Professor Stewart's Hoard of Mathematical Treasures by Ian Stewart

put the top one on the second one, so that its centre was poised exactly on the edge. So it poked outa unit. Then it was obvious that the centre of mass of the top two boxes was in the middle, so I placed them with the centre of mass exactly over the edge of the third box. If you do the sums, that makes it poke out anotherof a unit. Then I placed the three of them so that their combined centre of mass was right on the edge of the table, and that turned out to add a furtherof a unit to the overhang.’
    ‘And,’ Luigi said. ‘You’re right, it does poke out almost 1 unit.’
    Alert readers will observe that Angelina and Luigi are assuming the boxes are identical and they are uniform, that is,
the mass is evenly distributed. Real pizza boxes, full or empty, are not like that, but for this puzzle you should pretend that they are.
    ‘What happens if you add more boxes?’ Luigi asked.
    ‘I think the pattern continues. I could replace the table by a fourth box, and then slide the pile out until it is just about to topple, adding a furtherto the overhang. Then the top box does poke out over the edge of the table: the overhang is. And with more boxes still, I could do the same again, adding, and so on.’
    ‘So you’re saying,’ said Luigi, ‘that with n boxes you can get an overhang of
units. Which I instantly recognise as H n , where H n is the n th harmonic number:
    Isn’t that right?’
    Angelina agreed that it was. As you do.
    This is a time-honoured puzzle, and the biggest overhang you can get with n boxes using this method is indeed H n , so Angelina and Luigi are right. You can find the details nicely worked out in many sources, and I would have included them, but for one thing: this traditional answer is valid only with the extra assumption that exactly one box occurs at each level. And that raises a very interesting question: what happens without that assumption?
    In 1955, R. Sutton noticed that, even with just three boxes, you can do better than Angelina: an overhang of 1 instead of. With four boxes, the biggest possible overhang is

    Sutton discovered how to make the top one poke out 1 unit with three boxes

    With four boxes, the biggest overlap involves leaving a gap in the second layer.
    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?
     
    Answers on page 297
    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