Changing my Clock
up vote
2
down vote
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I was given a problem in class. Let's say I bought a clock with no numbers on it (has only the hands). Because of the recent time change, I had to change my clock one hour back. In order to do this, I can rotate the clock.
Is it possible to rotate it exactly one hour?
puzzle angle local-time
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
add a comment |
up vote
2
down vote
favorite
I was given a problem in class. Let's say I bought a clock with no numbers on it (has only the hands). Because of the recent time change, I had to change my clock one hour back. In order to do this, I can rotate the clock.
Is it possible to rotate it exactly one hour?
puzzle angle local-time
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
I don't think so.
– WesleyGroupshaveFeelingsToo
Nov 13 at 10:30
At $2$ AM the angle between the hands is $60^{circ}.$ At $3$ AM the angle is $90^{circ}$ at $1$ AM the angle is $30^{circ}.$ Since rotation of the clock preserves the angle between the hands, the time can't be adjusted by rotating the clock.
– saulspatz
Nov 13 at 10:39
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was given a problem in class. Let's say I bought a clock with no numbers on it (has only the hands). Because of the recent time change, I had to change my clock one hour back. In order to do this, I can rotate the clock.
Is it possible to rotate it exactly one hour?
puzzle angle local-time
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
I was given a problem in class. Let's say I bought a clock with no numbers on it (has only the hands). Because of the recent time change, I had to change my clock one hour back. In order to do this, I can rotate the clock.
Is it possible to rotate it exactly one hour?
puzzle angle local-time
puzzle angle local-time
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
edited Nov 13 at 11:10
WesleyGroupshaveFeelingsToo
773217
773217
asked Nov 13 at 10:26
user614096
I don't think so.
– WesleyGroupshaveFeelingsToo
Nov 13 at 10:30
At $2$ AM the angle between the hands is $60^{circ}.$ At $3$ AM the angle is $90^{circ}$ at $1$ AM the angle is $30^{circ}.$ Since rotation of the clock preserves the angle between the hands, the time can't be adjusted by rotating the clock.
– saulspatz
Nov 13 at 10:39
add a comment |
I don't think so.
– WesleyGroupshaveFeelingsToo
Nov 13 at 10:30
At $2$ AM the angle between the hands is $60^{circ}.$ At $3$ AM the angle is $90^{circ}$ at $1$ AM the angle is $30^{circ}.$ Since rotation of the clock preserves the angle between the hands, the time can't be adjusted by rotating the clock.
– saulspatz
Nov 13 at 10:39
I don't think so.
– WesleyGroupshaveFeelingsToo
Nov 13 at 10:30
I don't think so.
– WesleyGroupshaveFeelingsToo
Nov 13 at 10:30
At $2$ AM the angle between the hands is $60^{circ}.$ At $3$ AM the angle is $90^{circ}$ at $1$ AM the angle is $30^{circ}.$ Since rotation of the clock preserves the angle between the hands, the time can't be adjusted by rotating the clock.
– saulspatz
Nov 13 at 10:39
At $2$ AM the angle between the hands is $60^{circ}.$ At $3$ AM the angle is $90^{circ}$ at $1$ AM the angle is $30^{circ}.$ Since rotation of the clock preserves the angle between the hands, the time can't be adjusted by rotating the clock.
– saulspatz
Nov 13 at 10:39
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
If every specific time on a clock has a certain distance between the two hands, it cannot be that two subsequent hours have the same distance between the hands.
let $alpha$ be the angle of the short hand with respect to $12$ o 'clock and $beta$ the angle of the long hand. Then we know that for clocks the rate of change (the speed at which the hands move is different) of $beta$ is greater than that of $alpha$, so consider $alpha '$ and $beta'$ to be the angles an hour later, we know that $$ beta - alpha neq beta' - alpha'$$
Since the angles between the hands are different there is no way to solve this by rotating the clock, as this preserves the angle between hands, and we need the angle to change.
As a concrete example, suppose it is 12 o 'clock. The two hands must be aligned, an hour later we have that the two hands are no longer aligned, we can not fix this by rotating.
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
add a comment |
up vote
0
down vote
No. If you let the clock run for exactly an hour, the angle between the hands changes (by $30°$ for an ordinary clock).
answered Nov 14 at 19:52
Yves Daoust
121k668217
add a comment |
up vote
-1
down vote
why is this hard to solve?
just keep rolling the minutes handle backwards until it reaches the same position
edit for super specific people :
put your finger on the tip of the minutes hand, change the minutes hand counter clock wise until it reaches the tip of your finger again
answered Nov 14 at 19:44
bigworld12
1024
why was my solution downvoted without a reason?
– bigworld12
Nov 14 at 20:58
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If every specific time on a clock has a certain distance between the two hands, it cannot be that two subsequent hours have the same distance between the hands.
let $alpha$ be the angle of the short hand with respect to $12$ o 'clock and $beta$ the angle of the long hand. Then we know that for clocks the rate of change (the speed at which the hands move is different) of $beta$ is greater than that of $alpha$, so consider $alpha '$ and $beta'$ to be the angles an hour later, we know that $$ beta - alpha neq beta' - alpha'$$
Since the angles between the hands are different there is no way to solve this by rotating the clock, as this preserves the angle between hands, and we need the angle to change.
As a concrete example, suppose it is 12 o 'clock. The two hands must be aligned, an hour later we have that the two hands are no longer aligned, we can not fix this by rotating.
