Professor Stewart's Hoard of Mathematical Treasures (55 page)

With five boxes, the maximal overhang is 1.30455. With six boxes, it is 1.4367. The stacks look like this:

The paper by Paterson and Zwick is ‘Overhang’, American Mathematical Monthly, vol. 116, January 2009, pp. 19-44.

The areas of the three pies are (mini) 9π, (midi) 16π, and (maxi) 25π, so the maxi-pie has the same area as the other two put together. If the maxi-pie is split in half, Alvin and Brenda can each have a piece (25π/2). That leaves the other two pies to share between Casimir and Desdemona. This can be done by slicing 7π/2 off the midi-pie and
giving that piece plus the mini-pie to Casimir (9π + 7π/2 = 25π/2). Desdemona gets the larger piece of the midi-pie (16π - 7π/2 = 25π/2).

There are lots of ways to slice the midi-pie. The traditional one is to place the mini-pie over the middle of the midi-pie, trace round half of its circumference, and then join the ends to the edge of the midi-pie. But you could slice off any shape with area 7π/2. And the cut that divides the maxi-pie in half need not be a diameter - it could be curved.

Cut the three pies as on the left, and share as on the right.

There are ten basically different answers, where, for example, swapping the ace and seven in the right-hand side doesn’t count as different since it’s a simple transformation that obviously keeps the sums the same. There are two with a sum of 18, four with 19, two with 20, and two with 22. Here’s one of them.

One of the ten answers: sums are all 18.

You can solve this by trial and error, or by listing all the possible states and moves and finding a path from the initial state to the desired final one. Here is one solution, which takes nine moves (two are combined in the second picture). There’s a shorter solution, which I’ll derive in a moment.

One way to divide the water.

However, there is a more systematic method, which seems to have first been published by M. C. K. Tweedie in 1939. It uses a grid of equilateral triangles, which engineers call isometric paper and mathematicians call trilinear coordinates.

Representing the possible states of the jugs.

Here the triples of numbers indicate how much each jar holds, in the order (3-litre jug, 5-litre jug, 8-litre jug). So for example 251 means that the 3-litre jug holds 2 litres, the 5-litre jug holds 5 litres, and the 8-litre jug holds 1 litre. If you look at the first number, then the lowest line in my diagram always starts 0, the line above starts 1, and so on. Similarly, the second number reads 0, 1, 2, 3, ... from left to right in each row. So the two arrows in the diagram are ‘coordinate axes’ for the amount in the 3-litre jug and the amount in the 5-litre jug.

What of the 8-litre jug? Because the total amount of water is always 8 litres, the third number is always determined by the first two. Just add them and subtract from 8. But there is a nice pattern here. Thanks to the geometry of isometric paper, the third number is constant along lines that slope up and to the left - that is, the third system of lines in the picture. For example, look at the line through 035, 125, 215, 305.

If we represent the possible ‘states’ of the jugs - how much each contains - in this way, then the allowable moves from any state to any other take on a simple geometric form, as I’ll now explain.

First, observe that the states in which some jug is either full or empty - which are the states allowed by the moves - are precisely those on the boundary of the picture.

The allowable moves, starting from some state (necessarily on the boundary) amount to moving along a line until you next hit the boundary. If you start at a corner, and move along the boundary (say from 008 to 053), then you can’t stop part way along, but have to go all the way to the next corner.

One solution.

So we can solve the puzzle by starting from 008 (lower left corner) and bouncing around the parallelogram like a billiard ball. The arrowed path shows what happens: we visit the states
and notice that this is the final state we want, so we stop.

This is precisely the solution given above. But there’s another one:

Now the sequence is
which uses seven moves instead of nine. You may have found that one instead.

How to ensure the greatest mat.

Remember, the avatars ‘watch over’ every other cushion. Otherwise you could make the mat bigger.

Tuck hit the outer ring (light grey), while Robin hit the three inner rings (dark grey).

The rings Robin and Tuck hit.

The areas of successive circles are π
r
²
, where
r
= 1, 2, 3, 4, 5, respectively: