A usenet thread from rec.puzzles, alt.math.recreational, sci.math in December 1998 recovered using www.dejanews.com
The question is in effect the first part of the KotaniKnuth Ant Problem , thought one of the threads raises the question in Jim Propp's Surface Distance Conjecture. Many of the suggested solutions are wrong, and because of the posting to three groups, some later responders may not have seen all earlier responses. (Some names have been changed to protect the guilty; all email addresses have been removed.)
Author: Barrie Snell
Date:12/19/1998
Latest from New Scientist.
I have a rectangular box with a Christmas present inside. The dimensions of the box are 2 units x 1 unit x 1 unit. Firmly attached to one corner of the box is a length of red ribbon. The other end of the ribbon (when swung every which way) will reach any point of the box. If the ribbon were (was) any shorter then this would not be possible.
To the nearest one thousandth of a unit, what is the length of the ribbon?
(A wrong solution which equals 1+ sqrt(5))
Author: Chris
Date: 12/19/19983.236 units
Author: Barrie Snell
Date: 12/19/1998Maybe Yes, Maybe No. The traditional reply is to give an inkling of the method. Please share it with the rest of us Chris.
This IS a competion question, but I (and others) am not interested in entering the competition. Please explain your method so that the rest of us can maybe learn some Math(s). Personally, I know a little bit of fundamental numbertheory, calculus, and so on and so on. I rely on clever people like yourself to give me some insights on problemsolving and the method(s) with which you arrived at a solution. Surely parts of your method can be applied to other problems??
Yours very sincerely
Barrie
(the intuitively obvious, but wrong solution)
Author: MD
Date: 12/20/1998____
/ /
A/____/ 
  /D
____/
B Cunfolding the box into a net
<2> 2 2
A ____ A ____ ____
  1    1
B____C or ________D
  B C
____ 1
DThe shortest surface distance to the opposite corner is 2 sqrt(2)
(again the intuitively obvious, but wrong solution)
Author: JB
Date: 12/20/1998sqrt(8)
2.828I think that you must unfold box and longest distance is ab opposite corners
a ______
 

  b 

 
______
(rejects the intuitively obvious, but still wrong)
Author: BA
Date: 12/20/1998Unfolding the box is certainly the way forward. But the mistake is in believeing the longest distance is to the opposite corner (this is sqrt(8)).
I believe it is 2/3 of the way along the bottom of the opposite face giving a distance of sqrt(85/9) or 3.073.
It is easy to prove that the point must be somewhere along the opposite bottom edge since any movement off this edge shortens the distance.
Let P be a point at a distance k from one corner of this bottom edge. We can find three useful routes to P. One of these can be discrded as always being longer that another.
Consider the other two routes and look at the squares of the length of the ribbon. (It is easier to focus on the squares  it make the notation easier too.)
Of the other two routes one increases with k, with a value of 9+k^2 and the other decreases with a value of 13+k^6k. The point we require is the intersection of these two functions. Solving gives k=2/3.
This gives the distance^2 as 85/9
(wrong...)
Author: Martin
Date: 12/20/1998Here's my guess (but I'm most probably wrong)
root 10 ( about 3.1623 )
Martin.
(...and corrected)
Author: Martin
Date: 12/26/1998Well... I was wrong. My answer, sqrt(10) was at least better than the sqrt(8) others have proposed, in that a ribbon of length root ten would reach any point on the box!
I'm now convinced that the answer is sqrt(65 / 8) or about 2.85044. I think Ilan Mayer was the first to post this.
(a correct solution)
Author: ilan
Date: 12/20/1998The farthest point from a corner is on the opposite 1x1 face at (0.25, 0.25), where (0, 0) indicates the corner farthest away from the original corner. The distance to this point along any straight path is sqrt(8.125) = 2.8504.
__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\

