Conference dypss1::brain_bogglers

    
    	Can the string be cut?
    
    	And is it longer than the perimeter of the original field?
    
    	And can we use any/all tools in the solution?
    

    Sell the piece of string and use the proceeds to hire a lawyer to
    dispute the boundary with your neighbours. Better hope it was a very
    long piece of string!
    
    It is not always possible to locate the field without further
    information. For instance, if the four posts are themselves at the
    corners of a square, then they could have been at the mid-points of the
    four boundary lines of a field _/2 times as wide, or they could each
    have been, say, 6 inches to the left of one of the corners of the
    original square, or anywhere in between.
    
    Other cases will take some more thought ...
    
    Andy.

    Some background. I got this from a newspaper and there it was given as
    a pencil and paper exercise. I reworded it to make it more of an
    open-ended practical exercise.
    
    I can think of a way to do it with a long piece of string that you may
    need to cut, or a shorter length of string and an assistant. But I
    don't want to make it any more precise, which might stifle lateral
    thinking on your part. Let's say the winner is the person who does it
    with the least "equipment".
    
    You have a handful of pegs, and you must end up with one staked in each
    corner of your field.
    
    Dick

    Andy
    
    Sorry, I replied without reading your reply. Consider that case as an
    opportunity as Lawyer/Mathematician to get an even bigger field than
    was intended.
    
    Dick

    
    Not positive about this, but wouldn't the string connecting the four
    gateposts into a four sided object be exactly half of the size of the
    perimeter of your property.  Thus establishing the size of the property
    and each side.  Then it would be a simple case of geometric proofs to
    determine the exact location of the property.  At least I think so...
    
    
    Hmm, on second thought it isn't half the size.  Put I'll bet there is a
    sqrt 2  involved.
    
    Marc
    

    
    	The length of string required to encircle all 4 posts is not cons-
    	istent.  It depends on the location of the post.
    
    	The only reason I know is that I tried it yesterday.
    

    	Andy (in .2) suggests that the problem may not be well-defined.
    This is true for one special case, but usually the answer is
    well-defined.
    
    	All I've got is some algebra, I'm afraid. I'm not very gifted in
    the geometric construction space.
    
    	Our four posts are ABCD. They form a convex quadrilateral. Join A
    to C, and B to D [using string, if you will :-)]. The two lines do
    meet, at X. Say the angles these two lines meet at are k+pi/2 and
    -k+pi/2, where k lies in [0,pi/2). We can draw two lines through X, one
    parallel to each of the undetermined walls of the field. These make
    angles p and q with AC and BD, where p and q are =< pi/4. Then:
    
    	|AC|cos(p) = |BD|cos(q) [= l, the side length of the square field.]
    
    	Now |p-q| = k, so we can rearrange the above as:
    
    	tan(q) = cot(k) - (|BD|/|AC|)cosec(k)
    
    which gives us a unique solution for q. Now, Andy's special case is
    where AC & BD are perpendicular, and the same length, and the cos
    equation does not give us any information.
    
    	Good luck with the geometry.
    
    Thanks,
    Andrew.

    Actually the geometrical construction is very simple and lends itself
    to a string-only method.
    
    Things you can do with string:
    
        1. Bisect a line.
        2. Construct a perpendicular from a point to a line.
        ...
    
    There, I've given you a clue.
    
    Dick

    1. Bisect AC at E
    2. Drop a perpendicular from E to BD at F
    3. Extend EF so that E is the mid-point of a line GH of length BD
    4. AG is a boundary of the field, so is CH.
    
    To prove this, use vectors. Perhaps someone can check for me?
    
    Thanks,
    Andrew.

I had a go at this problem, too.  My solution involved realizing that
where AC intersects BD isn't important.  Obviously the shortest AC or
BD can get is the length of a side of the square; and the longest is 
the length of a diagonal.  So if I have a square like this (break out 
your monospaced fonts!) 

        +--B-------+
        A          |
        |          C
        |          |
        X          |
        |          Y
        +----D-----+

then line AC should give the same answer if it was replaced by 
line XY.  (Intuitively this disproves the answer in .-1.  If I slide
line AC toward line XY then line AG would change and this shouldn't
happen.  Or, if AC is perpendicular to BD then where is point E?)

