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The projection of a barycenter onto a plane is the barycenter of the projected triangle?
Clash Royale CLAN TAG#URR8PPP
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When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?
I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.
geometry affine-geometry
edited Jul 20 at 19:18
Arnaud Mortier
19k22159
asked Jul 20 at 13:37
Alexandre Tourinho
1305
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up vote
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When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?
I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.
geometry affine-geometry
edited Jul 20 at 19:18
Arnaud Mortier
19k22159
asked Jul 20 at 13:37
Alexandre Tourinho
1305
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?
I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.
geometry affine-geometry
edited Jul 20 at 19:18
Arnaud Mortier
19k22159
asked Jul 20 at 13:37
Alexandre Tourinho
1305
When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?
I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.
geometry affine-geometry
edited Jul 20 at 19:18
Arnaud Mortier
19k22159
asked Jul 20 at 13:37
Alexandre Tourinho
1305
edited Jul 20 at 19:18
Arnaud Mortier
19k22159
edited Jul 20 at 19:18
Arnaud Mortier
19k22159
edited Jul 20 at 19:18
Arnaud Mortier
19k22159
19k22159
asked Jul 20 at 13:37
Alexandre Tourinho
1305
asked Jul 20 at 13:37
Alexandre Tourinho
1305
asked Jul 20 at 13:37
Alexandre Tourinho
1305
1305
add a comment |Â
add a comment |Â
1 Answer
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The answer is yes, and the reason is both deep and straightforward.
A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).
An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The answer is yes, and the reason is both deep and straightforward.
A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).
An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
add a comment |Â
up vote
3
down vote
The answer is yes, and the reason is both deep and straightforward.
A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).
An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The answer is yes, and the reason is both deep and straightforward.
A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).
An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
The answer is yes, and the reason is both deep and straightforward.
A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).
An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
answered Jul 20 at 13:48
Arnaud Mortier
19k22159
19k22159
add a comment |Â
add a comment |Â
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