(a correct solution, which goes on to ask the second part of the KotaniKnuth Ant Problem)
Author: David M Einstein
Date: 12/20/1998\sqrt(65/8) I believe.
Where must we attach the ribbon to make the distance from the point of attachment to the farthest point as large as possible?
Will the most distant pair of points be symmetrically opposite each other?
Author: BS
Date: 12/21/1998Can you generalize your solution to any other color of ribbon?
(a summary of previous responses, though missing Ilan Mayer's, and plumping for a wrong one)
Author: Barrie Snell
Date: 12/21/1998Well it all seems to be a matter of interpretation doesn't it?
Unfolding the box is necessary to visualise the problem, but there are lots of ways to draw some rectangles with attached squares in the plane so that when folded and glued, they form the required box. Personally I'm of the opinion that the longest distance to be reached is from the attached corner to the diametrically opposite corner (though this could possibly be in error).
For those who only look at 1 newsgroup, here is a summary of all the different replies received so far (20 Dec 98) from <sci.math>, <rec.puzzles>, and <alt.math.recreational>.
BA gave 3.073 = sqrt(85/9)
David Einstein gave 2.850 = sqrt(65/8)
Chris gave 3.236 = 1+ sqrt(5), and the path was exactly diagonally across the long face then vertically (or horizontally) down the 1 unit straight edge.
JB & MD gave 2.828 = sqrt(8), and the path was across the long face, and continuing across the adjoining long face, to the opposite corner from the starting one.
Martin gave 3.162 = sqrt(10), and the path was across the long face over to the 1 unit side, then continuing across the square face to the required opposite corner.So who is correct? I agree with 3.162 = sqrt(10) as being the optimal length of ribbon, because anything less than this will not reach the corner by the above corresponding path, and anything more than this is superfluous.
(correcting the error, and extending the question towards Jim Propp's Surface Distance Conjecture)
Author: David M Einstein
Date: 12/21/1998It is [in error]. The diametrically opposite corner is \sqrt(8) away (across two 2x1 squares) and all the points on the 2x1 faces and the adjacent 1x1 face are all within of the corner. Unfortunately, the opposite 1x1 face has points more than sqrt(8) away and the point (1/4,1/4) away from the opposite corner is furthest away. You must consider many possible unfoldings since points close to each other may take very different paths.
The fact that the opposite corner is not the most distant point is very counterintuitive (at least to me), which is why I am curious as to whether or not the most distant pair of points on a box are opposite each other, I believe that they are but to show it for a general box will be messy, and will put it off till I have more time for grinchliness.
If it is true for the box is it true for all centrally symettric solids? how would one even start to prove this?
(explaining why Jim Propp's Surface Distance Conjecture requires some kind of convexity requirement)
Author: George Russell
Date: 12/21/1998It can't be true for all centrally symmetric solids. Imagine a polyhedron in the shape of a torus with very ridged faces making travel from the outer circle to the inner circle very expensive; then the most distant pair of points would have to be on the outer and inner circle.
Perhaps it might be true for centrally symmetric solids which are also convex.
(a correct solution, though misses that the result (1/4)*sqrt(130) = sqrt(65/8) has already been given)
Author: Ken Starks
Date: 12/22/1998AS it happens, my solution os different from all those you quote. It is also smaller than all of them apart from sqrt(8) which makes the error of assuming that if the ribbon can reach the 'far' corner it can reach every point on the surface of the parcel.

Let the faces of the parcel be:
North 1 x 1
South 1 x 1
East 2 x 1
West 2 x 1
Top 2 x 1
Bottom 2 x 1Give the North face a coordinate system with the origin at BottomEast, and point P = ( x,y)
What is the 'ribbondistance' from P to Q, the TopWestSouth corner ?

Rather than taking a purely geometric approach, let us take a geometrical one.
We can label and define the possible routes. according to the faces they cross, as follows:
A North / East / Top
B North / Top
C North / West
D North / Bottom / West
E North / Bottom / South
F North / Esat / SouthIt is not difficult to draw a plane diagram with the 'North' face as previously defined, and six points ABCDEF at the correct distances from variable point P. ( eg distance PA = 'ribbon distance' from P to Q using route A.
For example A = ( 2, 2 )
B = ( 1, 3 )
C = ( 3, 1 )
D = ( 2, 2)
E = ( 1, 3)
F = ( 3, 1)Points E and F can be quickly discounted as too far. Each of the other routes is 'shortest' for a certain region of the 'North' face.
As it happens, for this particular cuboid, points ABCD is a cyclic quadrilateral, and the radius of its circumcircle is the length of ribbon needed.
In fact the circumcentre is at the point ( 1/4, 1/4 ) and the radius R
R = Pythag( 1.75, 2.25 ) = (1/4) * sqrt(130) = 2.850 ( 3 decimal places)

The general case with a parcel of dimensions a,b,c is much harder, because you don't get a cyclic quadrilateral. You have to find the smallest circle to cover a noncyclic quadrilateral.
Ken, __O
_\<,_
(_)/ (_) Virtuale Saluton.
(again a correct result)
Author: Andreas Gieriet
Date: 12/25/1998I get the same solution:
R = SQRT(65/8) = 2.850Approach:
1) the problem is symetrical, i.e., only half of the box has
to be considered
2) unfold the considered half of the box
3) let the ribbonend move along the far "boundary" AB+++A
  \
+++B
 
