This should be a piece of cake… right?
Clash Royale CLAN TAG#URR8PPP
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You probably know the following problem:
We have two circular cakes of the same height but unknown and potentially different radii, and we want to cut them into two equal shares. Each cut can only cut one piece of one cake. What is the minimal number of cuts required?
The solution is easily found:
It's $1$ cut. Stack the two cakes and center them, then cut the big one around the small one :
Here you go.
While I was thinking about this easy problem, the following question logically arose:
We have $ninmathbbN^*$ circular cakes of unknown and potentially different radii (hmm yummy) but the same height, and we want to cut them into $pgt n$ equal shares. Each cut can only go through one piece of cake. What is the minimal number $c$ of cuts required? How do you cut the cakes?
The cuts can be done however you want (i.e. lines, curves, etc) and stacking cakes etc is allowed. You are of course not allowed to measure the exact radius. To make it simple, let's say you have a straightedge (unlabelled ruler), an arbitary-precision protractor and a compass.
Any help/thoughts are appreciated.
recreational-mathematics puzzle
edited Sep 2 at 11:13
JonathanDavidArndt
21613
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
 |Â
show 8 more comments
up vote
24
down vote
favorite
You probably know the following problem:
We have two circular cakes of the same height but unknown and potentially different radii, and we want to cut them into two equal shares. Each cut can only cut one piece of one cake. What is the minimal number of cuts required?
The solution is easily found:
It's $1$ cut. Stack the two cakes and center them, then cut the big one around the small one :
While I was thinking about this easy problem, the following question logically arose:
We have $ninmathbbN^*$ circular cakes of unknown and potentially different radii (hmm yummy) but the same height, and we want to cut them into $pgt n$ equal shares. Each cut can only go through one piece of cake. What is the minimal number $c$ of cuts required? How do you cut the cakes?
The cuts can be done however you want (i.e. lines, curves, etc) and stacking cakes etc is allowed. You are of course not allowed to measure the exact radius. To make it simple, let's say you have a straightedge (unlabelled ruler), an arbitary-precision protractor and a compass.
Any help/thoughts are appreciated.
recreational-mathematics puzzle
edited Sep 2 at 11:13
JonathanDavidArndt
21613
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
Initial results: The largest cake must always be cut into $p$ pieces, so $p-1$ is a lower bound. Otherwise, there don't seem to be any answers as elegant as the $p = 2, n = 2$ case. I've made a few drawings.
– Paul Z
Jun 26 '14 at 19:26
@PaulZ $p-1$ is indeed a lower bound, but I doubt we can always do it in $p-1$ cuts. (can we?) - What's more, the same problem exists with 3 cakes that have to be shared between 4 people, so even if the solution isn't as nice as the simple problem's one, a good answer should exist.
– Hippalectryon
Jun 26 '14 at 20:04
In your diagram, you are trying to cut the total into $2$ equal shares. The words seem to indicate that each cake must be cut into $p$ shares. Do I just have to get $frac 1p$ of the total volume into each share?
– Ross Millikan
Jun 26 '14 at 20:09
When you stack the two cakes - you can simply cut BOTH at the same time in two equal pieces... Or is that considered as 2 cuts... Ok that are 2 cuts as I see now...
– johannesvalks
Jun 26 '14 at 20:09
4
@johannesvalks: OP said that each cut could only cut one cake.
– Ross Millikan
Jun 26 '14 at 20:10
 |Â
show 8 more comments
up vote
24
down vote
favorite
up vote
24
down vote
favorite
You probably know the following problem:
We have two circular cakes of the same height but unknown and potentially different radii, and we want to cut them into two equal shares. Each cut can only cut one piece of one cake. What is the minimal number of cuts required?
The solution is easily found:
It's $1$ cut. Stack the two cakes and center them, then cut the big one around the small one :
While I was thinking about this easy problem, the following question logically arose:
We have $ninmathbbN^*$ circular cakes of unknown and potentially different radii (hmm yummy) but the same height, and we want to cut them into $pgt n$ equal shares. Each cut can only go through one piece of cake. What is the minimal number $c$ of cuts required? How do you cut the cakes?
