Mixing
What is the travelled distance of the red mark on the upper surface of the rotating cube?
$begingroup$
Each side of a cube is 2 unit in length. This cube is kept on a
table such a way that one surface (i.e., 4 vertices) of it
completely touches the table. At this position, a red point is
drawn on the center of the upper surface. Now the cube is being
rotated along a straight line towards a certain direction. At the time
of rotation, at least two vertices of the cube are in touch with
the table. Rotation is stopped when the red mark reached its initial
position. Total distance traveled by the red mark is
$$(sqrt{{b}}+1)pi$$
What is the value of b?
Source: Bangladesh Math Olympiad 2017
I am facing trouble to find the distance travelled by the red mark. Is it a straight line from start to end or a curved line (almost a circle) which is travelled by the red mark at every rotation of the cube?
geometry contest-math pi
asked Jan 10 at 14:54
ShromiShromi
1219
$endgroup$
add a comment |
$begingroup$
Each side of a cube is 2 unit in length. This cube is kept on a
table such a way that one surface (i.e., 4 vertices) of it
completely touches the table. At this position, a red point is
drawn on the center of the upper surface. Now the cube is being
rotated along a straight line towards a certain direction. At the time
of rotation, at least two vertices of the cube are in touch with
the table. Rotation is stopped when the red mark reached its initial
position. Total distance traveled by the red mark is
$$(sqrt{{b}}+1)pi$$
What is the value of b?
Source: Bangladesh Math Olympiad 2017
I am facing trouble to find the distance travelled by the red mark. Is it a straight line from start to end or a curved line (almost a circle) which is travelled by the red mark at every rotation of the cube?
geometry contest-math pi
asked Jan 10 at 14:54
ShromiShromi
1219
$endgroup$
add a comment |
$begingroup$
Each side of a cube is 2 unit in length. This cube is kept on a
table such a way that one surface (i.e., 4 vertices) of it
completely touches the table. At this position, a red point is
drawn on the center of the upper surface. Now the cube is being
rotated along a straight line towards a certain direction. At the time
of rotation, at least two vertices of the cube are in touch with
the table. Rotation is stopped when the red mark reached its initial
position. Total distance traveled by the red mark is
$$(sqrt{{b}}+1)pi$$
What is the value of b?
Source: Bangladesh Math Olympiad 2017
I am facing trouble to find the distance travelled by the red mark. Is it a straight line from start to end or a curved line (almost a circle) which is travelled by the red mark at every rotation of the cube?
geometry contest-math pi
asked Jan 10 at 14:54
ShromiShromi
1219
$endgroup$
Each side of a cube is 2 unit in length. This cube is kept on a
table such a way that one surface (i.e., 4 vertices) of it
completely touches the table. At this position, a red point is
drawn on the center of the upper surface. Now the cube is being
rotated along a straight line towards a certain direction. At the time
of rotation, at least two vertices of the cube are in touch with
the table. Rotation is stopped when the red mark reached its initial
position. Total distance traveled by the red mark is
$$(sqrt{{b}}+1)pi$$
What is the value of b?
Source: Bangladesh Math Olympiad 2017
I am facing trouble to find the distance travelled by the red mark. Is it a straight line from start to end or a curved line (almost a circle) which is travelled by the red mark at every rotation of the cube?
