Trees On A Complex Plain

Here's an interesting old problem.  Pirates sail to a deserted island
to bury their treasure, and they decide to place it in a spot that
they can find later without using a map.  They mark three trees A,
B, and C, and then from B they pace off the distance to A, turn left
90 degrees and pace off the same distance, at which point they place
a stake.  Then they go back to B and pace off the distance to C, turn
right 90 degrees, and place another stake at this location.  They
bury the treasure mid-way between the two stakes.  Then they remove
the stakes and sail away.  Unfortunately, when they return to dig up
the treasure they discover that a storm has washes away the "B" tree.
The problem is to locate the treasure.

Say the left tree is at 0 and right tree is at 1 in the complex plane.  
The central tree is at some arbitrary point Ae^(ai), so it's position 
relative to the righthand tree is Be^(bi) = 1 - Ae^(ai).  The lefthand 
stake is placed at Ae^((a-pi/2)i), and the righthand stake is placed 
at 1 + Be^((b-pi/2)i).  The midpoint between the stakes is the average 
of these two points, given by

   (1/2) [ Ae^((a-pi/2)i) + 1 + Be^((b-pi/2)i) ]
             
      =  (1/2) [ 1 + {Ae^(ai) + Be^(bi)} e^(-i*pi/2) ]

Note that the quantity in the curly brackets is just 1 (by the 
definition of Be^(bi)), so the location of the midpoint is simply

               [ 1 + e^(-i*pi/2) ] / 2

regardless of the location of the central tree.  Therefore, starting 
at the "left" tree, walk halfway to the right tree, make a 90 degree 
right turn, and walk the same distance to the treasure.

Of course there's also a "test-taker's solution" to this problem.  
If we assume the problem has a unique answer, then it must give 
the same answer regardless of where the central tree is located.  
Hence we can place the central tree at any convenient point, such 
as the midpoint of the line connecting the left and right trees,
and locate the treasure.  This gives the same answer as above,
but it might not satisfy the traditional pirate's requirement for
mathematical rigor.  

One could probably create an entire sequence of problems like 
this, i.e., problems that have a trivial solution on the assumption 
that they actually have A solution.  For example, we might encode
the longitude and latitude of a treasure in the weight of golden
sphere with a cylindrical hole of length L=2s through its center.
We're not given the radius R of the sphere, but if we were, the 
volume of this object could be computed by integrating the 
incremental annular slices at a height y relative to the center.  
These slices have an area given by  pi [ (R^2 - y^2) - r^2) ] 
where r is the radius of the cylindrical hole, so we can integrate 
this function from y=-s to +s.  But of course r^2 = R^2 - s^2, so 
the parameter R drops out, and the volume integral is simply

                 s
                 /
          V = pi | (s^2 - y^2) dy    =  (4/3)pi s^3
                 /
               y=-s

The "test taker's solution of this problem is to reason that,
since we haven't been given the radius of the sphere, the answer
must be the same for any radius.  In particular, if we take R=s
then the cylindrical hole has zero diameter, and the volume of
the sphere of radius s is just (4/3)pi s^3.