Homotheties: Let $A$ and $B$ be distinct points of a circle $o$. What is the set of possible centroids of...












2












$begingroup$


Question: Let $A$ and $B$ be distinct points of a circle $o$. What is the set of possible centroids of triangles $ABC$ with $Cin o$?



Here is what I have:
The angle at $C$ will always be the same as it is always subtended by the same arc as $A$ and $B$ are fixed.



There are 2 cases: Either $C$ lies on the small arc of $AB$ or $C$ lies on the big arc of $AB$.



In the case where $C$ lies on the large arc, by looking at the possible positions of $C$ one can observe that at some point $C$ and $A$ are on the same diameter and at another point $C$ and $B$ are on the same diameter. Also, it is worth mentioning that the midpoint at which $c$ intersects on the chord $AB$ does not depend on $C$ and is thus always the same.



One can observe, by determining various points that satisfy the criteria in a drawing, that the possible points $C$ all seem to lie on a smaller circle contained in the original circle $o$.



One can then guess that the center of this smaller circle is a possible center of homothethy mapping the smaller circle to the larger circle with scale $r_1/r_2$ where $r_1$ is the radius of the larger circle and $r_2$ is the radius of the smaller circle(I am really not sure about this).



I am not sure what to say about the case where $c$ lies on the small arc and am not sure where to continue with the problem.



Any help is appreciated.






geometry euclidean-geometry







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Question: Let $A$ and $B$ be distinct points of a circle $o$. What is the set of possible centroids of triangles $ABC$ with $Cin o$?



    Here is what I have:
    The angle at $C$ will always be the same as it is always subtended by the same arc as $A$ and $B$ are fixed.



    There are 2 cases: Either $C$ lies on the small arc of $AB$ or $C$ lies on the big arc of $AB$.



    In the case where $C$ lies on the large arc, by looking at the possible positions of $C$ one can observe that at some point $C$ and $A$ are on the same diameter and at another point $C$ and $B$ are on the same diameter. Also, it is worth mentioning that the midpoint at which $c$ intersects on the chord $AB$ does not depend on $C$ and is thus always the same.



    One can observe, by determining various points that satisfy the criteria in a drawing, that the possible points $C$ all seem to lie on a smaller circle contained in the original circle $o$.



    One can then guess that the center of this smaller circle is a possible center of homothethy mapping the smaller circle to the larger circle with scale $r_1/r_2$ where $r_1$ is the radius of the larger circle and $r_2$ is the radius of the smaller circle(I am really not sure about this).



    I am not sure what to say about the case where $c$ lies on the small arc and am not sure where to continue with the problem.



    Any help is appreciated.






    geometry euclidean-geometry







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Question: Let $A$ and $B$ be distinct points of a circle $o$. What is the set of possible centroids of triangles $ABC$ with $Cin o$?



      Here is what I have:
      The angle at $C$ will always be the same as it is always subtended by the same arc as $A$ and $B$ are fixed.



      There are 2 cases: Either $C$ lies on the small arc of $AB$ or $C$ lies on the big arc of $AB$.



      In the case where $C$ lies on the large arc, by looking at the possible positions of $C$ one can observe that at some point $C$ and $A$ are on the same diameter and at another point $C$ and $B$ are on the same diameter. Also, it is worth mentioning that the midpoint at which $c$ intersects on the chord $AB$ does not depend on $C$ and is thus always the same.



      One can observe, by determining various points that satisfy the criteria in a drawing, that the possible points $C$ all seem to lie on a smaller circle contained in the original circle $o$.



      One can then guess that the center of this smaller circle is a possible center of homothethy mapping the smaller circle to the larger circle with scale $r_1/r_2$ where $r_1$ is the radius of the larger circle and $r_2$ is the radius of the smaller circle(I am really not sure about this).



      I am not sure what to say about the case where $c$ lies on the small arc and am not sure where to continue with the problem.



      Any help is appreciated.






      geometry euclidean-geometry







      share|cite|improve this question











      $endgroup$




      Question: Let $A$ and $B$ be distinct points of a circle $o$. What is the set of possible centroids of triangles $ABC$ with $Cin o$?



      Here is what I have:
      The angle at $C$ will always be the same as it is always subtended by the same arc as $A$ and $B$ are fixed.



      There are 2 cases: Either $C$ lies on the small arc of $AB$ or $C$ lies on the big arc of $AB$.



      In the case where $C$ lies on the large arc, by looking at the possible positions of $C$ one can observe that at some point $C$ and $A$ are on the same diameter and at another point $C$ and $B$ are on the same diameter. Also, it is worth mentioning that the midpoint at which $c$ intersects on the chord $AB$ does not depend on $C$ and is thus always the same.



