endgroup$



                            Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$



                            Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)



                            Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.



                            Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 4 '18 at 16:32









                            liaombroliaombro

                            18017




                            18017






























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