Where are the extra coins?
I am a manager of a coin casting foundry. We produce perfectly round coins with some (fixed) thickness and a diameter of exactly 1 inch. The working room is well-secured such that if any coin tries to leave the room, the alarm goes off. To allow workers to safely carry coins out of the room, a special container that is 10" * 10" square is available. A maximum of 1 layer of coins may be spread in the container and when it passes the door, the alarm won't go off. Normally, up to 100 coins can be carried with one such container.
Yesterday we encountered a theft. A container (with coins) was taken out of the working room without triggering the alarm. But what confused me is that 106 coins were lost. I don't know how the thief took the extra 6 coins without setting off the alarm.
Can you help me solve this mystery?
geometry
asked Dec 1 '18 at 12:21
iBugiBug
711219
add a comment |
I am a manager of a coin casting foundry. We produce perfectly round coins with some (fixed) thickness and a diameter of exactly 1 inch. The working room is well-secured such that if any coin tries to leave the room, the alarm goes off. To allow workers to safely carry coins out of the room, a special container that is 10" * 10" square is available. A maximum of 1 layer of coins may be spread in the container and when it passes the door, the alarm won't go off. Normally, up to 100 coins can be carried with one such container.
Yesterday we encountered a theft. A container (with coins) was taken out of the working room without triggering the alarm. But what confused me is that 106 coins were lost. I don't know how the thief took the extra 6 coins without setting off the alarm.
Can you help me solve this mystery?
geometry
asked Dec 1 '18 at 12:21
iBugiBug
711219
This must have been asked before...
– Dr Xorile
Dec 1 '18 at 21:38
add a comment |
I am a manager of a coin casting foundry. We produce perfectly round coins with some (fixed) thickness and a diameter of exactly 1 inch. The working room is well-secured such that if any coin tries to leave the room, the alarm goes off. To allow workers to safely carry coins out of the room, a special container that is 10" * 10" square is available. A maximum of 1 layer of coins may be spread in the container and when it passes the door, the alarm won't go off. Normally, up to 100 coins can be carried with one such container.
Yesterday we encountered a theft. A container (with coins) was taken out of the working room without triggering the alarm. But what confused me is that 106 coins were lost. I don't know how the thief took the extra 6 coins without setting off the alarm.
Can you help me solve this mystery?
geometry
asked Dec 1 '18 at 12:21
iBugiBug
711219
I am a manager of a coin casting foundry. We produce perfectly round coins with some (fixed) thickness and a diameter of exactly 1 inch. The working room is well-secured such that if any coin tries to leave the room, the alarm goes off. To allow workers to safely carry coins out of the room, a special container that is 10" * 10" square is available. A maximum of 1 layer of coins may be spread in the container and when it passes the door, the alarm won't go off. Normally, up to 100 coins can be carried with one such container.
Yesterday we encountered a theft. A container (with coins) was taken out of the working room without triggering the alarm. But what confused me is that 106 coins were lost. I don't know how the thief took the extra 6 coins without setting off the alarm.
Can you help me solve this mystery?
geometry
geometry
asked Dec 1 '18 at 12:21
iBugiBug
711219
asked Dec 1 '18 at 12:21
iBugiBug
711219
asked Dec 1 '18 at 12:21
iBugiBug
711219
asked Dec 1 '18 at 12:21
iBugiBug
711219
asked Dec 1 '18 at 12:21
iBugiBug
711219
711219
This must have been asked before...
– Dr Xorile
Dec 1 '18 at 21:38
add a comment |
This must have been asked before...
– Dr Xorile
Dec 1 '18 at 21:38
This must have been asked before...
– Dr Xorile
Dec 1 '18 at 21:38
This must have been asked before...
– Dr Xorile
Dec 1 '18 at 21:38
add a comment |
1 Answer
1
active
oldest
votes
The coins were
packed in a (partially) hexagonal packing. This is more efficient than a 'square' packing, so it allows for some extra coins above the 100.
It looks like this:
(only the black circles; the gray circles are just for comparing measurements)
SVG source code available here
A proof that
this configuration fits is as follows: the distance between $a_1$ and $b_2$ is $1$; the distance between $a_2$ and $b_2$ is $frac{1}{2}$. By Pythagoras, the distance between $a_1$ and $a_2$ is $sqrt{1-frac{1}{2}^2}$ = $frac{1}{2}sqrt{3}$. So the distance between $a_1$ and $a_3$ is $sqrt{3}$, between $a_1$ and $a_5$ it's $2sqrt{3}$, between $a_1$ and $a_7$ it's $2sqrt{3}+2$ and between $a_1$ and $a_{11}$ it's $4sqrt{3}+2 ≈ 8.9282 < 9$, so this configuration fits in a $10 times 10$ square.
Note that 106
is the currently known maximum number of coins that would fit; see this site for more details about minimal circle packings. For $N=107$, the 'ratio' between the circle radius and the length of the square is $20.1995... > 20$, which means the container size needs to be $0.5 ,text{inch} times , 20.1995... > 10 ,text{inch}$.
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
The coins were
packed in a (partially) hexagonal packing. This is more efficient than a 'square' packing, so it allows for some extra coins above the 100.