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
add a comment |
up vote
1
down vote
If every specific time on a clock has a certain distance between the two hands, it cannot be that two subsequent hours have the same distance between the hands.
let $alpha$ be the angle of the short hand with respect to $12$ o 'clock and $beta$ the angle of the long hand. Then we know that for clocks the rate of change (the speed at which the hands move is different) of $beta$ is greater than that of $alpha$, so consider $alpha '$ and $beta'$ to be the angles an hour later, we know that $$ beta - alpha neq beta' - alpha'$$
Since the angles between the hands are different there is no way to solve this by rotating the clock, as this preserves the angle between hands, and we need the angle to change.
As a concrete example, suppose it is 12 o 'clock. The two hands must be aligned, an hour later we have that the two hands are no longer aligned, we can not fix this by rotating.
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
add a comment |
up vote
1
down vote
up vote
1
down vote
If every specific time on a clock has a certain distance between the two hands, it cannot be that two subsequent hours have the same distance between the hands.
let $alpha$ be the angle of the short hand with respect to $12$ o 'clock and $beta$ the angle of the long hand. Then we know that for clocks the rate of change (the speed at which the hands move is different) of $beta$ is greater than that of $alpha$, so consider $alpha '$ and $beta'$ to be the angles an hour later, we know that $$ beta - alpha neq beta' - alpha'$$
Since the angles between the hands are different there is no way to solve this by rotating the clock, as this preserves the angle between hands, and we need the angle to change.
As a concrete example, suppose it is 12 o 'clock. The two hands must be aligned, an hour later we have that the two hands are no longer aligned, we can not fix this by rotating.
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
If every specific time on a clock has a certain distance between the two hands, it cannot be that two subsequent hours have the same distance between the hands.
let $alpha$ be the angle of the short hand with respect to $12$ o 'clock and $beta$ the angle of the long hand. Then we know that for clocks the rate of change (the speed at which the hands move is different) of $beta$ is greater than that of $alpha$, so consider $alpha '$ and $beta'$ to be the angles an hour later, we know that $$ beta - alpha neq beta' - alpha'$$
Since the angles between the hands are different there is no way to solve this by rotating the clock, as this preserves the angle between hands, and we need the angle to change.
As a concrete example, suppose it is 12 o 'clock. The two hands must be aligned, an hour later we have that the two hands are no longer aligned, we can not fix this by rotating.
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
edited Nov 13 at 10:45
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
answered Nov 13 at 10:32
WesleyGroupshaveFeelingsToo
773217
773217
add a comment |
add a comment |
up vote
0
down vote
No. If you let the clock run for exactly an hour, the angle between the hands changes (by $30°$ for an ordinary clock).
answered Nov 14 at 19:52
Yves Daoust
121k668217
add a comment |
up vote
0
down vote
No. If you let the clock run for exactly an hour, the angle between the hands changes (by $30°$ for an ordinary clock).
answered Nov 14 at 19:52
Yves Daoust
121k668217
add a comment |
up vote
0
down vote
up vote
0
down vote
No. If you let the clock run for exactly an hour, the angle between the hands changes (by $30°$ for an ordinary clock).
answered Nov 14 at 19:52
Yves Daoust
121k668217
No. If you let the clock run for exactly an hour, the angle between the hands changes (by $30°$ for an ordinary clock).
answered Nov 14 at 19:52
Yves Daoust
121k668217
answered Nov 14 at 19:52
Yves Daoust
121k668217
answered Nov 14 at 19:52
Yves Daoust
121k668217
answered Nov 14 at 19:52
Yves Daoust
121k668217
121k668217
add a comment |
add a comment |
up vote
-1
down vote
why is this hard to solve?
just keep rolling the minutes handle backwards until it reaches the same position
edit for super specific people :
put your finger on the tip of the minutes hand, change the minutes hand counter clock wise until it reaches the tip of your finger again
answered Nov 14 at 19:44
bigworld12
1024
why was my solution downvoted without a reason?
– bigworld12
Nov 14 at 20:58
add a comment |
up vote
-1
down vote
why is this hard to solve?
just keep rolling the minutes handle backwards until it reaches the same position
edit for super specific people :
put your finger on the tip of the minutes hand, change the minutes hand counter clock wise until it reaches the tip of your finger again
answered Nov 14 at 19:44
bigworld12
1024
why was my solution downvoted without a reason?
– bigworld12
Nov 14 at 20:58
add a comment |
up vote
-1
down vote
up vote
-1
down vote
why is this hard to solve?
just keep rolling the minutes handle backwards until it reaches the same position
edit for super specific people :
put your finger on the tip of the minutes hand, change the minutes hand counter clock wise until it reaches the tip of your finger again
answered Nov 14 at 19:44
bigworld12
1024
why is this hard to solve?
just keep rolling the minutes handle backwards until it reaches the same position
edit for super specific people :
put your finger on the tip of the minutes hand, change the minutes hand counter clock wise until it reaches the tip of your finger again
answered Nov 14 at 19:44
bigworld12
1024
answered Nov 14 at 19:44
bigworld12
1024
answered Nov 14 at 19:44
bigworld12
1024
answered Nov 14 at 19:44
bigworld12
1024
1024
why was my solution downvoted without a reason?
– bigworld12
Nov 14 at 20:58
add a comment |
why was my solution downvoted without a reason?
– bigworld12
Nov 14 at 20:58
why was my solution downvoted without a reason?
– bigworld12
Nov 14 at 20:58
why was my solution downvoted without a reason?
– bigworld12
Nov 14 at 20:58
add a comment |
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I don't think so.
– WesleyGroupshaveFeelingsToo
Nov 13 at 10:30
At $2$ AM the angle between the hands is $60^{circ}.$ At $3$ AM the angle is $90^{circ}$ at $1$ AM the angle is $30^{circ}.$ Since rotation of the clock preserves the angle between the hands, the time can't be adjusted by rotating the clock.
– saulspatz
Nov 13 at 10:39