Not too far away from earth, a new planet has been discovered. It is
called Vee-Ess, and seems to be exclusively inhabited by VS models. An
expedition is sent there, and ACD is dropped together with his team at
the north pole of that planet.

They have three airplanes (called Harem 1, Harem 2 and Harem 3) and
plenty of fuel, but they discover two problems: For the time being,
the only place they can start or land is at their base at the north
pole, but the VS models all live at the south pole, where the sunny
beaches are.

In order to establish contact, they have to fly over the south pole
with at least one airplane. A fully fuelled airplane can only fly the
distance from the north pole to the south pole. However, the planes
can exchange fuel in mid-air. All planes fly at a constant speed of
1000 mph.

So you have to find a way to fly over the south pole with at least one
plane, and bring back all three planes safely to the north pole. The
exchange of fuel and the starting/landing can be done in no time, so
you don't have to worry about time used for that.

Hello,

Here is the answer, found after a few trials.

(Sorry for the english, i'm not a native english speaker)

There is a table below describing the positions and content of the
fuel tank for each of the planes.

The rules are as follow:

The position is given in %, on a path from the north pole to the south
pole, 0% being the north pole, 100% being the south pole, 50% being in
between.  The fuel content of the tank is also in %, 0% meaning empty,
100% meaning full.
Two planes can only tranfert fuel when at the same position (indicated
in the "action" column by a "T") and this action is instantaneous.  A
plane can only refuel at the "0%" position (indicated by a "R") and
this action is also instantaneous.  "M" is for movement.  When a plane
is a the 100% position, contact is established with the south pole
denizens.

Here is the table.

Harem 1               Harem 2                Harem 3            Action done  
Position  Fuel        Position  Fuel         Position  Fuel
0         100         0         100          0         100      Start
25        75          25        75           25        75       M
25        25          25        100          25        100      T1 to 2 & 3
0         0           50        75           50        75       M
0         100         50        50           50        100      R1, T2 to 3
25        75          25        25           75        75       M
25        25          25        75           75        75       T1 to 2
0         0           50        50           100       50       M, contact!
0         100         50        50           100       50       R1
50        50          0         0            50        0        M
50        25          0         100          50        25       R2, T1 to 3
25        0           25        75           25        0        M
25        25          25        25           25        25       T2 to 1 & 3
0         0           0         0            0         0        M, back to base



Maybe there is other good answers...

Wezel-ga

Here we go lets assume the diameter of the planet is 1000 miles
at time=0 all planes (h1, h2, h3) at north pole with gas full
at t=15 min - all planes 1/2 to equator with tanks 3/4 full
at t=15 - all planes 1/2 to equator (h1=F, h2=1/2, h3=3/4)
t=30 - h1 @ equator with tank=3/4, h2 home with tank full, h3 with tank = 1/2
t=30 - h1 @ equat & tank = full, h2 home & tank full, h3 at equator tank = 1/4
t=45 - h1 @ 1/2 to SP & tank 3/4, h2 1/4 to equator  tank 3/4, h3 1/2
to NP & tank E
t=45 - h1 still the same, h2 give h3 1/4 tank of gas
t=60 - h1 over SP 1/2 tank, h2 & h3 @ NP full tanks
t=75 - h1 1/4 to equator on south side & tank 1/4, h2 & h3 1/4 to
equator on north side and tanks 3/4 full
t=75 h1 stays the same, h3 gives h2 1/4 tanks so h2 is full and h3 has 1/2
t=90 - h1 @ equator empty, h2 @ equat 3/4 tank, h3 home tank full
t=90 - h1 @ equator tank 1/2, h2 @ equator tank 1/4, h3 home tank full
t=105- h1 @ 1/4 to NorthP tank 1/4, h2 1/4 to NP tank E, h3 1/4 from NP tank 3/4
t=105 - h1 @ 1/4 to NorthP tank 1/4, h2 1/4 to NP tank 1/4 h3 1/4 from NP tank 1/2
t=120 - all three back at NP, h1 and h2 on empty

I assume that they go to empty and refuel in the same instant

Here is one way.

Given: 
3 planes with plenty of fuel in their tanks dropped at the North Pole
(NP). Atleast one plane to go to the South Pole (SP). All 3 planes to
go back and land safely to the NP.


-----------------
(Not specifically mentioned), so, ....

1. (Initially, at the NP drops, all fuel tanks are full.) So I assume
the other two planes can fill up the tank of the one plane to go all
the way to the SP, and then these 2 planes still have enough fuel left
in their tanks to land at the NP.