The cuts can be done however you want (i.e. lines, curves, etc) and stacking cakes etc is allowed. You are of course not allowed to measure the exact radius. To make it simple, let's say you have a straightedge (unlabelled ruler), an arbitary-precision protractor and a compass.
Any help/thoughts are appreciated.
recreational-mathematics puzzle
edited Sep 2 at 11:13
JonathanDavidArndt
21613
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
You probably know the following problem:
We have two circular cakes of the same height but unknown and potentially different radii, and we want to cut them into two equal shares. Each cut can only cut one piece of one cake. What is the minimal number of cuts required?
The solution is easily found:
It's $1$ cut. Stack the two cakes and center them, then cut the big one around the small one :
While I was thinking about this easy problem, the following question logically arose:
We have $ninmathbbN^*$ circular cakes of unknown and potentially different radii (hmm yummy) but the same height, and we want to cut them into $pgt n$ equal shares. Each cut can only go through one piece of cake. What is the minimal number $c$ of cuts required? How do you cut the cakes?
The cuts can be done however you want (i.e. lines, curves, etc) and stacking cakes etc is allowed. You are of course not allowed to measure the exact radius. To make it simple, let's say you have a straightedge (unlabelled ruler), an arbitary-precision protractor and a compass.
Any help/thoughts are appreciated.
recreational-mathematics puzzle
recreational-mathematics puzzle
edited Sep 2 at 11:13
JonathanDavidArndt
21613
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
edited Sep 2 at 11:13
JonathanDavidArndt
21613
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
edited Sep 2 at 11:13
JonathanDavidArndt
21613
edited Sep 2 at 11:13
JonathanDavidArndt
21613
edited Sep 2 at 11:13
JonathanDavidArndt
21613
21613
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
asked Jun 26 '14 at 19:06
Hippalectryon
4,30122553
4,30122553
Initial results: The largest cake must always be cut into $p$ pieces, so $p-1$ is a lower bound. Otherwise, there don't seem to be any answers as elegant as the $p = 2, n = 2$ case. I've made a few drawings.
– Paul Z
Jun 26 '14 at 19:26
@PaulZ $p-1$ is indeed a lower bound, but I doubt we can always do it in $p-1$ cuts. (can we?) - What's more, the same problem exists with 3 cakes that have to be shared between 4 people, so even if the solution isn't as nice as the simple problem's one, a good answer should exist.
– Hippalectryon
Jun 26 '14 at 20:04
In your diagram, you are trying to cut the total into $2$ equal shares. The words seem to indicate that each cake must be cut into $p$ shares. Do I just have to get $frac 1p$ of the total volume into each share?
– Ross Millikan
Jun 26 '14 at 20:09
When you stack the two cakes - you can simply cut BOTH at the same time in two equal pieces... Or is that considered as 2 cuts... Ok that are 2 cuts as I see now...
– johannesvalks
Jun 26 '14 at 20:09
4
@johannesvalks: OP said that each cut could only cut one cake.
– Ross Millikan
Jun 26 '14 at 20:10
 |Â
show 8 more comments
Initial results: The largest cake must always be cut into $p$ pieces, so $p-1$ is a lower bound. Otherwise, there don't seem to be any answers as elegant as the $p = 2, n = 2$ case. I've made a few drawings.
– Paul Z
Jun 26 '14 at 19:26
@PaulZ $p-1$ is indeed a lower bound, but I doubt we can always do it in $p-1$ cuts. (can we?) - What's more, the same problem exists with 3 cakes that have to be shared between 4 people, so even if the solution isn't as nice as the simple problem's one, a good answer should exist.
– Hippalectryon
Jun 26 '14 at 20:04
In your diagram, you are trying to cut the total into $2$ equal shares. The words seem to indicate that each cake must be cut into $p$ shares. Do I just have to get $frac 1p$ of the total volume into each share?
