# Unsolved Problems: Part I

There are many unsolved problems in mathematics. (Most too complicated for me to even understand, let alone explain in a blog post!) However, this week I am going to talk about 4 simple unsolved problems in mathematics in a two part series.

### Moving Sofa Problem

In 1966, mathematician Leo Moser posed the following mathematical problem:

“What is the shape of largest area in the plane that can be moved around a right-angled corner in a two-dimensional hallway of width 1?”

In other words, this problem asks what is the rigid two-dimensional shape with largest area A that can be moved through and L shaped region with unit width. It is an idealisation of real-life furniture moving problem, and thus the area A obtained is referred to as the sofa constant.

Let us look at few different examples.

Unit Square:

The area of this square is one, and therefore very low. It is easy to find a better shape.

Semi-Circle:

The area of this shape is greater than that of the unit square: it is ᴨ/2. Additionally, it is more interesting because it order for it to move around the corner, it must rotate, whereas the unit square merely translates.

Now, can we find a shape that can rotate and translate around the corner in order for the area to be maximised? Mathematician John Hammersley noticed that cutting a semicircle into two and filling the gap with a rectangular block would create a shape, which could be moved around the corner if only a smaller semicircular hole is also removed from the rectangular block. In doing so, he found a shape with A = 2/ᴨ+ᴨ/2 that can move around the L-shape.

Hammersley thought that this construction was optimal, but it turned out that it was not. In 1992, Joseph Gerver found a better shape with a slightly bigger area of around 2.2195.

Furthermore, Hammersley found an upper bound on the sofa constant, showing that it is at most , however the shape corresponding to this value has not been found.

### Perfect Cuboid Problem

This problem is an extension of the Pythagorean Theorem – A2 + B2 = C– into 3 dimensions.

In three dimensions there are four numbers, which correspond to A, B, C and G in the image above; A, B, C are the dimensions of the cuboid, and G is the diagonal running from one of the top corners to the opposite bottom corner. But, there are three more diagonals on the three surfaces, labelled D, E and F, which raises the question of whether there can be a box where all seven of these lengths are integers.

In other words, the goal is to find a cuboid such that A2 + B2 + C2 = G2 and where all 7 lengths are integers. Mathematicians have still not been able to find an example of such a cuboid, but they have also not been able to show that it is not possible. So, it remains unsolved.

Stay tuned for part 2 on Monday! M x