Problem
On Bagshot Island, there is an airport. The airport is the homebase 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?
Notes:
(a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.
(b) Each airplane must have enough fuel to return to airport.
(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)
(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)
Solution
The fewest number of airplane is 3!Imagine 3 airplane: A (black), B (red) and C (green). A is going to fly round the world. All three aircraft start at the same time in the same direction.
Check out the pics (where blue arrow shows fuel transfer):
After 1/6 of the earth's circumference, C passes 1/3 of its fuel to B and returns home, where it is refueled and starts immediately again to tail A and B.
A = 2/3 fuel left
B = 2/3 (fuel left) + 1/3 (refueled by C) = Full tank
C = 2/3 (fuel left) - 1/3 (refueling B) = 1/3 fuel left
Then C backs to the airport with 1/3 fuel left.
Now A and B go to the 1/4 earth's circumference, meaning that they're gonna spend another 1/6 fuel in their tanks to reach that point. At this point, B transfer fuel to A until A's tank is full.
A = 1/2 (fuel left) + 1/2 (refueled by B) = Full tank
B = 5/6 (fuel left) - 1/2 (refueling A) = 1/3 fuel left.
After that B returns to 1/6 circumference, which consumes B fuel until only 1/6 tank left and doesn't enough to bring it back to the airport.
Luckily, C already departed at the point with fuel of 2/3 tank and now awaits to refuel B.
B and C then return to the airport with 1/3 fuel each.
A (now with full tank) can go 1/2 world's length, meaning it will reach 3/4 earth's distance.
With the same scheme as above, B now goes to the opposite direction from the airport and awaits A at point 3/4 circumference with 5/6 tank of fuel. A who runs out of fuel is saved by B and received 1/2 tank of fuel.
A = 0 (fuel left) + 1/2 (refueled by B) = 1/2 fuel left.
B = 5/6 (fuel left) - 1/2 (refueling A) = 1/3 fuel left.
A now has the enough fuel to go straight home to the airport.
B (with only 1/3 fuel left on the tank) then refueled by C at 5/6 circumference, and both plane now have enough fuel to go back to the airport.
References
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