2a. (There is fuel pump at the SP.) So I assume the chosen plane can
refuel to a full tank there at SP.

So the chosen plane flies to SP, make the necesarry contact there,
take pictures of some of the models at the beach or whatever, refuel,
then head back to the NP.

Easy.

-----
2b.) (The chosen plane cannot land and refuel at the SP. Which, maybe,
is the idea of the question as posted. That the chosen plane only fly
over, not land on, the SP and then head back to the NP with the fuel
still in its tank to start with.) So it is not easy afterall.

But,...if that is the case, how come there seems to be a fuel pump at
the far away and desolate NP and no fuel pump at the civilized SP? The
VS models only walk or fly with their wings? And/or why can't the
plane land at SP? Why fly over only? If the contact can be made only
when the plane is vertically over the SP, how is that possible? The
transmitters at SP and/or the receiver at the plane really that
handicapped? Unbelievable. Can't the message be sent and be receive at
least 1/4 the distance from SP to NP? How can they build super modern
planes like that and only walkie-talkies for their receivers?
Oh, the transmitters at the NP are the walkie-talkies? That figures.
If VS models have wings, why they like the beach? For diving for fish
away from the beach?
What is VS anyway?

More questions than an answer....

For simpliciy, assume that it takes an hour to fly from NP to SP. Then, 
the amount of fuel translates into travel time as follows:

1/6 = 10 min
1/4 = 15 min
1/3 = 20 min
1/2 = 30 min
2/3 = 40 min
3/4 = 45 min
5/6 = 50 min

Here is one possible solution:

min   H1 fuel  H2 fuel  H3 fuel  Action
===   =======  =======  =======  ===============================================
000   full     full     full     All ships leave NP (towards SP)

015   3/4      3/4      3/4      H3 gives both H1 and H2 1/4 of its fuel
      full     full     1/4      H3 heads back to NP

030   3/4      3/4               H2 gives 1/4 of its fuel to H1 
      full     1/2      full     H2 heads back to NP, H3 refuels at NP

060   1/2                        H1 reaches SP, H2 refuels at NP
      1/2      full     full     H1 heads back to NP, H2 and H3 leave NP

080   1/6      2/3      2/3      H3 gives 1/3 of its fuel to H2
      1/6      full     1/3      H3 heads back to NP

090   0        5/6      1/6      H1 and H2 meet, H2 gives 1/3 of its fuel to H1
      1/3      1/2      1/6      H1 and H2 head back to NP

100   1/6      1/3      full     H3 refuels and leaves NP

110   0        1/6      5/6      H3 meets H1 and H2, H3 gives 1/6 of its fuel to H1
      1/6      1/6      2/3      All ships head back to NP

120   0        0        1/2      Misson accomplished!


Let me know if you need more details.

Hello,

Basic Assumptions:  The planet is a sphere.
             A fully fueled plane can travel only half the distance of
any great circle  (equator).

Additional assumption: Take-off and landing require no time.
                       Making a U-turn requires no time.
                       Knowledge is gained about the time of journey. 


Solution.

1.  Three planes leave the North Pole.
2.  When there fuel reaches 25% of journey, each has 75% fuel.
Arbitrarily, one plane refuels the other two and returns base.
3. Two plane reach the equator, each has 75% fuel. Arbitrarily, one
plane refuels the other and returns base.  In the, meanwhile, the next
plane arrives at base.
4.  The fully fueled plane at equator continues for South pole, the
50% fueled plane returns to base.
5.  When the plane arrive at base, fully fuel, 100%, plane departs for
equator on other side of planet, next plane flies over South pole,50%.
6.  When planes reach equator,(25%,25%), third plane departs from
base, (Base on assumption that we now know time to reach equator).
7.  When at 25% of journey, plane fuel (25%,25%,25%).
8.  Planes return to base.

The maybe other results...

Fly all three airplanes to the halfway mark.

At the halfway mark, have two of the airplanes each give the third
airplane 25% of the maximum amount of fuel. At this point, one
airplane will have 100% fuel and the other two will have, 50%.

Send the fully-fueled plane to the south pole, and have the other two
go back to the north pole to refuel back to 100%.

The plane that reached the south pole will now be at 50%, enough to
reach the halfway mark. The other two planes will expend 50% of their
fuel to meet the third plane at the halfway point. They will again
each give the third plane 25% of the maximum amount of fuel. At this
point all three planes will have 50% of the maximum amount of fuel,
which they will use to return to the north pole.