Solution
Around the World by Relay
Show the problem again
An airport sits on a great circle around a small planet. Every plane has a tank that holds exactly enough fuel to fly half the way around, flies at the same constant speed, may fly in either direction and turn round at any point, and burns fuel at a constant rate whenever it is airborne. Fuel can be passed between two planes instantly when they are at the same point, any amount up to tank capacity. The only place to land or take on fuel from the ground is the airport, where planes may wait and may refuel instantly, as often as they like. A tank may become empty only at the very instant a plane lands or receives fuel; no plane may ever fly on an empty tank. Every plane must be back at the airport at the end. You want one plane to fly once all the way around the planet, and you may schedule the flights of the others freely. What is the smallest number of planes (including the one that circles the world) that makes this possible?
Worked solution
The answer is 3. Measure distance and fuel in eighths of the circle, so a tank holds 4, and let P circle clockwise with helpers Q and R. All three take off together. At 1/8, Q gives R 1 unit (P 3, Q 2, R 4) and Q turns home, landing with 1 to spare and refuelling. At 2/8, R tops P up to a full 4 and turns home with 1. Q, out again, meets R at 1/8 just as R runs dry and gives it 2; both land. P flies on its full tank and reaches 6/8 exactly empty, where R, having taken off anticlockwise with a full tank, meets it holding 2 and gives it 1. At 7/8, Q, also out anticlockwise with a full tank, meets both and gives each 1, so all three reach the airport together. Two planes are not enough. Call the circling plane P and the helper Q, and let a full tank be half the circumference. Any flight of Q that gets at least a quarter of the way from the airport lasts at least half the circumference (a quarter out, a quarter back), so Q by itself burns a whole tank on it. Fuel enters the air only at the airport, so during such a flight P and Q together burn Q's whole tank plus everything P flies: P must already hold, at Q's take-off, enough fuel for all of its own flying until Q lands. Such a flight can lend P fuel temporarily but never extends how far P can go. Now let A and B be the points a quarter and three quarters of the way round, and take the last moment P is at A before it first reaches B. From then until B, P stays at least a quarter from the airport and flies at least half the circumference, so only such useless flights of Q can meet it. Hence P needs a full tank at A (or, if the flight that meets it took off earlier, more than a full tank at that take-off, which is impossible). Since P has been burning fuel, a full tank at A means a top-up at A at that very instant, by a flight of Q that reached A; that flight is useless too, so P would have needed more than a full tank when it took off. So 2 planes fail, 1 plane obviously fails, and the answer is 3.
Source: Classic around-the-world airplane puzzle in wide circulation; no traceable original source. Statement written for AxiomIQ.