Professor Stewart's Hoard of Mathematical Treasures (20 page)

What happens for n boxes, with as many on each layer as you wish? (There is an even more general question, where the boxes can be tilted, but let’s restrict ourselves to layers, like the courses of a brick wall.)

You might like to try your hand at this puzzle before reading any further. What is the biggest overhang you can get with 5 or 6 boxes?

 

To avoid misunderstandings, let me make the conditions clear. All boxes are identical and uniform, and everything is idealised to exact rectangles and all the usual stuff we assume in Euclidean geometry. The problem is posed in the plane, because in three-dimensional space you could also rotate boxes, without violating the ‘layers’ condition. The arrangement must be in equilibrium: that is, if you work out all the forces that act on any box, they all balance each other out. Boxes must be arranged in layers, but you can leave gaps. And one other important condition: you do not have to be able to build the arrangement by adding one box at a time. Intermediate stages might topple if left unsupported. Only the final arrangement must be in equilibrium. (This equilibrium condition turns out not to be terribly intuitive; it can be turned into equations and checked by computer. When there aren’t too many boxes, though, it should be intuitive enough for you to tackle this puzzle.)

The answers for 4, 5 and 6 boxes were worked out by J. F. Hall
in 2005. In fact, he proposed some general patterns, and suggested that they should always maximise the overhang. But, in 2009, Mike Paterson and Uri Zwick showed that Hall’s stacks maximise the overhang only for 19 boxes or fewer (see page 297 for the reference). Finding exact arrangements with a lot of boxes is extremely complicated, but they proposed some near-optimal arrangements for up to 100 boxes.

One very interesting question is: how fast can the biggest overhang grow as the number n of boxes increases? For the classic ‘one box per layer’ solution, the answer is
H
_n
. There doesn’t seem to be a simple formula for this number, but H
_n
is very closely approximated by the natural logarithm log n. So the ‘asymptotic’ size of the largest overhang is
log
n
.

Paterson and Zwick proved that, when layers can contain many boxes, the maximal overhang is approximately proportional to the cube root of n. More precisely, there are constants c and C for which the maximal overhang always lies between
and
. They exhibited explicit arrangements with an overhang of at least
units, using what they call ‘parabolic stacks’. The picture shows such a stack with 111 boxes and an overhang of exactly 3 units. (The approximate formula gives only 2.50069 instead of 3 when n = 111, but it still gives the best-known overhang for very large n.)

Early in 2009, Peter Winkler, Yuval Peres and Mikkel Thorup joined the team, and took the question further. They proved that C is at most 6: the overhang can never be greater than
. Their proof uses the probability theory of ‘random walks’, in which a person takes a step forward or backward with specified probabilities. Each new brick spreads the forces that act in a similar manner to the way probabilities spread as a random walk proceeds.

A parabolic stack with 111 boxes, and overhang 3.

Alvin, Brenda and Casimir went to the pie-shop and bought three of PieThagoras’s world-famous perfectly circular mince pies. They bought one mini-pie with diameter 6 centimetres, one midi-pie with diameter 8 centimetres, and one maxi-pie with diameter 10 centimetres, because those were the only pies left.

Three pies.

They could have settled for one pie each, but they wanted to share the pies fairly. Now, as everyone knows, PieThagoras’s world-famous mince pies consist of two flat layers of pastry, of uniform thickness, with a uniform layer of mince sandwiched in between. The thicknesses of the pastry and the mince are the
same for all sizes of pie. So ‘fair’ means ‘having equal area’ when viewed from above as in my picture.

They decided that sharing the pies fairly would be quite complicated, and had just settled on dividing each pie separately into thirds when Desdemona turned up and demanded her fair share too. Fortunately they had not started cutting the pies. After some thought, they discovered that now they could divide the pies more easily, by cutting two of them into two pieces each, and leaving the third pie uncut. How?

 

Innumeratus’s attempt at a magic frame.

Innumeratus had taken the ace to 10 of diamonds from a pack of cards, and was arranging them to make a rectangular frame.

‘Look!’ he shouted to Mathophila. ‘I’ve arranged them so that the total number of pips along each side of the frame is the same!’

Mathophila had learned to take such statements with a pinch of salt, and she quickly pointed out that the sums concerned were 19 (top), 20 (left), 22 (right) and 16 (bottom).

‘Well, I’ve arranged them so that the total number of pips along each side of the frame is different, then.’