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.
The area of this square is one, and therefore very low. It is easy to find a better shape.
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 = C2 – 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