With one straight cut, and only the knife as a tool, halve a slice of wedding cake iced on top and on one...

You mean slicing _plane_, right?  Of course, if one assumes that the
knife is straight, then it's trivial to use it as a straightedge and
mark lines in the icing...

As stated, the only true solutions are ones which cut the piece in the
obvious way (cutting perpendicular to both icing faces, through the
middle of the cake piece).  In order for the cut to give the same amount
of top icing, it must pass through the center of the top face.  In
order for the cut to give the same amount of side icing, it must pass
through the center of that side face.  In order to give the same amount
of cake, the cut must pass through the center of the cake piece.  This
gives three non-colinear points which the cut plane must pass through,
so uniquely determines that plane.

I don't see any "elegant" ways to "construct" that plane at the moment.

-- Isaac Kuo <isaackuo@math.berkeley.edu>

[ Editor: no construction was among the collected postings. ]

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 20); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
You are a guest at a wedding and have been given the usual piece of
wedding cake - a rectangular slice with icing on one side and on the
top.  The icing on the top is a bit thicker than that on the side.

You want to cut it into two pieces, each with the same amount of cake
and the same amount of icing.  You must do so with one straight cut.

Note also that a description of the cut is not sufficient - you must
*construct* the slicing line.  Needless to say you don't have a
graduated ruler, protractor, compasses, etc. at the wedding feast -
only the knife edge.

-- Barry Harridge <barry@torps.apana.org.au>


Use the knife to mark the diagonals on the top and the side with the
icing, ending up with a cross on the top and a cross on the side.

Join the centre of the crosses and you have a line that divides the top
icing layer, the side icing layer, and the cake, in half.

All of this is on the assumption that the knife isn't bent and is long
enough to go from one corner to the opposite corner.

-- Pete Calvert <pcalvert@economics.adelaide.edu.au>

[ Editor: Barry Harridge's own solution, posted in April 1994 a month
after the problem, and two replies.  In the original the three steps
were separated by spoiler space. ]

I'm disappointed that in a whole month, no one has found the solution.
So here it is.  To give you one more chance for an AHA! experience, the
solution is gradually revealed below in 3 steps.

The Problem :

   F--------------------QG
   |                  .  |
  D+----------------.-E  |    ABED(A) is a slice of cake with
   |              .   |  |    icing BCGFDE(B). Construct (with
   |            .     |  |    straight knife edge as the only
   |          .       |  |    geometrical instrument) a slicing
   |        .         |  |    line PQ which gives equal shares of
   |      .           |  |    cake and icing.
   |    .             |  |
   |  .               |  |
   AP-----------------B--C

Solution:

1 Any straight line which passes through the centre of the rectangle
  will halve the area of the rectangle.

2 To halve the cake, the line goes through R, the centre of ABED (found
  as the intersection of lines AE and BD).

3 To halve (cake + icing), the line goes through S, the centre of ACGF
  (found as the intersection of AG and CF).

  Carefully put the knife edge along RS and you have the required
  slicing line.

-- Barry Harridge <barry@torps.apana.org.au>

> 3 To halve (cake + icing), the line goes through S, the centre of ACGF
>   (found as the intersection of AG and CF).

Excuse me?  How did you connect AG?  They are in a different plane.
And while F & C are in the same plane, there is cake in the way to
create this line.  Finally, I was under the impression the there was
only icing on one side and the top.  And further, even if you could
construct this line, I dont see where it is certain to divide the icing
in half, wasn't it of varying widths on different sides?

-- Scott Purdy <spurdy@pomona.claremont.edu>

Very clever, Barry!  I like it.  One thing you may have wanted to point
out in your diagram is that it is a SIDE view.  That confused me the
first time, and apparently Scott, too (see his post).

So, going back over your solution, what we do is tip the RECTANGULAR
slice of cake over on its side, so that we see on top a rectangle of
cake augmented on two adjacent sides (not necessarily by the same
amount) by some icing.  So, we make the two X's, to find the centroid
of the cake and the centroid (as far as volume is concerned) of
{cake+icing}, and cut on the vertical plane through them, thus giving
each piece half the volume of cake and half the volume of
{cake+icing}.

Note that one piece has icing mostly from the top of the cake, and the
other mostly from the side of the cake.

-- Joel Evan Rosenberg <joelr@tucson.Princeton.EDU>