Covering a circle with rectangular labels
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I've got an applied math problem. Suppose that you have a circle of diameter $1$ (we can scale units so that this is the case) and labels of width $d_1, d_2, ..., d_n$ (the labels can be as long as you wish, but their width is fixed) where $d_1 + dots + d_n < 1.$
Can you cover the circle with these labels? I suspect that it's impossible, but I can't prove this for $n>2.$ If it's possible, how low can you make the sum?
Edit: I have placed a bounty to be awarded for full resolution (either a proof or disproof, as the possibility that the problem is independent of ZFC seems to be highly unlikely here) of the earlier conjecture.
circle rectangles
asked Nov 28 at 22:00
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789313
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up vote
2
down vote
favorite
I've got an applied math problem. Suppose that you have a circle of diameter $1$ (we can scale units so that this is the case) and labels of width $d_1, d_2, ..., d_n$ (the labels can be as long as you wish, but their width is fixed) where $d_1 + dots + d_n < 1.$
Can you cover the circle with these labels? I suspect that it's impossible, but I can't prove this for $n>2.$ If it's possible, how low can you make the sum?
Edit: I have placed a bounty to be awarded for full resolution (either a proof or disproof, as the possibility that the problem is independent of ZFC seems to be highly unlikely here) of the earlier conjecture.
circle rectangles
asked Nov 28 at 22:00
Display name
789313
I am pretty sure I have seen this problem before and there is a beautiful solution that uses a clever embedding of the original circle into a tridimensional sphere of radius 1.
– Daniel
Dec 1 at 3:02
1
Searching on mathstackexchange I found essentially the argument on this post
– Daniel
Dec 1 at 3:03
1
To explain it briefly, imagine your circle dividing a diameter $1$ sphere in $mathbb{R}^3$ in half and suppose that the rectangles indeed cover the original circle. For each rectangle, consider its projection on the sphere, which is a ring of area $2pi d_j cdot frac12$ (radius of sphere is $frac12$) where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere. But the total area covered by the rings is $sum_{j=1}^n pi d_j < pi$ while the surface area of the sphere is $4pi (1/2)^2 = pi$.
– Daniel
Dec 1 at 3:11
@Daniel Since your comment answers it, maybe make it an answer? [I get it if you don't want to since it has the basic answer you mention in previous comment.]
– coffeemath
Dec 1 at 4:03
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've got an applied math problem. Suppose that you have a circle of diameter $1$ (we can scale units so that this is the case) and labels of width $d_1, d_2, ..., d_n$ (the labels can be as long as you wish, but their width is fixed) where $d_1 + dots + d_n < 1.$
Can you cover the circle with these labels? I suspect that it's impossible, but I can't prove this for $n>2.$ If it's possible, how low can you make the sum?
Edit: I have placed a bounty to be awarded for full resolution (either a proof or disproof, as the possibility that the problem is independent of ZFC seems to be highly unlikely here) of the earlier conjecture.
circle rectangles
asked Nov 28 at 22:00
Display name
789313
I've got an applied math problem. Suppose that you have a circle of diameter $1$ (we can scale units so that this is the case) and labels of width $d_1, d_2, ..., d_n$ (the labels can be as long as you wish, but their width is fixed) where $d_1 + dots + d_n < 1.$
Can you cover the circle with these labels? I suspect that it's impossible, but I can't prove this for $n>2.$ If it's possible, how low can you make the sum?
Edit: I have placed a bounty to be awarded for full resolution (either a proof or disproof, as the possibility that the problem is independent of ZFC seems to be highly unlikely here) of the earlier conjecture.
circle rectangles
circle rectangles
asked Nov 28 at 22:00
Display name
789313
asked Nov 28 at 22:00
Display name
789313
edited Dec 1 at 2:36
asked Nov 28 at 22:00
Display name
789313
asked Nov 28 at 22:00
Display name
789313
asked Nov 28 at 22:00
Display name
789313
789313
I am pretty sure I have seen this problem before and there is a beautiful solution that uses a clever embedding of the original circle into a tridimensional sphere of radius 1.
– Daniel
Dec 1 at 3:02
1
Searching on mathstackexchange I found essentially the argument on this post
– Daniel
Dec 1 at 3:03
1
To explain it briefly, imagine your circle dividing a diameter $1$ sphere in $mathbb{R}^3$ in half and suppose that the rectangles indeed cover the original circle. For each rectangle, consider its projection on the sphere, which is a ring of area $2pi d_j cdot frac12$ (radius of sphere is $frac12$) where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere. But the total area covered by the rings is $sum_{j=1}^n pi d_j < pi$ while the surface area of the sphere is $4pi (1/2)^2 = pi$.
