Trigonometry/For Enthusiasts/Hilbert's third Problem

The haberdasher's problem, created by Henry Dudeney.

The haberdasher's problem (right), proposed in 1907 by the puzzle composer Henry Dudeney is a dissection of a triangle to a square, with only three cuts.

Any two 2D polygons of the same area can be cut into the pieces so that one can be reassembled as the other. It is unusual to have a dissection as elegant as this. It is not hard to find dissections when you are not worried about the number of pieces.

Prove that we can dissect any two polygons of the same area into the same set of pieces
  • Notice that we can always cut polygons into triangles. We can use this fact so as to just concern ourselves with whether we can convert one triangle into another of the same area by dissection.
  • If you are stuck trying to convert one triangle into another, here is a possible solution. Only look it it after you've given the problem a try.


Hilbert's third problem is whether two tetrahedra of the same base area and height, and therefore the same volume, can be dissected into tetrahedra and reassembled one into the other. It is possible for some tetrahedra pairs, but not all. A very closely related problem is whether a cube can be cut up into a finite number of pieces and reassembled as a regular tetrahedron. It turns out that it can't.

To prove that we need some mathematical machinery.


Vector SpacesEdit

For this page we will be using concepts about vector spaces.

Vector Space over the RealsEdit

One way to treat 3D space mathematically is as 3 dimensional Euclidean space \mathbb{R}\times\mathbb{R}\times\mathbb{R} also written as \mathbb{R}^3. A point in space is represented by three numbers where each number is a real number like \displaystyle 3 or \displaystyle 6.18239013... or \displaystyle 1.414...

Distance calculation using Pythagoras

For \mathbb{R}^3 we can use the Pythagorean theorem to calculate distances - sum the squares of each coordinate and take the square root.

There are also formulas for working out angles between different directions, and for various kinds of transformation of position in 3D space, such as rotating about the origin. \mathbb{R}^3 is a space in which we have a lot of intuition about how things should be.

We don't have to restrict ourselves to three dimensions. We can have vector spaces over \mathbb{R}^{109}. Many formulas about vector spaces over \mathbb{R} work whether we are in 1,2,3 or some higher number of dimensions. In those cases it is more economical to write the results in terms of \mathbb{R}^n, where it is understood that we are free to choose what n should be. Even so, we generally use \mathbb{R}^3 and not \mathbb{R}^{109} for our mental pictures of what is going on.


Vector Space over the RationalsEdit

We can also form a vector space over the rationals, \mathbb{Q}.

\mathbb{Q} includes numbers like \displaystyle 1.3 (it can be expressed as a rational fraction \displaystyle 13/10) but does not include numbers like \sqrt{2} = 1.4142... or \displaystyle 6.18239013... we're assuming that the '...' means it continues with arbitrary numbers, 6.18239013... \notin \mathbb{Q}.

In \mathbb{Q}^3 a point with coordinates \displaystyle(4.5,-7.1,9/7) is allowed. A point with coordinates \displaystyle(9.2, 6.18239013..., 8.1) is not allowed since all three coordinates must be in \mathbb{Q}

The fact that \mathbb{Q} 'misses out some values' is going to be very important to us in showing that we can't dissect an arbitrary tetrahedron and reassemble to make a cube.

Whether we use \mathbb{R}^3 or \mathbb{Q}^3, we use the term 'point' for a position relative to the origin and vector for a relative displacement between two points. We can add vectors. We can subtract vectors. We can multiply a vector by a number, provided the number is of the correct kind (in \mathbb{R} or \mathbb{Q} respectively).


For investigating Hilbert's third problem, we will need a vector space over \mathbb{Q}, the rational numbers. Numbers like \displaystyle 3/4 and \displaystyle -1.125 are allowed. Numbers like \displaystyle 6.18239013... are not.


Quotient SpaceEdit

Explain what a quotient space is


InvariantEdit

PolyominosEdit

Polyominos

A very useful tool for showing that one figure cannot be subdivided up a particular way is the idea of an invariant. They can be used for example in polyomino problems to prove impossibility.

Here is a simple dissection or subdivision puzzle where an invariant shows why a solution is impossible.

The board to be tiled
A first attempt at a tiling
A second attempt at a tiling

In this puzzle the objective is to tile the chosen area with 23 dominos, polyominos with just two squares.

It can't be done.


Why the tiling can't work (spoiler)

Spoiler Warning: There is a simple and elegant proof which we are about to give here. Stop reading if you want to have a try yourself first. If not, keep going. If we colour the board as in a chess board, say with black in the top left, then there are two more black squares than white squares. However each domino covers one white square and one black square.

The difference between the number of black squares covered and the number of white squares covered is an invariant. It starts off on an empty board as zero. As we place more domino tiles the invariant is unchanged. It really is an invariant. It starts off as zero and it stays at zero. However for all squares to be covered it needs to reach two - a contradiction.


Other InvariantsEdit

In the polyominos problem the invariant is just a number. To solve Hilbert's third problem we need a more complex invariant. We need something that captures information about angle and about length at the same time. You might think a complex number would fit the bill, but we actually need more because there is more than one length involved. We construct a domain for our invariant to come from in a way tailor made to the problem we are solving. The domain isn't R, it isn't C, instead it's a rather strange mathematical object that we construct to have just the properties we need.

Tensor ProductEdit

Explain what a tensor product is

Dehn InvariantEdit

\operatorname{D}(P) = \sum_{e} \ell(e)\otimes (\theta(e)+\mathbb{Q}\pi)

It is invariantEdit

Show that, because we've constructed it in the way we have, it does not change when we cut a polytope into pieces.

Last modified on 4 January 2011, at 19:52