S++(B)r_1(u) = SQRT(2u^2  4u + 10) for 0 <= u <= 1
4) let the ribbonend move along the far "boundary" BA
y
^

++(A)
 
+++A
  /
S+++B > xr_2(u) = SQRT(2u^2 + 4u + 4) for 0 <= u <= 1
5) merge the two trajections
6) the maximum of the minimal ribbon length is at the intersection of the two trajectionsr_1(3/4) = r_2(3/4)
==> r = SQRT(65/8) = 2.850
==> the "farthest" point is on line AB, SQRT(2)/4 away from point A.Andreas Gieriet
(incorrect)
Author: BS
Date: 12/22/1998That's what I got [3.162 = sqrt(10)]. Assuming the ribbon to be of zero width, of course!
Author: Martin
Date: 12/22/1998Yeah, I like that! If the ribbon is wide enough, say greater than twice the square root of ten, then the length can be as small as you like! I suppose someone's going to ask about the thickness of the ribbon now! I assume the original question related to a ribbon of negligible width and thickness.
By the way, what's all this stuff in this thread about root 8 ???
Sometimes, it's like trying to push a piece of string uphill.
"The unnamed Martin"... aka Martin Round (I thought my email address would give it away)
(again a correct result)
Author: Rainer Thonnes
Date: 12/26/1998It is obvious that root 2 is enough to reach any point on the near small face, that root 5 is enough to reach any point on the near large faces, and that root 8 will get you to anywhere on the far large faces, including the diagonally opposite corner of the box. Unfortunately this length will fail to reach some parts of the far small face.
This is easy to visualise by unfolding the box in such a way that the four large faces remain parallel and attached to each other, forming a vertical 2x4 rectangle. Let the fixed end of the ribbon be at one of the Tjunctions on your left, and the far small face be attached to the right of each of the long faces in turn. It gets a bit complicated to imagine because as you move that square to a different rectangle you also have to rotate it.
To get an idea of the size and shape of the area of inaccessibility, it's a good idea to consider the far small face to be fixed in a cartesian xy coordinate system and to rotate the rest of the unfolded box around it. Let the far square be the unit square, and the origin be the corner of the box spacediagonally opposite the ribbon's fixed point. Then the nogo zone is a convex kite, symmetric about the line y=x, with its tail end at the origin. It is bounded by the arcs of four circles, all of radius root 8, with centres at (2,2), (1,3), (3,1), and (2,2).
Now, in order to reduce the area of this kite to zero, we simply need to beef up the radii until the arcs intersect at the "inaccessibility pole". Because of symmetry, this pole will lie on the y=x diagonal, and we can concentrate on just the first two circles. Now, given that their radii will remain equal, it is clear that the points at which the circles intersect will lie on the line which passes through the point midway between their centres, and perpendicular to the line joining the centres.
This midpoint has coordinates (1/2,5/2). The centrejoining line has slope 1/3, so the perpendicular has slope 3. This gives us an equation for the line as (y5/2) = 3*(x(1/2)). To find where it intersects y=x we substitute x for y and get: x5/2 = 3x3/2, or 4x=1, or x=y=1/4.
It remains only to work out the radius length, which is root((9/4)^2 + (7/4)^2) for the first circle, which is root(81+49)/4, and root((11/4)^2 + (3/4)^2) for the second circle, which is root(121+9)/4. Fortunately they give the same result of root(130)/4.
This is about 2.85044 and therefore the required answer is 2.851, and *not* 2.850; although the latter is the *nearest* whole thousandth to the exact answer, it will be too short, so we *must* round up rather than down.
Author: Martin
Date: 12/26/1998Nice explanation... but I think you mean a CONCAVE kite.
Martin.
Author: Rainer Thonnes
Date: 12/27/1998Heh, heh, just testing to see if anyone's paying attention.
OK, I goofed, you're quite right, I just picked the wrong word.
Author: AGa
Date: 12/22/1998Asuming the ribbon is of negligable thickness.. isn't the answer that the ribbon is infinitely long?
(again the intuitively obvious, but wrong solution)
Author: DS
Date: 12/23/19982*sqrt(2)
Other pages: Usenet thread "2x1x1 cuboid" Surface distances on a cube or cuboid