circumcenter coincides with center of mass
Clash Royale CLAN TAG#URR8PPP
up vote
-1
down vote
favorite
I am trying to prove the following statement. Any suggestions or references are highly appreciated.
Consider $n$ points in $R^2$, i.e., $x_iin R^2, i=1,ldots, n.$ Suppose the centroid (or center of mass with unit mass) denoted as $barx=frac1nsum_k=1^n x_k$ coincides with the circumcenter (namely we assume there exists a circumcircle for the points $x_i, i=1,ldots,n$), then there exists $alphain R$ such that
beginalign
x_i+1-x_i + x_i-1 - x_i & = alpha (barx-x_i), forall i=2,ldots,n-1, \
x_2-x_1 + x_n - x_1 & = alpha (barx-x_1), \
x_1-x_n + x_n-1 - x_n & = alpha (barx-x_n).
endalign
plane-geometry
asked Aug 21 at 9:27
Chandler
31
add a comment |Â
up vote
-1
down vote
favorite
I am trying to prove the following statement. Any suggestions or references are highly appreciated.
Consider $n$ points in $R^2$, i.e., $x_iin R^2, i=1,ldots, n.$ Suppose the centroid (or center of mass with unit mass) denoted as $barx=frac1nsum_k=1^n x_k$ coincides with the circumcenter (namely we assume there exists a circumcircle for the points $x_i, i=1,ldots,n$), then there exists $alphain R$ such that
beginalign
x_i+1-x_i + x_i-1 - x_i & = alpha (barx-x_i), forall i=2,ldots,n-1, \
x_2-x_1 + x_n - x_1 & = alpha (barx-x_1), \
x_1-x_n + x_n-1 - x_n & = alpha (barx-x_n).
endalign
plane-geometry
asked Aug 21 at 9:27
Chandler
31
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 21 at 9:51
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I am trying to prove the following statement. Any suggestions or references are highly appreciated.
Consider $n$ points in $R^2$, i.e., $x_iin R^2, i=1,ldots, n.$ Suppose the centroid (or center of mass with unit mass) denoted as $barx=frac1nsum_k=1^n x_k$ coincides with the circumcenter (namely we assume there exists a circumcircle for the points $x_i, i=1,ldots,n$), then there exists $alphain R$ such that
beginalign
x_i+1-x_i + x_i-1 - x_i & = alpha (barx-x_i), forall i=2,ldots,n-1, \
x_2-x_1 + x_n - x_1 & = alpha (barx-x_1), \
x_1-x_n + x_n-1 - x_n & = alpha (barx-x_n).
endalign
plane-geometry
asked Aug 21 at 9:27
Chandler
31
I am trying to prove the following statement. Any suggestions or references are highly appreciated.
Consider $n$ points in $R^2$, i.e., $x_iin R^2, i=1,ldots, n.$ Suppose the centroid (or center of mass with unit mass) denoted as $barx=frac1nsum_k=1^n x_k$ coincides with the circumcenter (namely we assume there exists a circumcircle for the points $x_i, i=1,ldots,n$), then there exists $alphain R$ such that
beginalign
x_i+1-x_i + x_i-1 - x_i & = alpha (barx-x_i), forall i=2,ldots,n-1, \
x_2-x_1 + x_n - x_1 & = alpha (barx-x_1), \
x_1-x_n + x_n-1 - x_n & = alpha (barx-x_n).
endalign
plane-geometry
asked Aug 21 at 9:27
Chandler
31
asked Aug 21 at 9:27
Chandler
31
asked Aug 21 at 9:27
Chandler
31
asked Aug 21 at 9:27
Chandler
31
31
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 21 at 9:51
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 21 at 9:51
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 21 at 9:51
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 21 at 9:51
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
This does not look true as stated
Consider the six points $(5,0),(4,3),(-4,3),(-5,0),(-4,-3),(4,-3)$ which clearly have a centroid and circumcentre at $barx=(0,0)$. Consider $x_i=(-5,0)$ and then $x_i=(-4,-3)$:
$(-4,-3)-(-5,0)+(-4,3)-(-5,0) = (2,0) qquad$ $= frac25big((0,0)-(-5,0)big)$
$(4,-3)-(-4,-3)+(-5,0)-(-4,-3) = (7,3) qquad$ $ not= frac25big((0,0)-(-4,-3)big)$
answered Aug 21 at 10:03
Henry
93.4k471149
Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_i+1-x_i+x_i-1-x_i$ is on the same line as $barx-x_i$. So the parameter $alpha$ should be changed to $alpha_i$ which can be different for different $x_i$.
– Chandler
Aug 21 at 10:14
@Chandler - my second example has $x_i+1-x_i+x_i-1-x_i$ having a different direction from $barx-x_i$
– Henry
Aug 21 at 10:15
I see it now. Thank you so much about this counterexample.
– Chandler
Aug 21 at 10:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This does not look true as stated
Consider the six points $(5,0),(4,3),(-4,3),(-5,0),(-4,-3),(4,-3)$ which clearly have a centroid and circumcentre at $barx=(0,0)$. Consider $x_i=(-5,0)$ and then $x_i=(-4,-3)$:
$(-4,-3)-(-5,0)+(-4,3)-(-5,0) = (2,0) qquad$ $= frac25big((0,0)-(-5,0)big)$
$(4,-3)-(-4,-3)+(-5,0)-(-4,-3) = (7,3) qquad$ $ not= frac25big((0,0)-(-4,-3)big)$
answered Aug 21 at 10:03
Henry
93.4k471149
Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_i+1-x_i+x_i-1-x_i$ is on the same line as $barx-x_i$. So the parameter $alpha$ should be changed to $alpha_i$ which can be different for different $x_i$.
