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.
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.