Aeroplane

[Guest]

My initial thought on this was that it could simply be done with 2 planes. Remove gas tank of plane #2 and retrofit to plane #1.

[Guest]

This puzzle is very similar to the army trucks crossing the dessert puzzle but I like that added twist of it circumnavigating a planet so that planes can meet up with the original plane from the other side of the destination as well. Good puzzle rookie1. Maybe add the stipulation that planes can land wherever they need to refuel but there is no other plane fuel anywhere else on the planet. That way there's no confusion about whether or not the planes need to hover or burn fuel while waiting for refueling.

Nice job PAUL in explaining the solution with a detailed diagram.

[Guest]

that i kinda weird

[Guest]

I believe this is a solution (that doesn't require planes to hover):

Let the fraction of fuel-tank-filled for planes A, B, and C (respectively) be represented as:

[1, 1, 1]

<!-- ia0 -->airplane_puzzle.gif<!-- ia0 -->[/attachment:6a5ea]

1. All 3 planes go 1/4 the way toward the south pole. [3/4, 3/4, 3/4]

2. At that point plane C gives 1/4 tank to EACH of the other planes, leaving them full, and plane C with 1/4 tank to return to the north pole. [1, 1, 1/4]

3. At the equator, plane B gives plane A (the "full-circle plane") 1/4 tank, thus filling plane A; plane B has 1/2 tank left to return to the north pole. (Plane C arrives at airport) [1, 1/2, 1]

(Plane A now has enough fuel to pass the south pole and reach the equator on the other side.)

4. When plane B arrives at the airport, both B and C must instantly refuel and leave going the other direction. [1/2, 1, 1]

5. At 1/4 the way from the north pole, plane C gives plane B 1/4 tank, filling it up, while leaving itself with 1/2 tank to get back with (plenty). [1/4, 1, 1/2]

6. Plane B meets plane A at the equator as plane A is running out of fuel. Plane B, which has 3/4 tank left, gives half its fuel to plane A, leaving 3/8 tank in each plane. Plane C reaches the airport at this same time. [3/8, 3/8, 1]

7. Plane C instantly refuels and goes back to meet planes A and B at 1/4 the way from the north pole, with plenty of fuel for all three to return safely. [1/8, 1/8, 3/4] --> [1/3, 1/3, 1/3]

It sounds a bit messy, and I assume things happen instantly, but it works, doesn't it??

good explanation. i 2 solved the puzzle this way.

[Guest]

Couldnt you just fly all 3 to the south pole then have a C47 with enough fuel to fully resupply the planes at the s pole, leave the C47 at the s pole then fly the 3 that HAVE to make it around back up to the N pole. Easy, assuming the army doesnt care about C47s.

[Guest]

since theres lots of fuel at the airport... put lots of tanks of fuel on an airplane and the 3 pilots in that plane. 1 pilot pilots the plane while the other 2 refuel the plane. With all the fuel, the plane would be able to reach the south pole then go back up to the north pole. (it says nothing against carrying extra fuel tanks)

[Guest]

I love poking holes in the question. This was a good question.

In the original solution it's saying "thirds", but people (I) could assume you're cutting the world (circle) into thirds. It probably would have been better to say cut the world into sixths, number them from start (North Pole), and refer back to the numbers. The diagram certainly helped out. Good explanation.

[preflop]

I believe this is a solution (that doesn't require planes to hover):

Let the fraction of fuel-tank-filled for planes A, B, and C (respectively) be represented as:

[1, 1, 1]

1. All 3 planes go 1/4 the way toward the south pole. [3/4, 3/4, 3/4]

2. At that point plane C gives 1/4 tank to EACH of the other planes, leaving them full, and plane C with 1/4 tank to return to the north pole. [1, 1, 1/4]

3. At the equator, plane B gives plane A (the "full-circle plane") 1/4 tank, thus filling plane A; plane B has 1/2 tank left to return to the north pole. (Plane C arrives at airport) [1, 1/2, 1]

(Plane A now has enough fuel to pass the south pole and reach the equator on the other side.)

