On Bagshot Island, there is an airport. The airport is the home base of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel.
What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?
(1) Ignore extra fuel consumption as a result of acceleration, evaporation of fuel, bleeding-heart-liberal fiscal policies, etc.
(2) All the planes have to make it back safely, so you can’t give all your fuel away to another plane.
(3) Assume that refueling is an extremely fast process.
As per the puzzle given above The fewest number of aircraft is 3!
Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refueled and starts immediately again to follow A and C.
C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first ‘auxiliary’ aircraft reaches it in time in order to refuel it, and both ‘auxiliary’ aircraft are the able to return safely to the home base.
Now in the same manner as before both B and C fully refueled fly towards A. Again B refuels C and returns home to be refueled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refueling process is applied analogously as for the first phase of the flight.