I have just watched this old riddle video from Ted-Ed. It basically asks how can three pilots coordinate their flights such that the professor (one of the three pilots) can circumnavigate the earth. I would recommend you watch the video before answering this question. The question is: If the planes were to only hold 90 kiloliters of fuel, how many pilots would they need to allow the professor to circumnavigate the earth. I do not know the number, nor do I know if it is even possible.

Summarised rules of the original problem:


And rules for this problem is as follows.

N pilots, one of them is our professor, want to circumnavigate the earth (360 longitudinal degrees) with the following conditions:

  1. Professors plane cannot turn/land (must be continuous)
  2. Each plane travels 1deg/min using 1kL fuel in the process. Each plane has 90kL fuel capacity.
  3. Refueling can be done in either at airport or between planes. Refueling is instant and the two planes must be at the same position for fuel transfer to happen.
  4. All refuelers can turn around instantly (but not the professor's, see rule 1)
  5. Refuelers cannot crash; all must land on the airport.

How big must N be for this to be possible?

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    $\begingroup$ Please provide enough specifications/rules from the video so that this puzzle is self-contained in the question. For example, we don't know how fuel restricts the pilots just by reading your question. $\endgroup$
    – bobble
    Commented Jun 14, 2021 at 4:42
  • 4
    $\begingroup$ Could someone please transcribe the image text? Text-as-images damage accessibility, as well as make something harder to search for. $\endgroup$
    – bobble
    Commented Jun 15, 2021 at 4:16
  • $\begingroup$ @bobble I just transcribed and adapted to fit the proble, $\endgroup$
    – Ankit
    Commented Jun 15, 2021 at 5:12
  • 2
    $\begingroup$ AFAIU, the problem in this question is well define, and the image with the rules of this original problem is just confusing stuff. Maybe you should just acknowledge that your problem is based on another one (with link) but remove unneeded details about it and give only the specifications that are relevant for the puzzle ? $\endgroup$
    – Evargalo
    Commented Jun 15, 2021 at 8:25

3 Answers 3


As pointed out by @loopy wait, this answer is wrong. Just leaving it up here for inspiration.

The answer is

11 refuelers

Corridor Property:

A plane can budget its fuel as follows: 18L for flying away from port, 36L donated to the next furthest plane, 18L for flying towards port. This creates a "corridor"... imagine a plane at 0deg, 2 at 36deg, 2 at 72 deg, etc. From here, 2 planes fly towards each other to 18deg, 2 planes to 54 deg, two planes to 90 etc. Then they fly back to previous positions. The fuel is given by the previous plane in the loop. It should look like the picture below: enter image description here As you could calculate, the planes are only using 72L of their tank. This allows them to carry 18L, which it can use to escort another plane and donate 18L when turning back 18deg later. Hence you can fly the main plane 18*numRefuelers degrees.

Sustainance property:

It's basically the same thing as before, but without having to escort the main plane, the planes can seperate 22.5 degrees instead of 18deg, allowing

Starting the main plane:

We are going to use 8 refuelers with the corridor property, but the 8th will leave the main plane at 135deg, so it can fly solo till 225.

Retriving the refuelers:

When the main , we have a refueler at 9,27,45,63,81,99,117,135. Since there is no longer an escort, the fuel can be proportioned as 22.5L West, 22.5L East, 45L donated, forming a sustainance loop, freeing up the closest 2 planes to land ASAP (because 22.5*6=135 in a similar loop to the corridor) plus one plane clear to land every 22.5 minutes.

Recieving the plane:

27 minutes used, 63 till the solo is complete. No plane can fly to -135deg from the airport in 63 minutes. Therefore, when the first western refueler lands (81 minutes left), 3 refuelers must be in sustainance in -22.5, -45 and -67.5. With each western refueler landing and takes off to the east, the 3 planes continue the sustainance paths, but the paths themselves move further to the east by 22.5 degrees. Once the range of furthest eastern refueler reaches 135, the planes on the East (of which there should be 6 now) should do "corridor". However, note that because there are 7 planes (the main plane has been recieved) and 6 half-full planes able to provide fuel, the system has 240 L of fuel, holding the sustainance paths for 240/7~34 minutes before the planes run out of extra fuel and the last plane must be ditched.