      One can observe, by determining various points that satisfy the criteria in a drawing, that the possible points $C$ all seem to lie on a smaller circle contained in the original circle $o$.



      One can then guess that the center of this smaller circle is a possible center of homothethy mapping the smaller circle to the larger circle with scale $r_1/r_2$ where $r_1$ is the radius of the larger circle and $r_2$ is the radius of the smaller circle(I am really not sure about this).



      I am not sure what to say about the case where $c$ lies on the small arc and am not sure where to continue with the problem.



      Any help is appreciated.






      geometry euclidean-geometry




      geometry euclidean-geometry






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      share|cite|improve this question













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      edited Dec 11 '18 at 14:26









      Asaf Karagila♦

      305k33435766




      305k33435766










      asked Dec 11 '18 at 7:06









      ricorico

      756




      756






















          2 Answers
          2






          active

          oldest

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          6












          $begingroup$

          [This is just a translation of Carl Schildkraut's answer into synthetic language.]



          Note that the centroid of $ABC$ can be constructed by taking the midpoint $M$ of $AB$ and then taking the point $G$ which is $1/3$ of the way along $MC$ (closer to $M$). This means exactly that it is the image of $C$ under the homothety $T$ with center $M$ and scale factor $1/3$. Since homotheties preserve circles and their centers, this homothety maps a circle with center $D$ and radius $r$ to a circle with center $T(D)$ and radius $r/3$.






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            Here's a moderately obnoxious idea:



            If you use complex numbers and set your circle to be the unit circle, then the centroid of the triangle determined by $a,b,c$ is simply $frac{a+b+c}{3}$. Thus, the locus of possible centroids, as $A$ and $B$ are fixed and $C$ varies, is simply the circle with radius $frac{1}{3}$ centered at $frac{a+b}{3}$ (I think you need to get rid of two of the points because $Cneq A,B$, but oh well).



            There should be a geometric interpretation of this idea as well, if you essentially try to phrase everything with homotheties.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties.
              $endgroup$
              – rico
              Dec 11 '18 at 7:15










            • $begingroup$
              In terms of homotheties, the transformation $T:cmapsto frac{a+b+c}{3}$ is a homothety with center $frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$.
              $endgroup$
              – Eric Wofsey
              Dec 11 '18 at 7:20











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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

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            6












            $begingroup$

            [This is just a translation of Carl Schildkraut's answer into synthetic language.]



            Note that the centroid of $ABC$ can be constructed by taking the midpoint $M$ of $AB$ and then taking the point $G$ which is $1/3$ of the way along $MC$ (closer to $M$). This means exactly that it is the image of $C$ under the homothety $T$ with center $M$ and scale factor $1/3$. Since homotheties preserve circles and their centers, this homothety maps a circle with center $D$ and radius $r$ to a circle with center $T(D)$ and radius $r/3$.






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              [This is just a translation of Carl Schildkraut's answer into synthetic language.]



              Note that the centroid of $ABC$ can be constructed by taking the midpoint $M$ of $AB$ and then taking the point $G$ which is $1/3$ of the way along $MC$ (closer to $M$). This means exactly that it is the image of $C$ under the homothety $T$ with center $M$ and scale factor $1/3$. Since homotheties preserve circles and their centers, this homothety maps a circle with center $D$ and radius $r$ to a circle with center $T(D)$ and radius $r/3$.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                [This is just a translation of Carl Schildkraut's answer into synthetic language.]



                Note that the centroid of $ABC$ can be constructed by taking the midpoint $M$ of $AB$ and then taking the point $G$ which is $1/3$ of the way along $MC$ (closer to $M$). This means exactly that it is the image of $C$ under the homothety $T$ with center $M$ and scale factor $1/3$. Since homotheties preserve circles and their centers, this homothety maps a circle with center $D$ and radius $r$ to a circle with center $T(D)$ and radius $r/3$.






                share|cite|improve this answer









                $endgroup$



                [This is just a translation of Carl Schildkraut's answer into synthetic language.]



                Note that the centroid of $ABC$ can be constructed by taking the midpoint $M$ of $AB$ and then taking the point $G$ which is $1/3$ of the way along $MC$ (closer to $M$). This means exactly that it is the image of $C$ under the homothety $T$ with center $M$ and scale factor $1/3$. Since homotheties preserve circles and their centers, this homothety maps a circle with center $D$ and radius $r$ to a circle with center $T(D)$ and radius $r/3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 11 '18 at 7:26









                Eric WofseyEric Wofsey

                187k14215344




                187k14215344























                    4












                    $begingroup$

                    Here's a moderately obnoxious idea:



                    If you use complex numbers and set your circle to be the unit circle, then the centroid of the triangle determined by $a,b,c$ is simply $frac{a+b+c}{3}$. Thus, the locus of possible centroids, as $A$ and $B$ are fixed and $C$ varies, is simply the circle with radius $frac{1}{3}$ centered at $frac{a+b}{3}$ (I think you need to get rid of two of the points because $Cneq A,B$, but oh well).



