I'm interested in a generalization to the following riddle:
You're the director at an airport on the equator, and each plane at your airport can hold enough fuel to fly half way around the world. Your planes can also meet up in midair to share fuel. Your task is to use as few planes as possible to successfully fly one plane around the world, but all planes that you use must safely land back at your airport. How many planes do you need?
The answer to the riddle is $3$ (you can try to figure this out, or do a quick search of the internet for a solution).
Here is my formulation of a generalization, which I have yet to solve and would love for the community here to think about.
Let $n$ be the number of planes at your airport. What is the minimum distance $d(n)$ each plane can fly on a full tank (measured as a proportion of the whole distance around the world) to ensure that it is possible to successfully fly one plane around the world with all planes landing safely at your airport?
Here is what I think I know so far. If $n=1$, we need the only plane to be able to fly all the way around the world, so $d(1)=1$. If I haven't made any mistakes, then I believe that $d(2)=\frac{3}{5}$. I also believe that $d(3)=\frac{1}{2}$, or in other words, the original riddle is "sharp" in some sense. I don't know $d(n)$ for $n\ge 4$, but I have reason to believe that
$$\lim_{n\to\infty}d(n)=\frac{1}{4}$$
so that if our planes can travel one fourth of the way around the world (or less) on a full tank, then we have no hope of flying one around the world no matter how many planes we have.
I intentionally leave out justification for this information about $d(n)$, because I don't want to pollute someone else's thinking, and I'd love independent confirmation.
For a further challenge, one could also define $f(D,n)$ to be the maximum total amount of fuel "left over" among $n$ planes that can travel a distance $D$ on a full tank upon successfully flying one plane around the world. Intuition might suggest that $f(d(n),n)=0$ (i.e. if we have any left-over fuel at the end of the flights, then our planes must be able to fly a longer distance than necessary), but I don't think this is true. In fact, I think that $f(\frac{1}{2},3)=\frac{1}{4}$, that is, there is a solution to the original riddle in which the planes will be left with enough fuel to still fly one plane one fourth of the way around the world.
I have more to say, but I'll end here because the question is already quite lengthy. Perhaps in a few days after giving the community time to think, and if my results haven't been verified (or proven incorrect), I'll post my justification.
Addition: As pointed out by Fimpellizieri below, I should state that I'm assuming planes all travel at the same constant speed at all times. The end of his answer provides a better solution to the case $n=3$ if we don't make these assumptions and rid ourselves of time constraints. This time-free problem is also very interesting to me.
Also, I'd like to note that this question (of which I was unaware when I originally posted) and its answers are relevant. In particular, Brian Trial's answers claim that $d(8)\le \frac{1}{3}$ and provide a great way to visualize the planes' trips graphically.