– Chandler
Aug 21 at 10:14
@Chandler - my second example has $x_i+1-x_i+x_i-1-x_i$ having a different direction from $barx-x_i$
– Henry
Aug 21 at 10:15
I see it now. Thank you so much about this counterexample.
– Chandler
Aug 21 at 10:25
add a comment |Â
up vote
0
down vote
accepted
This does not look true as stated
Consider the six points $(5,0),(4,3),(-4,3),(-5,0),(-4,-3),(4,-3)$ which clearly have a centroid and circumcentre at $barx=(0,0)$. Consider $x_i=(-5,0)$ and then $x_i=(-4,-3)$:
$(-4,-3)-(-5,0)+(-4,3)-(-5,0) = (2,0) qquad$ $= frac25big((0,0)-(-5,0)big)$
$(4,-3)-(-4,-3)+(-5,0)-(-4,-3) = (7,3) qquad$ $ not= frac25big((0,0)-(-4,-3)big)$
answered Aug 21 at 10:03
Henry
93.4k471149
Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_i+1-x_i+x_i-1-x_i$ is on the same line as $barx-x_i$. So the parameter $alpha$ should be changed to $alpha_i$ which can be different for different $x_i$.
– Chandler
Aug 21 at 10:14
@Chandler - my second example has $x_i+1-x_i+x_i-1-x_i$ having a different direction from $barx-x_i$
– Henry
Aug 21 at 10:15
I see it now. Thank you so much about this counterexample.
– Chandler
Aug 21 at 10:25
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This does not look true as stated
Consider the six points $(5,0),(4,3),(-4,3),(-5,0),(-4,-3),(4,-3)$ which clearly have a centroid and circumcentre at $barx=(0,0)$. Consider $x_i=(-5,0)$ and then $x_i=(-4,-3)$:
$(-4,-3)-(-5,0)+(-4,3)-(-5,0) = (2,0) qquad$ $= frac25big((0,0)-(-5,0)big)$
$(4,-3)-(-4,-3)+(-5,0)-(-4,-3) = (7,3) qquad$ $ not= frac25big((0,0)-(-4,-3)big)$
answered Aug 21 at 10:03
Henry
93.4k471149
This does not look true as stated
Consider the six points $(5,0),(4,3),(-4,3),(-5,0),(-4,-3),(4,-3)$ which clearly have a centroid and circumcentre at $barx=(0,0)$. Consider $x_i=(-5,0)$ and then $x_i=(-4,-3)$:
$(-4,-3)-(-5,0)+(-4,3)-(-5,0) = (2,0) qquad$ $= frac25big((0,0)-(-5,0)big)$
$(4,-3)-(-4,-3)+(-5,0)-(-4,-3) = (7,3) qquad$ $ not= frac25big((0,0)-(-4,-3)big)$
answered Aug 21 at 10:03
Henry
93.4k471149
answered Aug 21 at 10:03
Henry
93.4k471149
answered Aug 21 at 10:03
Henry
93.4k471149
answered Aug 21 at 10:03
Henry
93.4k471149
93.4k471149
Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_i+1-x_i+x_i-1-x_i$ is on the same line as $barx-x_i$. So the parameter $alpha$ should be changed to $alpha_i$ which can be different for different $x_i$.
– Chandler
Aug 21 at 10:14
@Chandler - my second example has $x_i+1-x_i+x_i-1-x_i$ having a different direction from $barx-x_i$
– Henry
Aug 21 at 10:15
I see it now. Thank you so much about this counterexample.
– Chandler
Aug 21 at 10:25
add a comment |Â
Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_i+1-x_i+x_i-1-x_i$ is on the same line as $barx-x_i$. So the parameter $alpha$ should be changed to $alpha_i$ which can be different for different $x_i$.
– Chandler
Aug 21 at 10:14
@Chandler - my second example has $x_i+1-x_i+x_i-1-x_i$ having a different direction from $barx-x_i$
– Henry
Aug 21 at 10:15
I see it now. Thank you so much about this counterexample.
– Chandler
Aug 21 at 10:25
Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_i+1-x_i+x_i-1-x_i$ is on the same line as $barx-x_i$. So the parameter $alpha$ should be changed to $alpha_i$ which can be different for different $x_i$.
– Chandler
Aug 21 at 10:14
Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_i+1-x_i+x_i-1-x_i$ is on the same line as $barx-x_i$. So the parameter $alpha$ should be changed to $alpha_i$ which can be different for different $x_i$.
– Chandler
Aug 21 at 10:14
@Chandler - my second example has $x_i+1-x_i+x_i-1-x_i$ having a different direction from $barx-x_i$
– Henry
Aug 21 at 10:15
@Chandler - my second example has $x_i+1-x_i+x_i-1-x_i$ having a different direction from $barx-x_i$
– Henry
Aug 21 at 10:15
I see it now. Thank you so much about this counterexample.
– Chandler
Aug 21 at 10:25
I see it now. Thank you so much about this counterexample.
– Chandler
Aug 21 at 10:25
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2889669%2fcircumcenter-coincides-with-center-of-mass%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 21 at 9:51