My solution involves translating line AC "up and to the right" so 
that point A is coincident with point B.  Now I have a trianglular
figure with point AB on one corner and points C and D on opposite
sides of the square.  I solve for this congruent square and then 
translate "down and to the left" to get the actual answer.  
To solve for the square:

       BA----------+
        |          C
        |          |
        S          |
        |          |
        |          |
        +-D--------+

Construct a perpendicular to BD from point C of length BD.  The
left endpoint of this line, point S, is on the left side of the square.
Join point S to point (AB).  Now construct a perpendicular from S(AB) to
point C. This line is the length of a side of the square and parallel
to to two of them. Constructing the square from here is straight forward.

Regards,
Chuck

    
    I've been thinking about this and it is more difficult than I first
    thought.  For example, none of the "proofs" in here that I've seen can
    handle the following:  ABCD (the four points on the sides) make a
    perfect square themselves.  Now there is TWO solutions.  Either they
    are the midpoints (bisects) of each side, or they are the corners.
    
    Marc
    

>    I've been thinking about this and it is more difficult than I first
>    thought.  For example, none of the "proofs" in here that I've seen can
>    handle the following:  ABCD (the four points on the sides) make a
>    perfect square themselves.  Now there is TWO solutions.  Either they
>    are the midpoints (bisects) of each side, or they are the corners.
    
Marc,

Sorry.  My answer in .10 isn't a proof at all, it is a construction.
Give me any four points and I'll construct a square which passes through
them using only pegs and a string.  Why doesn't it work?

For the degenerate case where ABCD form a square as given, please reread
.2.  There are more than two solutions: any square with a side length from AB 
to AC can be an answer.  I believe that my construction in .10 will give 
you the short-side answer where the resulting square has sides with
length AB.

Regards,
Chuck

1. From point C draw a line CS perpendicular to BD of length BD.
2. Point S and point A are on one side of the square.
3. The distance from point C to line CS is the length of sides of the square.

My original soluton (.10) had that translation business but that isn't 
really needed.

Chuck

Andrew,

I re-read .9 and I believe that you are correct.  In Step 2 you wanted
a perpendicular to line BD and not AC as I had thought.  Indeed AG
and CH are the desired sides.

Chuck

    
    Maybe I'm missing something here Chuck.  Explain to me how finding "A
    square" that fits the 4 posts tells you exactly where the property
    boundaries are located.  As far as I can tell it can't be done and .2
    (as you pointed out) has the only relevant piece of information in
    here.  We need a bare minimum of one other piece of information in
    order to solve the puzzle in the base note.
    
    I don't think we are disagreeing.
    
    Marc
    

    Re -.1
    
    	Andy Strangeways in .2 suggested that the problem was flawed, by
    presenting a case where there was no unique answer.
    
    	In a later reply, I showed that in most cases, there *is* a unique
    answer. The problem is ill-defined only in the case where the two
    diagonals meet at right-angles. Note that this is a superset of the
    original observation, that there was a problem if the four points form
    a square.
    
    	In a further reply, I gave a constructive solution, without proof.
    Of course, this constructive solution does not work in the pathological
    case. It gives one more confidence in the solution to see that it
    breaks down at this point.
    
    	However, there is a hole in my construction. It will not work in
    the special case where the two diagonals bisect one another. I don't see any
    obvious way to patch the construction to cover this case. This is a
    different issue from the pathology in the original problem statement. 
    
    	I have not yet studied the alternative construction proposed in
    the earlier notes, but it seems plausible to me that such an approach
    might work. It would not suffer from the minor defect present in my
    construction, since it takes a different approach.
    
    	It's actually very common in geometry problems for the basic
    argument to break down in certain special cases. Often, the problem
    statement excludes such possibilities, e.g. by specifying that "the
    points are all in general position". I think it's valuable to approach
    this problem in that spirit and not get hung up on the pathological
    case of orthogonal diagonals.
    
    Cheers,
    Andrew.  

Gosh, I was so slow. It's really very simple:

	Let E lie on BD such that AE is orthogonal to BD.
	Let F lie on AE such that |AF| = |BD| (& such that E is *between* A&F)
	Then CF lies on a boundary line of the field.

Once the first boundary line is established, it's easy to get the others,
and figure the size of the field. Note that the only way this algorithm fails
is if C&F are the same, ie, if AC & BD are orthogonal. Exactly as it should
be. This illustrates the proverb: "The exception proves the rule".

Cheers,
Andrew.

    Sorry I haven't revisited this for a while, but you have cleaned up
    nicely, Andrew, in <.17>. That is basically the solution I was looking for.
    
    For accuracy's sake you would drop the perpendicular from whichever of
    A or C was further from BD. I think we all realise how to drop the
    perpendicular with string only.
    
    Congrats
    
    Dick