                    There should be a geometric interpretation of this idea as well, if you essentially try to phrase everything with homotheties.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties.
                      $endgroup$
                      – rico
                      Dec 11 '18 at 7:15










                    • $begingroup$
                      In terms of homotheties, the transformation $T:cmapsto frac{a+b+c}{3}$ is a homothety with center $frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$.
                      $endgroup$
                      – Eric Wofsey
                      Dec 11 '18 at 7:20
















                    4












                    $begingroup$

                    Here's a moderately obnoxious idea:



                    If you use complex numbers and set your circle to be the unit circle, then the centroid of the triangle determined by $a,b,c$ is simply $frac{a+b+c}{3}$. Thus, the locus of possible centroids, as $A$ and $B$ are fixed and $C$ varies, is simply the circle with radius $frac{1}{3}$ centered at $frac{a+b}{3}$ (I think you need to get rid of two of the points because $Cneq A,B$, but oh well).



                    There should be a geometric interpretation of this idea as well, if you essentially try to phrase everything with homotheties.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties.
                      $endgroup$
                      – rico
                      Dec 11 '18 at 7:15










                    • $begingroup$
                      In terms of homotheties, the transformation $T:cmapsto frac{a+b+c}{3}$ is a homothety with center $frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$.
                      $endgroup$
                      – Eric Wofsey
                      Dec 11 '18 at 7:20














                    4












                    4








                    4





                    $begingroup$

                    Here's a moderately obnoxious idea:



                    If you use complex numbers and set your circle to be the unit circle, then the centroid of the triangle determined by $a,b,c$ is simply $frac{a+b+c}{3}$. Thus, the locus of possible centroids, as $A$ and $B$ are fixed and $C$ varies, is simply the circle with radius $frac{1}{3}$ centered at $frac{a+b}{3}$ (I think you need to get rid of two of the points because $Cneq A,B$, but oh well).



                    There should be a geometric interpretation of this idea as well, if you essentially try to phrase everything with homotheties.






                    share|cite|improve this answer









                    $endgroup$



                    Here's a moderately obnoxious idea:



                    If you use complex numbers and set your circle to be the unit circle, then the centroid of the triangle determined by $a,b,c$ is simply $frac{a+b+c}{3}$. Thus, the locus of possible centroids, as $A$ and $B$ are fixed and $C$ varies, is simply the circle with radius $frac{1}{3}$ centered at $frac{a+b}{3}$ (I think you need to get rid of two of the points because $Cneq A,B$, but oh well).



                    There should be a geometric interpretation of this idea as well, if you essentially try to phrase everything with homotheties.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 11 '18 at 7:08









                    Carl SchildkrautCarl Schildkraut

                    11.5k11441




                    11.5k11441












                    • $begingroup$
                      Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties.
                      $endgroup$
                      – rico
                      Dec 11 '18 at 7:15










                    • $begingroup$
                      In terms of homotheties, the transformation $T:cmapsto frac{a+b+c}{3}$ is a homothety with center $frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$.
                      $endgroup$
                      – Eric Wofsey
                      Dec 11 '18 at 7:20


















                    • $begingroup$
                      Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties.
                      $endgroup$
                      – rico
                      Dec 11 '18 at 7:15










                    • $begingroup$
                      In terms of homotheties, the transformation $T:cmapsto frac{a+b+c}{3}$ is a homothety with center $frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$.
                      $endgroup$
                      – Eric Wofsey
                      Dec 11 '18 at 7:20
















                    $begingroup$
                    Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties.
                    $endgroup$
                    – rico
                    Dec 11 '18 at 7:15




                    $begingroup$
                    Wow that is really interesting way of thinking about it, thanks. Unfortunately, I am trying to find a solution using homotheties.
                    $endgroup$
                    – rico
                    Dec 11 '18 at 7:15












                    $begingroup$
                    In terms of homotheties, the transformation $T:cmapsto frac{a+b+c}{3}$ is a homothety with center $frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$.
                    $endgroup$
                    – Eric Wofsey
                    Dec 11 '18 at 7:20




                    $begingroup$
                    In terms of homotheties, the transformation $T:cmapsto frac{a+b+c}{3}$ is a homothety with center $frac{a+b}{2}$ and scale factor $1/3$. So, it maps a circle of radius $r$ and center $d$ to a circle of radius $r/3$ and center $T(d)$.
                    $endgroup$
                    – Eric Wofsey
                    Dec 11 '18 at 7:20


















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