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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.

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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. – D. Ungaretti Dec 01 '18 at 03:02
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    Searching on mathstackexchange I found essentially the argument on this post – D. Ungaretti Dec 01 '18 at 03:03
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    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 $2\pi d_j \cdot \frac12$ (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 $4\pi (1/2)^2 = \pi$. – D. Ungaretti Dec 01 '18 at 03:11
  • @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 01 '18 at 04:03
  • Cf. this post: math.stackexchange.com/questions/231051 – Rafi Mar 31 '23 at 13:05

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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 $$ A_j = 2\pi d_j \cdot \frac12 = \pi d_j, $$ 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.

D. Ungaretti
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