It looks like this:
(only the black circles; the gray circles are just for comparing measurements)
SVG source code available here
A proof that
this configuration fits is as follows: the distance between $a_1$ and $b_2$ is $1$; the distance between $a_2$ and $b_2$ is $frac{1}{2}$. By Pythagoras, the distance between $a_1$ and $a_2$ is $sqrt{1-frac{1}{2}^2}$ = $frac{1}{2}sqrt{3}$. So the distance between $a_1$ and $a_3$ is $sqrt{3}$, between $a_1$ and $a_5$ it's $2sqrt{3}$, between $a_1$ and $a_7$ it's $2sqrt{3}+2$ and between $a_1$ and $a_{11}$ it's $4sqrt{3}+2 ≈ 8.9282 < 9$, so this configuration fits in a $10 times 10$ square.
Note that 106
is the currently known maximum number of coins that would fit; see this site for more details about minimal circle packings. For $N=107$, the 'ratio' between the circle radius and the length of the square is $20.1995... > 20$, which means the container size needs to be $0.5 ,text{inch} times , 20.1995... > 10 ,text{inch}$.
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
add a comment |
The coins were
packed in a (partially) hexagonal packing. This is more efficient than a 'square' packing, so it allows for some extra coins above the 100.
It looks like this:
(only the black circles; the gray circles are just for comparing measurements)
SVG source code available here
A proof that
this configuration fits is as follows: the distance between $a_1$ and $b_2$ is $1$; the distance between $a_2$ and $b_2$ is $frac{1}{2}$. By Pythagoras, the distance between $a_1$ and $a_2$ is $sqrt{1-frac{1}{2}^2}$ = $frac{1}{2}sqrt{3}$. So the distance between $a_1$ and $a_3$ is $sqrt{3}$, between $a_1$ and $a_5$ it's $2sqrt{3}$, between $a_1$ and $a_7$ it's $2sqrt{3}+2$ and between $a_1$ and $a_{11}$ it's $4sqrt{3}+2 ≈ 8.9282 < 9$, so this configuration fits in a $10 times 10$ square.
Note that 106
is the currently known maximum number of coins that would fit; see this site for more details about minimal circle packings. For $N=107$, the 'ratio' between the circle radius and the length of the square is $20.1995... > 20$, which means the container size needs to be $0.5 ,text{inch} times , 20.1995... > 10 ,text{inch}$.
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
add a comment |
The coins were
packed in a (partially) hexagonal packing. This is more efficient than a 'square' packing, so it allows for some extra coins above the 100.
It looks like this:
(only the black circles; the gray circles are just for comparing measurements)
SVG source code available here
A proof that
this configuration fits is as follows: the distance between $a_1$ and $b_2$ is $1$; the distance between $a_2$ and $b_2$ is $frac{1}{2}$. By Pythagoras, the distance between $a_1$ and $a_2$ is $sqrt{1-frac{1}{2}^2}$ = $frac{1}{2}sqrt{3}$. So the distance between $a_1$ and $a_3$ is $sqrt{3}$, between $a_1$ and $a_5$ it's $2sqrt{3}$, between $a_1$ and $a_7$ it's $2sqrt{3}+2$ and between $a_1$ and $a_{11}$ it's $4sqrt{3}+2 ≈ 8.9282 < 9$, so this configuration fits in a $10 times 10$ square.
Note that 106
is the currently known maximum number of coins that would fit; see this site for more details about minimal circle packings. For $N=107$, the 'ratio' between the circle radius and the length of the square is $20.1995... > 20$, which means the container size needs to be $0.5 ,text{inch} times , 20.1995... > 10 ,text{inch}$.
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
The coins were
packed in a (partially) hexagonal packing. This is more efficient than a 'square' packing, so it allows for some extra coins above the 100.
It looks like this:
(only the black circles; the gray circles are just for comparing measurements)
SVG source code available here
A proof that
this configuration fits is as follows: the distance between $a_1$ and $b_2$ is $1$; the distance between $a_2$ and $b_2$ is $frac{1}{2}$. By Pythagoras, the distance between $a_1$ and $a_2$ is $sqrt{1-frac{1}{2}^2}$ = $frac{1}{2}sqrt{3}$. So the distance between $a_1$ and $a_3$ is $sqrt{3}$, between $a_1$ and $a_5$ it's $2sqrt{3}$, between $a_1$ and $a_7$ it's $2sqrt{3}+2$ and between $a_1$ and $a_{11}$ it's $4sqrt{3}+2 ≈ 8.9282 < 9$, so this configuration fits in a $10 times 10$ square.
Note that 106
is the currently known maximum number of coins that would fit; see this site for more details about minimal circle packings. For $N=107$, the 'ratio' between the circle radius and the length of the square is $20.1995... > 20$, which means the container size needs to be $0.5 ,text{inch} times , 20.1995... > 10 ,text{inch}$.
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
edited Dec 3 '18 at 8:45
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
answered Dec 1 '18 at 13:20
GlorfindelGlorfindel
13.6k34983
13.6k34983
add a comment |
add a comment |
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This must have been asked before...
– Dr Xorile
Dec 1 '18 at 21:38