– Daniel
Dec 1 at 3:11
@Daniel Since your comment answers it, maybe make it an answer? [I get it if you don't want to since it has the basic answer you mention in previous comment.]
– coffeemath
Dec 1 at 4:03
add a comment |
I am pretty sure I have seen this problem before and there is a beautiful solution that uses a clever embedding of the original circle into a tridimensional sphere of radius 1.
– Daniel
Dec 1 at 3:02
1
Searching on mathstackexchange I found essentially the argument on this post
– Daniel
Dec 1 at 3:03
1
To explain it briefly, imagine your circle dividing a diameter $1$ sphere in $mathbb{R}^3$ in half and suppose that the rectangles indeed cover the original circle. For each rectangle, consider its projection on the sphere, which is a ring of area $2pi d_j cdot frac12$ (radius of sphere is $frac12$) where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere. But the total area covered by the rings is $sum_{j=1}^n pi d_j < pi$ while the surface area of the sphere is $4pi (1/2)^2 = pi$.
– Daniel
Dec 1 at 3:11
@Daniel Since your comment answers it, maybe make it an answer? [I get it if you don't want to since it has the basic answer you mention in previous comment.]
– coffeemath
Dec 1 at 4:03
I am pretty sure I have seen this problem before and there is a beautiful solution that uses a clever embedding of the original circle into a tridimensional sphere of radius 1.
– Daniel
Dec 1 at 3:02
I am pretty sure I have seen this problem before and there is a beautiful solution that uses a clever embedding of the original circle into a tridimensional sphere of radius 1.
– Daniel
Dec 1 at 3:02
1
1
Searching on mathstackexchange I found essentially the argument on this post
– Daniel
Dec 1 at 3:03
Searching on mathstackexchange I found essentially the argument on this post
– Daniel
Dec 1 at 3:03
1
1
To explain it briefly, imagine your circle dividing a diameter $1$ sphere in $mathbb{R}^3$ in half and suppose that the rectangles indeed cover the original circle. For each rectangle, consider its projection on the sphere, which is a ring of area $2pi d_j cdot frac12$ (radius of sphere is $frac12$) where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere. But the total area covered by the rings is $sum_{j=1}^n pi d_j < pi$ while the surface area of the sphere is $4pi (1/2)^2 = pi$.
– Daniel
Dec 1 at 3:11
To explain it briefly, imagine your circle dividing a diameter $1$ sphere in $mathbb{R}^3$ in half and suppose that the rectangles indeed cover the original circle. For each rectangle, consider its projection on the sphere, which is a ring of area $2pi d_j cdot frac12$ (radius of sphere is $frac12$) where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere. But the total area covered by the rings is $sum_{j=1}^n pi d_j < pi$ while the surface area of the sphere is $4pi (1/2)^2 = pi$.
– Daniel
Dec 1 at 3:11
@Daniel Since your comment answers it, maybe make it an answer? [I get it if you don't want to since it has the basic answer you mention in previous comment.]
– coffeemath
Dec 1 at 4:03
@Daniel Since your comment answers it, maybe make it an answer? [I get it if you don't want to since it has the basic answer you mention in previous comment.]
– coffeemath
Dec 1 at 4:03
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Turning the above comment into an answer.
Consider a sphere of diameter $1$ in $mathbb{R}^3$ and identify the original circle of diameter $1$ with the intersection of the sphere with a plane that divides it precisely in half.
We prove by contradiction that it is impossible that $n$ rectangles of width $d_1, ldots, d_n$ and $sum_{j=1}^n d_j < 1$ cover the circle. Indeed, suppose that the rectangles cover the original circle. For each rectangle, consider its projection on the sphere. Since the sphere has radius $frac12$, the projection is a ring of area
$$
A_j = 2pi d_j cdot frac12 = pi d_j,
$$
where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere implying that
$$
text{Area of sphere} le sum_{j=1}^n A_j = pi sum_{j=1}^n d_j < pi.
$$
However, the surface area of the sphere is $4 pi (frac12)^2 = pi$ and we conclude $pi < pi$, reaching a contradiction.
answered 2 days ago
Daniel
1,516210
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Turning the above comment into an answer.
Consider a sphere of diameter $1$ in $mathbb{R}^3$ and identify the original circle of diameter $1$ with the intersection of the sphere with a plane that divides it precisely in half.