4. When plane B arrives at the airport, both B and C must instantly refuel and leave going the other direction. [1/2, 1, 1]

5. At 1/4 the way from the north pole, plane C gives plane B 1/4 tank, filling it up, while leaving itself with 1/2 tank to get back with (plenty). [1/4, 1, 1/2]

6. Plane B meets plane A at the equator as plane A is running out of fuel. Plane B, which has 3/4 tank left, gives half its fuel to plane A, leaving 3/8 tank in each plane. Plane C reaches the airport at this same time. [3/8, 3/8, 1]

7. Plane C instantly refuels and goes back to meet planes A and B at 1/4 the way from the north pole, with plenty of fuel for all three to return safely. [1/8, 1/8, 3/4] --> [1/3, 1/3, 1/3]

It sounds a bit messy, and I assume things happen instantly, but it works, doesn't it??

Nice solution but I would make one change to make it less messy.

At step 4 there is no need for B to leave with C. I would finish it like this

4. C leaves with a full tank of gas to meet A at the equator at that point C has 1/2 A has 0 we divide the fuel so both have 1/4.

5 C and A travel to the 1/4 point where they meet B who left the north pole with a full tank and now has 3/4. B gives 1/4 to both C and A leving all with 1/4 in their tanks which is enough to get them all home.

[Guest]

I haven't read all of the answers, but couldn't two airplanes go 1/4 of the earth, which would leave them with half and half fuel. One of them gives the other one all of their remaining fuel and it has full fuel again. Then it goes to 3/4 of the earth, and another plane travels from the north pole to the south pole but to where the plane that is already 3/4 from the complete way is (good thing the earth is round) and it gives it the remaining half of the fuel and it can travel the remaining 1/4 of the earth.

I'm sorry for my poor explaining skills, I'm bad at this.

Edit: Added the apology (I think it is necessary).

Edited August 19, 2008 by Trogdor

[Guest]

The suggested method by Paul could be adjusted slightly.

Take travel from North to South as 0 to 1 and return from South to North as 1 to 2. F and D denotes Fuel and Distance factors respectively.

When A travels past South Pole ( A > 1D ), then B could start and fly right to 1.5D (Equator) and fuel A with 0,25F, leaving 0.25F for plane B (which is not enough to return to base but enough to return to 1.75D) , enough for A and B to fly to 1.75D. At this refueling point, C could start to fly and meet both A and B at 1.75D, giving 0.25F to A and 0.25F to B leaving 0.25F to return ( as 0.25F was already used ), therefore, saving on Pauls method 0.5F fuel.

[Guest]

Actually it is possible to make it even for distance of 13/12 of globe (or flying distance of the planes equal to 6/13 of the way around planet) if planes can fly at 6/7 of their regular speed.

The plain model of the planet

Airport---A----B------------C----D---Airport.

Each "-" corresponds to one "Kmile" or 1000 miles

It is easier to show it if we assume distance to be 26 Kmiles and that plane can fly just 12K mile (less than half of the way)

3 planes starts together. At point A after 3 Kmile the planes redistribute fuel in a way 2 are full and first comes back having 1/4 of fuel. Then two other travel 4 Kmiles to point B. Then again one is full and another (second) has 1/3 to come back to point A while the first plane comes there to refuel the second plane. The third plane flies farther. The third travels 3+4+12=19 Kmiles until the moment his fuel is depleted (point C). That is 26-19=7 Kmiles away from North pole. It is the same distance as to the point B. It is possible to refuel the third plane at point C. First and second plane departure from airport together and first refuels second after 3 Kmiles at point D. Then the second flies 4 Kmile to the point C where it meets the third having 2/3 of fuel. The second gives 1/3 of fuel to the third so they can fly to point D where the first meets them with 3/4 of fuel and gives 1/4 to each so they all can reach airport.

The only problem of such an approach that the second plane has to travel 4+3+3+4=14 Kmiles to meet with the third while the third flies only 12 Kmiles. In order to avoid hovering we should put a condition that the third can fly at speed 12/14=6/7 of regular or assume possibility of hovering.

Actually it is not all. Now assume the distance is not 26 but 26.666 Kmiles. And actually we may move points B and C 0.33 Kmiles farther away from North pole. For example, While meeting the third plane, the first and the second ones can fly 4 miles together. The first fully refills second and is left with 1/3 to return. Travelling farther till point C the second spend only 3,333/12 of his fuel and is left with 1-3,333/12= 8,666/12 of fuel. Sharing with the third they both have fuel to fly 4,333 Kmiles. So they can reach point D where they will be met by the first.

And actually that further decreases speed of the third plane to 12/14.666=0.82 of normal speed.