Homotheties: Let $A$ and $B$ be distinct points of a circle $o$. What is the set of possible centroids of...
$begingroup$
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
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 7:06
ricorico
756
$endgroup$
add a comment |
$begingroup$
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
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 7:06
ricorico
756
$endgroup$
add a comment |
$begingroup$
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
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 7:06
ricorico
756
$endgroup$
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
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 7:06
ricorico
756
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 7:06
ricorico
756
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
edited Dec 11 '18 at 14:26
Asaf Karagila♦
305k33435766
305k33435766
asked Dec 11 '18 at 7:06
ricorico
756
asked Dec 11 '18 at 7:06
ricorico
756
asked Dec 11 '18 at 7:06
ricorico
756
756
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
[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 $r/3$.
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 7:08
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
2
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2 Answers
2
active
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active
oldest
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active
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votes
$begingroup$
[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 $r/3$.
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
[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 $r/3$.
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
[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 $r/3$.
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
[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 $r/3$.
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
answered Dec 11 '18 at 7:26
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 7:08
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 7:08
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 7:08
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
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.
answered Dec 11 '18 at 7:08
Carl SchildkrautCarl Schildkraut
11.5k11441
answered Dec 11 '18 at 7:08
Carl SchildkrautCarl Schildkraut
11.5k11441
answered Dec 11 '18 at 7:08
Carl SchildkrautCarl Schildkraut
11.5k11441
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
add a comment |
$begingroup$
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
add a comment |
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