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Create a cube from identical 3D objects
Fair share of a square watermelon?Create bridges to minimise distanceHexagonal minesweeperOld Indian ArtworkMeasuring Water with a Holey CubeEngineering drawing put to some good useSix pyramids in a cubeKnight's Tour on a cube surfaceLaser Beams in Helsinki SkiesRay reflection inside the cube
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$begingroup$
The diagram is the outline of the surface of a 3D object. Several objects, like the one created from this given surface, may be used to create a cube.
Let me know if you need more clarifications to solve this one
geometry three-dimensional
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
The diagram is the outline of the surface of a 3D object. Several objects, like the one created from this given surface, may be used to create a cube.
Let me know if you need more clarifications to solve this one
geometry three-dimensional
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
$endgroup$
$begingroup$
Is this the unwrapped surface? I first read this as saying that this is the projection of the 3D object onto a 2D surface.
$endgroup$
– Acccumulation
Jun 10 at 22:21
add a comment |
$begingroup$
The diagram is the outline of the surface of a 3D object. Several objects, like the one created from this given surface, may be used to create a cube.
Let me know if you need more clarifications to solve this one
geometry three-dimensional
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
$endgroup$
The diagram is the outline of the surface of a 3D object. Several objects, like the one created from this given surface, may be used to create a cube.
Let me know if you need more clarifications to solve this one
geometry three-dimensional
geometry three-dimensional
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
asked Jun 10 at 3:28
MotiMoti
9135 silver badges18 bronze badges
9135 silver badges18 bronze badges
$begingroup$
Is this the unwrapped surface? I first read this as saying that this is the projection of the 3D object onto a 2D surface.
$endgroup$
– Acccumulation
Jun 10 at 22:21
add a comment |
$begingroup$
Is this the unwrapped surface? I first read this as saying that this is the projection of the 3D object onto a 2D surface.
$endgroup$
– Acccumulation
Jun 10 at 22:21
$begingroup$
Is this the unwrapped surface? I first read this as saying that this is the projection of the 3D object onto a 2D surface.
$endgroup$
– Acccumulation
Jun 10 at 22:21
$begingroup$
Is this the unwrapped surface? I first read this as saying that this is the projection of the 3D object onto a 2D surface.
$endgroup$
– Acccumulation
Jun 10 at 22:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's a silly looking animation with
3
copies of the piece:
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
$endgroup$
add a comment |
$begingroup$
I think we can do it with
3
copies of the given 3D object.
Note: I don't have a mathematical background, so this is mostly an intuitive "solve"and may not be very rigorous (or even correct). It also assumes that lengths that are visually equal, are (e.g. FD = EB = EJ, FA = EA = AB = BC = CD = DA = BH = JH).
Firstly fold the object on the red lines, like so:
Assuming (as stated above), edges FA = AB = AD, etc, then the resulting polyhedron is
a skewed square-based pyramid, which fits neatly inside a cube of side length
FAsuch that the baseABCDis one face, and the pointE/Fis one of the top corners. Thus,FADandEBAare both half faces of the cube, meaning we have two totally covered faces of the cube (so even intuitively we can see that three copies should cover all six faces).
If we call the remaining points of the cube, X (above B), Y (above C), and Z (above D), then we can
take another copy and place its base square on the
DCYZface and its apex atE. This covers the second half of theAEZDface and half the top (EXYZ). Then we take a third copy and slide it in such that its vertex is again atE, but with its base covering the whole of theXYCBface. This covers the second half of theEXYZ"top" and the second half of theEXBAface.
Another way to think of this is to take the original object and
duplicate it three times, each time rotating through the diagonal
ECaxis, moving the base to a new face of the cube each rotation.
As I said at the top, I'm not sure how to approach a rigorous proof, but it's hopefully intuitive that
because all copies share the central
ECaxis and each internal face is at 45°, the copies can't be overlapping each other. Similarly, the large internal 45° triangles they must be completely face to face, meaning we can't be left with any gaps, so we've filled the cube entirely.
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a silly looking animation with
3
copies of the piece:
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
$endgroup$
add a comment |
$begingroup$
Here's a silly looking animation with
3
copies of the piece:
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
$endgroup$
add a comment |
$begingroup$
Here's a silly looking animation with
3
copies of the piece:
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
$endgroup$
Here's a silly looking animation with
3
copies of the piece:
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
edited Jun 10 at 7:26
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
answered Jun 10 at 6:50
BassBass
33.4k4 gold badges79 silver badges198 bronze badges
33.4k4 gold badges79 silver badges198 bronze badges
add a comment |
add a comment |
$begingroup$
I think we can do it with
3
copies of the given 3D object.
Note: I don't have a mathematical background, so this is mostly an intuitive "solve"and may not be very rigorous (or even correct). It also assumes that lengths that are visually equal, are (e.g. FD = EB = EJ, FA = EA = AB = BC = CD = DA = BH = JH).
Firstly fold the object on the red lines, like so:
Assuming (as stated above), edges FA = AB = AD, etc, then the resulting polyhedron is
a skewed square-based pyramid, which fits neatly inside a cube of side length
FAsuch that the baseABCDis one face, and the pointE/Fis one of the top corners. Thus,FADandEBAare both half faces of the cube, meaning we have two totally covered faces of the cube (so even intuitively we can see that three copies should cover all six faces).
If we call the remaining points of the cube, X (above B), Y (above C), and Z (above D), then we can
take another copy and place its base square on the
DCYZface and its apex atE. This covers the second half of theAEZDface and half the top (EXYZ). Then we take a third copy and slide it in such that its vertex is again atE, but with its base covering the whole of theXYCBface. This covers the second half of theEXYZ"top" and the second half of theEXBAface.