geometry contest-math pi
geometry contest-math pi
asked Jan 10 at 14:54
ShromiShromi
1219
asked Jan 10 at 14:54
ShromiShromi
1219
asked Jan 10 at 14:54
ShromiShromi
1219
asked Jan 10 at 14:54
ShromiShromi
1219
asked Jan 10 at 14:54
ShromiShromi
1219
1219
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If I understand the problem correctly, the answer should be b=5 as seen from the following picture:
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
answered Jan 10 at 15:39
floxbrfloxbr
463
$endgroup$
add a comment |
$begingroup$
The cube is rolled keeping one edge on the table at each time. As it is rolled around each edge the red point traces a quarter circle with radius the distance to the edge of rotation. For the first edge the point is $sqrt 5$ away from the edge because we have a $1-2-sqrt 5$ triangle from the center of the edge to the point, so the point moves $frac pi 2 sqrt 5$. Then the point is $1$ from the edge of rotation for the next one, so it traces $frac pi 2$. The third edge has the point starting from the center bottom, again $1$ from the center, and we get another $frac pi 2$. The final one has the point $sqrt 5$ away again, so we get a total of
$$frac pi 2(2+2sqrt 5)=pi(1+sqrt 5)\b=5$$
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
$endgroup$
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
$begingroup$
If I understand the problem correctly, the answer should be b=5 as seen from the following picture:
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
answered Jan 10 at 15:39
floxbrfloxbr
463
$endgroup$
add a comment |
$begingroup$
If I understand the problem correctly, the answer should be b=5 as seen from the following picture:
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
answered Jan 10 at 15:39
floxbrfloxbr
463
$endgroup$
add a comment |
$begingroup$
If I understand the problem correctly, the answer should be b=5 as seen from the following picture:
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
answered Jan 10 at 15:39
floxbrfloxbr
463
$endgroup$
If I understand the problem correctly, the answer should be b=5 as seen from the following picture:
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
answered Jan 10 at 15:39
floxbrfloxbr
463
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
edited Jan 10 at 15:44
Jaap Scherphuis
4,119717
4,119717
answered Jan 10 at 15:39
floxbrfloxbr
463
answered Jan 10 at 15:39
floxbrfloxbr
463
answered Jan 10 at 15:39
floxbrfloxbr
463
463
add a comment |
add a comment |
$begingroup$
The cube is rolled keeping one edge on the table at each time. As it is rolled around each edge the red point traces a quarter circle with radius the distance to the edge of rotation. For the first edge the point is $sqrt 5$ away from the edge because we have a $1-2-sqrt 5$ triangle from the center of the edge to the point, so the point moves $frac pi 2 sqrt 5$. Then the point is $1$ from the edge of rotation for the next one, so it traces $frac pi 2$. The third edge has the point starting from the center bottom, again $1$ from the center, and we get another $frac pi 2$. The final one has the point $sqrt 5$ away again, so we get a total of
$$frac pi 2(2+2sqrt 5)=pi(1+sqrt 5)\b=5$$
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
The cube is rolled keeping one edge on the table at each time. As it is rolled around each edge the red point traces a quarter circle with radius the distance to the edge of rotation. For the first edge the point is $sqrt 5$ away from the edge because we have a $1-2-sqrt 5$ triangle from the center of the edge to the point, so the point moves $frac pi 2 sqrt 5$. Then the point is $1$ from the edge of rotation for the next one, so it traces $frac pi 2$. The third edge has the point starting from the center bottom, again $1$ from the center, and we get another $frac pi 2$. The final one has the point $sqrt 5$ away again, so we get a total of
$$frac pi 2(2+2sqrt 5)=pi(1+sqrt 5)\b=5$$
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
The cube is rolled keeping one edge on the table at each time. As it is rolled around each edge the red point traces a quarter circle with radius the distance to the edge of rotation. For the first edge the point is $sqrt 5$ away from the edge because we have a $1-2-sqrt 5$ triangle from the center of the edge to the point, so the point moves $frac pi 2 sqrt 5$. Then the point is $1$ from the edge of rotation for the next one, so it traces $frac pi 2$. The third edge has the point starting from the center bottom, again $1$ from the center, and we get another $frac pi 2$. The final one has the point $sqrt 5$ away again, so we get a total of
$$frac pi 2(2+2sqrt 5)=pi(1+sqrt 5)\b=5$$
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
$endgroup$
The cube is rolled keeping one edge on the table at each time. As it is rolled around each edge the red point traces a quarter circle with radius the distance to the edge of rotation. For the first edge the point is $sqrt 5$ away from the edge because we have a $1-2-sqrt 5$ triangle from the center of the edge to the point, so the point moves $frac pi 2 sqrt 5$. Then the point is $1$ from the edge of rotation for the next one, so it traces $frac pi 2$. The third edge has the point starting from the center bottom, again $1$ from the center, and we get another $frac pi 2$. The final one has the point $sqrt 5$ away again, so we get a total of
$$frac pi 2(2+2sqrt 5)=pi(1+sqrt 5)\b=5$$
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
answered Jan 10 at 15:23
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
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