Following the above reply, for instance that every airplane has a gas capability of 1, which is sufficient for it to journey a distance of 1, which means the journey around the globe covers a distance of 10. Since every airplane has the flexibility to share its gas with any variety of different planes mid-flight, we will preserve observe of the gas capability and consumption of a whole fleet of planes, with out worrying about how a lot gas any particular person airplane has in its tank.

A fleet of $N$ planes has a complete gas capability of $N$, and makes use of $N$ gas per unit distance traveled. To decrease gas consumption, the scale of the fleet ought to at all times be sufficiently big to carry all of the accessible gas, however no greater. One approach to accomplish that is to have, at any given time, a single airplane utilizing its fuel-sharing magic to energy your complete fleet; as soon as that airplane is out of gas, it crash-lands and one other takes its place. Using this technique, an $N$-plane fully-fueled fleet (FFF) can cowl a distance of $frac{1}{N}$ earlier than one airplane crashes and it turns into an $(N-1)$-plane FFF. Generalizing to a number of steps, the minimal fleet dimension $N(d)$ that may cowl a distance $d$ is
$$N(d)=textrm{minimal }Ntextrm{ such that }sum_{i=1}^Nfrac{1}{i}geq d$$

If all of the planes needed to circle the globe one-way, our greatest answer can be $N(10)=12367$. But we will do higher by beginning off with one fleet sufficiently big to get many of the means around the globe, then sending a second fleet to satisfy it from the opposite route. If the fleets meet up at a ways $d_m$, we’ve two necessities: the preliminary fleet dimension should be sufficiently big to cowl $d_m$, and the second fleet dimension should be sufficiently big to cowl the remaining distance $10-d_m$ as a spherical journey. Summing these to get the whole variety of planes:
$$N_{textrm{complete}}=N(d_m) + Nbig(2(10-d_m)huge)$$

I do not know of a great way to unravel these inequalities analytically, however with some numerical trial and error, I discovered an answer with 840 planes complete. We ship out 514 planes, which fly till one airplane with a virtually empty gas tank reaches the rendezvous level at $dapprox6.82$.Another fleet comes from the opposite route with 326 planes, 13 of which attain the rendezvous level. The remaining 13 planes can barely make it the remainder of the way in which to $d=10$.