The final stretch:

However, there are still planes on the west that are landing. If we can get 8 planes on the East within the 34 minutes we could land everyone safely. The 6th plane landed 13.5 minutes before the main solo flight ended, leaving 9 minutes till a 7th and 31.5 minutes till an 8th plane can be put on the East. This is less than 34 so we are saved! Now, we have the same corridor system as was used to start the professor, except in reverse. This allows the professor to land, as well as all their associates.

I apologise this is really confusing to read. I will try to make an animation soon, but I am very busy with school atm.

  • 1
    $\begingroup$ I'm pretty sure I either misunderstand or the corridor idea does not work: Fuel burned grows linearly with the number of planes in the cascade while fuel entering the system is constant as there is only one plane leaving base per cycle. $\endgroup$
    – loopy walt
    Commented Jun 15, 2021 at 12:01
  • $\begingroup$ @loopywalt, you are incorrect; corridor system works by inputting 18L and removing 18L at the same time. Essentially, what is happening is that only 72L are being used out of the 90L, so the 18L remaining is given to the escortee after they both travel 18 degrees. $\endgroup$
    – Ankit
    Commented Jun 15, 2021 at 17:52
  • 1
    $\begingroup$ Then explain to me the following: In the picture there are 8 planes; just to keep them in the air 8kL need to be burned every minute. But only planes leaving from base at 0 degrees bring new fuel into the system, that is 90kL every 36 minutes or 2.5kL per minute. Why does the configuration not run out of fuel eventually? $\endgroup$
    – loopy walt
    Commented Jun 15, 2021 at 18:42

Temporarily scrapping this partial idea as I found a better solution, but will leave it here to inspire others.

Essentially the strategy is

the further away from the airport you are, the fewer planes should be flying, which means to let the main plane fly from 3/8 to 5/8 alone.

I say this because

The further you are from the airport, the more fuel it takes to reach you, and after a while you will need several degrees of refuelers to reach the multiple planes far away.

Working backwards is the best way to do this:

Step 1:

P1 flies alone from 3/8 to 5/8

Step 2:

P2 flies from 7/24 to 3/8 with P1 (both have full tanks), refils P1 and goes back to 7/24


I believe that it takes a minimum of

20 planes, 19 refuelers plus the circumnavigating pilot, to complete the trip.

To do this, we rely on a range of alternate refueling steps.

1) Five planes start together at the airport, fly 15 degrees outwards, one transfers 15 kl fuel to each of the other four and returns to the airport. The other four can continue on with full tanks.

2) One plane starts from the airport, flies 15 degrees out to meet four returning planes just at the point of running out of fuel. The first one transfers 15 kl to each of the other four planes and all five have just enough fuel to return to the airport.

3) Two planes starting at point A (with full tanks) fly 30 degrees to point B. At point B, one transfers 30 kl to the other plane that can now continue onwards with a full tank. The first plane can return to point A (where it arrives empty and needs partial re-fueling from yet another plane).

3) One plane starting at point A (with full tanks) fly 30 degrees to point B to meet one returning plane just at the point of running out of fuel. The first one transfers transfers 30 kl to the other and both can now return to point A (where they arrive empty and need partial re-fueling from another plane/planes).

Using these strategies, as illustrated in the fairly self-explanatory diagram below, the circumnavigator can complete a trip that includes flying a 90 degree solo leg with refueling at both ends of that section of the trip. I believe that this solution not only uses the minimum number of pilots, but also uses the minimum quantity of fuel. Note that in the diagram, one or more re-fueling events (as described above) occurs at every intersection of two 'flight-lines'.

enter image description here

  • 1
    $\begingroup$ That doesn't work. You have only 1 plane crossing the 120 line forward, splitting into 2 at 135. You need to start with 20 planes. $\endgroup$
    – Florian F
    Commented Jun 17, 2021 at 18:28
  • $\begingroup$ is the fix to the flaw that Florian found to just use twice as many planes at the 0:00 mark? $\endgroup$
    – rhavelka
    Commented Jun 17, 2021 at 18:59
  • $\begingroup$ You need to double the planes on the left edge. The number of planes on the right are correct. In the middle it is in between. $\endgroup$
    – Florian F
    Commented Jun 17, 2021 at 19:26
  • $\begingroup$ are the number of planes on the right correct? at the 3:45 mark there are two planes merging into 1 plane $\endgroup$
    – rhavelka
    Commented Jun 18, 2021 at 14:13
  • $\begingroup$ yes - i need to fix this when I get back to a computer with graphics... $\endgroup$
    – Penguino
    Commented Jun 19, 2021 at 7:22

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