Last week we had a crisis at work — we misplaced the key to the Maths Arcade cupboard, in which we store the games (don’t ask!). So I was on the look out for something to do without opening the cupboard — i.e. on pen and paper — and I turned to Twitter for help. What suggestions did I get? What did we do in our Emergency Maths Arcade? Read on.
The next issue of the Carnival of Mathematics, rounding up blog posts from the month of September, and compiled by Evelyn Lamb, is now online at Roots of Unity.
The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.
Some people have expressed an interest in what I am teaching this year. Here it is.
Inspired by our Open Season post on the Perfect Cuboid earlier this year, Aperiodical reader Jos Schouten wrote to us describing his work on the problem over the past 20 years. He’s looking for someone to help take his work further. Are you up to the challenge?
This article is about my search for the Perfect Cuboid (PC), which started exactly on Wednesday April 15, 1987. At that time I was a young engineer with feelings for mathematics, and employed to write C-language programs on a UNIX platform. Since then I’ve written software and explored ideas to find the cuboid, at work and at home. I still haven’t found one!
This article is also hoping to find someone in the world community as sparring partner, who likes the subject, wants to propose additional solution methods, and can help to implement such a method. The attempt will be to find a perfect cuboid with an odd side less than a googol.
The problem is easy to state, extremely difficult to solve, but a solution, once found, would be easy to verify.
The problem in words:
Find a cuboid whose sides are integer lengths, whose face diagonals are integers and whose space diagonal (from corner to opposite corner) is integral too.
A006720
Somos-4 sequence: $a(0)=a(1)=a(2)=a(3)=1$; for $n \geq 4$, $a(n)=(a(n-1)a(n-3)+a(n-2)^2)/a(n-4)$.1, 1, 1, 1, 2, 3, 7, 23, 59, 314, 1529, 8209, 83313, 620297, 7869898, 126742987, 1687054711, 47301104551, 1123424582771, 32606721084786, 1662315215971057, 61958046554226593, 4257998884448335457, 334806306946199122193, ...
Puzzlebomb is a monthly puzzle compendium. Issue 22 of Puzzlebomb, for October 2013, can be found here:
Puzzlebomb – Issue 22 – October 2013
The solutions to Issue 22 can be found here:
Puzzlebomb – Issue 22 – October 2013 – Solutions
Previous issues of Puzzlebomb, and their solutions, can be found here.
The building where I work is named after Alexander Stewart Herschel. I suspect this is because it used to be the home of the physics department, since he was an astronomer, but it works for us too because he also has a pretty cool graph named after him.
An embedding of the Herschel graph in the plane
Helpfully, it’s called the Herschel graph. It’s the smallest non-Hamiltonian polyhedral graph – you can’t draw a path on it that visits each vertex exactly once, but you can make a polyhedron whose vertices and edges correspond with the graph exactly. It’s also bipartite – you can colour the vertices using two colours so that edges only connect vertices of different colours. The graph’s automorphism group – its symmetries – is $D_6$, the symmetry group of the hexagon. That means that there’s threefold rotational symmetry, as well as a couple of lines of reflection. It’s hard to see the threefold symmetry in the usual diagram of the graph, but it’s there!
Anyway, at the start of the summer, one of the lecturers here, Dr Michael White, told me about this graph and asked if we could work out how to construct the corresponding polyhedron. Making a polyhedron is quite simple – take the diagram on the Wikipedia page, pinch the middle and pull up – but it would be really nice if you could make a polyhedron which has the same symmetries as the graph.