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Posted 25 April 2008 - 02:50 AM
Posted 02 May 2008 - 03:33 PM
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.
Posted 07 August 2008 - 07:24 PM
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.
Posted 19 August 2008 - 05:42 PM
I'm sorry for my poor explaining skills, I'm bad at this.
Edit: Added the apology (I think it is necessary).
Edited by Trogdor, 19 August 2008 - 05:43 PM.
Posted 30 October 2008 - 02:36 PM
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.
Posted 17 October 2009 - 01:31 AM
The plain model of the planet
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.
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