Monday Math Madness #26 Winner - Around the World
27 Feb 2009 Quan Quach 9 comments 461 views
MMM #26 Winner

The winner for this edition of MMM is Tom Mayo. This problem is probably one of the top five problems that I have enjoyed the most out of all the MMMs that we have done. I hope you guys enjoyed it as much as I did. A lot of people wonder where these problems originate from, and all I can say is that I hear about these problems mostly from friends and colleagues, as they also love to solve these little brain teasers and riddles. So for the most part, these problems are not created by us folks.
The number of submissions were about the same, but the comments for this problem were up. It looks like we need to do a better job promoting MMM. We’ll try to continue to present more problems like this, which will contrast nicely with what Sol does over at wildaboutmath.com.
The Answer by Robert LaKamWa
3 total planes (traveller included) are required to allow one airplane to travel all around the world.
In order to minimize planes, we want to make sure that we limit the number of planes that are distant from the island. This means we want the traveler to be alone in covering the far half of the earth. The question is how to get him there (the 25% point) with a full tank of gas and how to fuel him again when he comes back around (the 75% point).
The most efficient way to fuel him at the quarter-way point is to send 2 extra planes, sending one of them 1/8 of the way around the world, donating 1/2 of its fuel to the others (1/4 in each of the others, who would be at 3/4 tank), then returning to the island. Both remaining planes would have a full tank at the 12.5% point. When they reach the 25% point, they will each have a 3/4 tank again, so the remaining extra plane will donate 1/4 of its fuel, leaving it enough to return, while also fueling the traveler plane to a full tank.
(It is clear that this fueling is not possible with only 1 extra plane because when it reaches the 25% point, it will not have anything to donate if it wishes to return to the island.)
In order to meet the plane at the 75% point we will have to send 2 extra planes travelling in the opposite direction of the traveler plane. Send them 1/8 of the way around, then let one of them donate a quarter of a tank to the other one (who should be at 3/4 tank) then let the donor return to the island. When the other extra plane meets the traveller, he will have 3/4 tank and the traveller will be at empty. The extra plane can donate 1/4 tank to the traveller and still have enough to return. The traveller will be able to make it to the 87.5% point now without needing to refuel. Luckily, the plane that returned from 1/8 of the way around will have time to come back around and donate fuel again at this point, giving the traveller enough fuel to make the trip and for all planes to return safely. (Again, it requires 2 extra planes for this part of the trip because a single tank is not big enough to hold the extra fuel required for the return of the traveler and itself.)
This may seem like it requires 4 extra planes, in addition to the traveler. However, while the traveler is going around the other side of the world, the other planes have the time to return to the island, fuel up and go in the opposite direction to make the meeting trip. Thus, the 2 extra planes from the first half of the trip can be used on the second half of the trip. This means only 2 extra planes are required.
Therefore,
3 total planes (traveler included) are required to allow one airplane to travel all around the world.
Post-refill tank statuses are in square brackets. [A,B,C]
Waypoint 0 (0): Planes A,B,C leave the island clockwise. [100%,100%,100%]
Waypoint 1 (pi/4): Plane B donates one quarter-tank to each of the others so B and C have full tanks. Plane B begins to return. [100%,25%,100%]
Waypoint 2 (pi/2): Plane B arrives at island. Plane C donates a quarter-tank to A, giving C a full-tank. Plane C begins to return. [100%,100%,50%]
Waypoint 3 (3pi/4): Plane A continues its travels. Plane B is sitting at the island. Plane C is still en route to island. [75%,100%, 25%]
Waypoint 4 (pi): Plane A is on other side of world from island. Plane C arrives at island. Planes B,C leave island counter-clockwise. [50%,100%, 100%]
Waypoint 5 (5pi/4): Plane B gives one quarter-tank to C so C has a full tank. Plane B begins to return. [25%,50%, 100%]
Waypoint 6 (3pi/2): Plane C gives a quarter-tank to Plane A, begins to return. Plane B arrives at island, immediately fuels up and leaves. [25%,100%, 50%]
Waypoint 7 (7pi/4): Plane B meets A and C, gives a quarter-tank to A, then all return. [25%, 50%, 25%]
Waypoint 8 (2pi): All planes arrive at island. [100%,100%,100%]
9 Responses to “Monday Math Madness #26 Winner - Around the World”
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Excellent. As an engineer, and with the dearth of planes, I tried to minimize trip tip, and came up with four planes. My seven year old son came up with 5 planes because he didn’t recognize that one of the planes could be used twice in this solution. I’ll diagram and blog on your solution.
Dearth? Did I say dearth of planes? I meant the opposite….abundance.
[...] Carnival of Mathematics is up at The Endeavor. And Blinkdagger has posted the answer to the last Monday Math Madness. Look for a new puzzle next Monday at Wild About [...]
I found the same solution as you did. I also sent a MATLAB-made graph of the planes position and fuel level. You can use it if you want.
David
[...] MMM #26 was a tough one. The answer: 3 planes suffice! Read more here. [...]
drats. I must have done my calculations wrong
Wouldn’t a complete answer also be required to show that two planes are insufficient to achieve the task?
Clueless
I have to find out the solution of similar puzzle and difference is that in my case plane have the fuel for 1/3rd of earth instead of 1/2 of the earth distance.
Please provide me the solution for the puzzle if plane have fuel to cover 1/3 of the full round instead of 1/2.
[...] MMM #26 was a tough one. The answer: 3 planes suffice! Read more here. [...]