– Ross Millikan
Jun 26 '14 at 20:09
When you stack the two cakes - you can simply cut BOTH at the same time in two equal pieces... Or is that considered as 2 cuts... Ok that are 2 cuts as I see now...
– johannesvalks
Jun 26 '14 at 20:09
4
@johannesvalks: OP said that each cut could only cut one cake.
– Ross Millikan
Jun 26 '14 at 20:10
Initial results: The largest cake must always be cut into $p$ pieces, so $p-1$ is a lower bound. Otherwise, there don't seem to be any answers as elegant as the $p = 2, n = 2$ case. I've made a few drawings.
– Paul Z
Jun 26 '14 at 19:26
Initial results: The largest cake must always be cut into $p$ pieces, so $p-1$ is a lower bound. Otherwise, there don't seem to be any answers as elegant as the $p = 2, n = 2$ case. I've made a few drawings.
– Paul Z
Jun 26 '14 at 19:26
@PaulZ $p-1$ is indeed a lower bound, but I doubt we can always do it in $p-1$ cuts. (can we?) - What's more, the same problem exists with 3 cakes that have to be shared between 4 people, so even if the solution isn't as nice as the simple problem's one, a good answer should exist.
– Hippalectryon
Jun 26 '14 at 20:04
@PaulZ $p-1$ is indeed a lower bound, but I doubt we can always do it in $p-1$ cuts. (can we?) - What's more, the same problem exists with 3 cakes that have to be shared between 4 people, so even if the solution isn't as nice as the simple problem's one, a good answer should exist.
– Hippalectryon
Jun 26 '14 at 20:04
In your diagram, you are trying to cut the total into $2$ equal shares. The words seem to indicate that each cake must be cut into $p$ shares. Do I just have to get $frac 1p$ of the total volume into each share?
– Ross Millikan
Jun 26 '14 at 20:09
In your diagram, you are trying to cut the total into $2$ equal shares. The words seem to indicate that each cake must be cut into $p$ shares. Do I just have to get $frac 1p$ of the total volume into each share?
– Ross Millikan
Jun 26 '14 at 20:09
When you stack the two cakes - you can simply cut BOTH at the same time in two equal pieces... Or is that considered as 2 cuts... Ok that are 2 cuts as I see now...
– johannesvalks
Jun 26 '14 at 20:09
When you stack the two cakes - you can simply cut BOTH at the same time in two equal pieces... Or is that considered as 2 cuts... Ok that are 2 cuts as I see now...
– johannesvalks
Jun 26 '14 at 20:09
4
4
@johannesvalks: OP said that each cut could only cut one cake.
– Ross Millikan
Jun 26 '14 at 20:10
@johannesvalks: OP said that each cut could only cut one cake.
– Ross Millikan
Jun 26 '14 at 20:10
 |Â
show 8 more comments
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Initial results: The largest cake must always be cut into $p$ pieces, so $p-1$ is a lower bound. Otherwise, there don't seem to be any answers as elegant as the $p = 2, n = 2$ case. I've made a few drawings.
– Paul Z
Jun 26 '14 at 19:26
@PaulZ $p-1$ is indeed a lower bound, but I doubt we can always do it in $p-1$ cuts. (can we?) - What's more, the same problem exists with 3 cakes that have to be shared between 4 people, so even if the solution isn't as nice as the simple problem's one, a good answer should exist.
– Hippalectryon
Jun 26 '14 at 20:04
In your diagram, you are trying to cut the total into $2$ equal shares. The words seem to indicate that each cake must be cut into $p$ shares. Do I just have to get $frac 1p$ of the total volume into each share?
– Ross Millikan
Jun 26 '14 at 20:09
When you stack the two cakes - you can simply cut BOTH at the same time in two equal pieces... Or is that considered as 2 cuts... Ok that are 2 cuts as I see now...
– johannesvalks
Jun 26 '14 at 20:09
4
@johannesvalks: OP said that each cut could only cut one cake.
– Ross Millikan
Jun 26 '14 at 20:10