We prove by contradiction that it is impossible that $n$ rectangles of width $d_1, ldots, d_n$ and $sum_{j=1}^n d_j < 1$ cover the circle. Indeed, suppose that the rectangles cover the original circle. For each rectangle, consider its projection on the sphere. Since the sphere has radius $frac12$, the projection is a ring of area
$$
A_j = 2pi d_j cdot frac12 = pi d_j,
$$
where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere implying that
$$
text{Area of sphere} le sum_{j=1}^n A_j = pi sum_{j=1}^n d_j < pi.
$$
However, the surface area of the sphere is $4 pi (frac12)^2 = pi$ and we conclude $pi < pi$, reaching a contradiction.
answered 2 days ago
Daniel
1,516210
add a comment |
up vote
3
down vote
accepted
Turning the above comment into an answer.
Consider a sphere of diameter $1$ in $mathbb{R}^3$ and identify the original circle of diameter $1$ with the intersection of the sphere with a plane that divides it precisely in half.
We prove by contradiction that it is impossible that $n$ rectangles of width $d_1, ldots, d_n$ and $sum_{j=1}^n d_j < 1$ cover the circle. Indeed, suppose that the rectangles cover the original circle. For each rectangle, consider its projection on the sphere. Since the sphere has radius $frac12$, the projection is a ring of area
$$
A_j = 2pi d_j cdot frac12 = pi d_j,
$$
where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere implying that
$$
text{Area of sphere} le sum_{j=1}^n A_j = pi sum_{j=1}^n d_j < pi.
$$
However, the surface area of the sphere is $4 pi (frac12)^2 = pi$ and we conclude $pi < pi$, reaching a contradiction.
answered 2 days ago
Daniel
1,516210
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Turning the above comment into an answer.
Consider a sphere of diameter $1$ in $mathbb{R}^3$ and identify the original circle of diameter $1$ with the intersection of the sphere with a plane that divides it precisely in half.
We prove by contradiction that it is impossible that $n$ rectangles of width $d_1, ldots, d_n$ and $sum_{j=1}^n d_j < 1$ cover the circle. Indeed, suppose that the rectangles cover the original circle. For each rectangle, consider its projection on the sphere. Since the sphere has radius $frac12$, the projection is a ring of area
$$
A_j = 2pi d_j cdot frac12 = pi d_j,
$$
where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere implying that
$$
text{Area of sphere} le sum_{j=1}^n A_j = pi sum_{j=1}^n d_j < pi.
$$
However, the surface area of the sphere is $4 pi (frac12)^2 = pi$ and we conclude $pi < pi$, reaching a contradiction.
answered 2 days ago
Daniel
1,516210
Turning the above comment into an answer.
Consider a sphere of diameter $1$ in $mathbb{R}^3$ and identify the original circle of diameter $1$ with the intersection of the sphere with a plane that divides it precisely in half.
We prove by contradiction that it is impossible that $n$ rectangles of width $d_1, ldots, d_n$ and $sum_{j=1}^n d_j < 1$ cover the circle. Indeed, suppose that the rectangles cover the original circle. For each rectangle, consider its projection on the sphere. Since the sphere has radius $frac12$, the projection is a ring of area
$$
A_j = 2pi d_j cdot frac12 = pi d_j,
$$
where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere implying that
$$
text{Area of sphere} le sum_{j=1}^n A_j = pi sum_{j=1}^n d_j < pi.
$$
However, the surface area of the sphere is $4 pi (frac12)^2 = pi$ and we conclude $pi < pi$, reaching a contradiction.
answered 2 days ago
Daniel
1,516210
answered 2 days ago
Daniel
1,516210
answered 2 days ago
Daniel
1,516210
answered 2 days ago
Daniel
1,516210
1,516210
add a comment |
add a comment |
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I am pretty sure I have seen this problem before and there is a beautiful solution that uses a clever embedding of the original circle into a tridimensional sphere of radius 1.
– Daniel
Dec 1 at 3:02
1
Searching on mathstackexchange I found essentially the argument on this post
– Daniel
Dec 1 at 3:03
1
To explain it briefly, imagine your circle dividing a diameter $1$ sphere in $mathbb{R}^3$ in half and suppose that the rectangles indeed cover the original circle. For each rectangle, consider its projection on the sphere, which is a ring of area $2pi d_j cdot frac12$ (radius of sphere is $frac12$) where $d_j$ is the width of the rectangle. Since the rectangles cover the circle, we conclude that the rings must cover the sphere. But the total area covered by the rings is $sum_{j=1}^n pi d_j < pi$ while the surface area of the sphere is $4pi (1/2)^2 = pi$.
– Daniel
Dec 1 at 3:11
@Daniel Since your comment answers it, maybe make it an answer? [I get it if you don't want to since it has the basic answer you mention in previous comment.]
– coffeemath
Dec 1 at 4:03