Another way to think of this is to take the original object and
duplicate it three times, each time rotating through the diagonal
ECaxis, moving the base to a new face of the cube each rotation.
As I said at the top, I'm not sure how to approach a rigorous proof, but it's hopefully intuitive that
because all copies share the central
ECaxis and each internal face is at 45°, the copies can't be overlapping each other. Similarly, the large internal 45° triangles they must be completely face to face, meaning we can't be left with any gaps, so we've filled the cube entirely.
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
$endgroup$
add a comment |
$begingroup$
I think we can do it with
3
copies of the given 3D object.
Note: I don't have a mathematical background, so this is mostly an intuitive "solve"and may not be very rigorous (or even correct). It also assumes that lengths that are visually equal, are (e.g. FD = EB = EJ, FA = EA = AB = BC = CD = DA = BH = JH).
Firstly fold the object on the red lines, like so:
Assuming (as stated above), edges FA = AB = AD, etc, then the resulting polyhedron is
a skewed square-based pyramid, which fits neatly inside a cube of side length
FAsuch that the baseABCDis one face, and the pointE/Fis one of the top corners. Thus,FADandEBAare both half faces of the cube, meaning we have two totally covered faces of the cube (so even intuitively we can see that three copies should cover all six faces).
If we call the remaining points of the cube, X (above B), Y (above C), and Z (above D), then we can
take another copy and place its base square on the
DCYZface and its apex atE. This covers the second half of theAEZDface and half the top (EXYZ). Then we take a third copy and slide it in such that its vertex is again atE, but with its base covering the whole of theXYCBface. This covers the second half of theEXYZ"top" and the second half of theEXBAface.
Another way to think of this is to take the original object and
duplicate it three times, each time rotating through the diagonal
ECaxis, moving the base to a new face of the cube each rotation.
As I said at the top, I'm not sure how to approach a rigorous proof, but it's hopefully intuitive that
because all copies share the central
ECaxis and each internal face is at 45°, the copies can't be overlapping each other. Similarly, the large internal 45° triangles they must be completely face to face, meaning we can't be left with any gaps, so we've filled the cube entirely.
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
$endgroup$
add a comment |
$begingroup$
I think we can do it with
3
copies of the given 3D object.
Note: I don't have a mathematical background, so this is mostly an intuitive "solve"and may not be very rigorous (or even correct). It also assumes that lengths that are visually equal, are (e.g. FD = EB = EJ, FA = EA = AB = BC = CD = DA = BH = JH).
Firstly fold the object on the red lines, like so:
Assuming (as stated above), edges FA = AB = AD, etc, then the resulting polyhedron is
a skewed square-based pyramid, which fits neatly inside a cube of side length
FAsuch that the baseABCDis one face, and the pointE/Fis one of the top corners. Thus,FADandEBAare both half faces of the cube, meaning we have two totally covered faces of the cube (so even intuitively we can see that three copies should cover all six faces).
If we call the remaining points of the cube, X (above B), Y (above C), and Z (above D), then we can
take another copy and place its base square on the
DCYZface and its apex atE. This covers the second half of theAEZDface and half the top (EXYZ). Then we take a third copy and slide it in such that its vertex is again atE, but with its base covering the whole of theXYCBface. This covers the second half of theEXYZ"top" and the second half of theEXBAface.
Another way to think of this is to take the original object and
duplicate it three times, each time rotating through the diagonal
ECaxis, moving the base to a new face of the cube each rotation.
As I said at the top, I'm not sure how to approach a rigorous proof, but it's hopefully intuitive that
because all copies share the central
ECaxis and each internal face is at 45°, the copies can't be overlapping each other. Similarly, the large internal 45° triangles they must be completely face to face, meaning we can't be left with any gaps, so we've filled the cube entirely.
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
$endgroup$
I think we can do it with
3
copies of the given 3D object.
Note: I don't have a mathematical background, so this is mostly an intuitive "solve"and may not be very rigorous (or even correct). It also assumes that lengths that are visually equal, are (e.g. FD = EB = EJ, FA = EA = AB = BC = CD = DA = BH = JH).
Firstly fold the object on the red lines, like so:
Assuming (as stated above), edges FA = AB = AD, etc, then the resulting polyhedron is
a skewed square-based pyramid, which fits neatly inside a cube of side length
FAsuch that the baseABCDis one face, and the pointE/Fis one of the top corners. Thus,FADandEBAare both half faces of the cube, meaning we have two totally covered faces of the cube (so even intuitively we can see that three copies should cover all six faces).
If we call the remaining points of the cube, X (above B), Y (above C), and Z (above D), then we can
take another copy and place its base square on the
DCYZface and its apex atE. This covers the second half of theAEZDface and half the top (EXYZ). Then we take a third copy and slide it in such that its vertex is again atE, but with its base covering the whole of theXYCBface. This covers the second half of theEXYZ"top" and the second half of theEXBAface.
Another way to think of this is to take the original object and
duplicate it three times, each time rotating through the diagonal
ECaxis, moving the base to a new face of the cube each rotation.
As I said at the top, I'm not sure how to approach a rigorous proof, but it's hopefully intuitive that
because all copies share the central
ECaxis and each internal face is at 45°, the copies can't be overlapping each other. Similarly, the large internal 45° triangles they must be completely face to face, meaning we can't be left with any gaps, so we've filled the cube entirely.
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
answered Jun 10 at 6:29
AlconjaAlconja
24.4k13 gold badges102 silver badges150 bronze badges
24.4k13 gold badges102 silver badges150 bronze badges
add a comment |
add a comment |
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$begingroup$
Is this the unwrapped surface? I first read this as saying that this is the projection of the 3D object onto a 2D surface.
$endgroup$
– Acccumulation
Jun 10 at 22:21