Some people have expressed an interest in what I am teaching this year. Here it is.
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Jos’ Perfect Cuboid
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?
Survey of the Perfect Cuboid
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.
Statement of the Problem
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.
Integer sequence review: A006720
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 – October 2013
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.
An enneahedron for Herschel
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.
‘Low barrier, high ceiling’ and the Maths Arcade
I’ve been catching up with the TES Maths Podcast. I just listened to episode 7, towards the end of which guest Brian Arnold shares ‘the Frogs puzzle’. You probably know this, but if not Brian points to the NRICH interactive version which explains:
Imagine two red frogs and two blue frogs sitting on lily pads, with a spare lily pad in between them. Frogs can slide onto adjacent lily pads or jump over a frog; frogs can’t jump over more than one frog. Can we swap the red frogs with the blue frogs?
You know the one? You can play it with coins or counters or people. Anyway, host Craig Barton refers to this as “low barrier, high ceiling”, in that
anyone can do a few moves. So there’s your low barrier, but you can take that, the maths that that goes into! You can extend it to different numbers on either side, everything’s in there.
Much as I dislike the term because it sounds jargony, I realise it describes something I’ve been explaining all week.
Matt Parker’s Twitter Puzzle – 24th Sept
Matt’s at it again, posting puzzles on that Twitter:
Fun fact: you can arrange all the numbers from 1 to 17 so that each adjacent pair adds to a square number. Off you go! #mathspuzzle
— Matt Parker (@standupmaths) September 24, 2013
Plus a clarification:
To clarify: “each adjacent pair” means all but the end numbers are used twice. It’s every pair of neighbours as you go down the row.
— Matt Parker (@standupmaths) September 24, 2013
No spoilers in the comments please. Reply to Matt on Twitter!