November’s MathsJam happened last night, and I’m totally pooped. We had a record twelve attendees and did absolutely loads of maths – so much that I ran out of space in my little notebook.

November MathsJam topics covered - christianp - Flickr

We were joined by Matt Parker (@standupmaths) who brought some of James Grime’s non-transitive dice (available from MathsGear) with him. They’re fun! Sadly, an attempt to demonstrate one die reliably beating another resulted in a draw. Undeterred, Matt did his best Ronco sales pitch about the many features of the dice. Maybe they should be called “Also!” dice.

We tried a few circle-geometry puzzles. The first one was called,

“The Pizza-Cutting Puzzle”

Can you cut a circle into congruent pieces so that at least one piece doesn’t touch the centre of the circle?

Here are some pictures to explain the kind of things that aren’t allowed:

IMAG0448-1 - christianp - Flickr

IMAG0447-1 - christianp - Flickr

The solution goes as follows, thanks to Eamonn:

  1. draw a circle using a set of compasses
  2. pick a point on the circle
  3. with the compasses set to the same width, put the pointy end on the point you picked and draw an arc connecting the centre of the circle to the edge
  4. put the pointy end where the arc you just drew touches the edge of the circle
  5. repeat steps 3 and 4 until you’ve gone all the way round the circle and divided it into six bits
  6. divide each bit in half along its line of symmetry

Now you’ve split the circle into twelve congruent pieces, of which six don’t touch the centre!

IMAG0451 - christianp - Flickr

It turns out that this solution has links to Reuleaux polygons! The solution I gave above uses Reuleaux triangles, but you can make other solutions by repeatedly tracing round just over half of any Reuleaux polygon, repeatedly. Matt Parker did this with a 50p piece (which is a Reuleaux heptagon)

The second circle puzzle didn’t have a name, I don’t think, so I’m going to call it

“The Unexpected Surd Identities Puzzle”

for reasons which will become apparent.

Inscribe four medium circles in a big circle so that they’re all tangent to each other and the big circle. Do the same thing for each of the four medium circles. Also draw as big a circle as you can in between the four medium circles. Is the circle in the middle bigger or smaller than the circles inside the medium circles?

IMAG0452 - christianp - Flickr

I got a solution to this one pretty quickly. It’s tempting to try to work everything out in terms of the radius of the big outer circle, but that’s making things hard. Instead, declare that the radius of the middle circles is $1$ unit.

Now, the square drawn by connecting the centres of the middle circles has sides of length $2$, because they’re two radii touching end-to-end.

IMAG0453 - christianp - Flickr

Straight away, using Pythagoras, you can work out that the length of the diagonals of the square is \[\sqrt{2^2+2^2} = \sqrt{8} = 2 \sqrt{2}.\]

The distance from the centre of the picture to the centre of one of the middle circles is half that, so $\sqrt{ 2 }$.

IMAG0454 - christianp - Flickr

If you carry on along that line to the edge of the big circle, you travel another unit, so the radius of the big circle is $\sqrt{2}+1$. That means that when you inscribe four circles in a bigger circle like we did, the ratio of the radius of the inscribed circles to the radius of the outer circle is $1:(1+\sqrt2)$.

IMAG0455 - christianp - Flickr

That means that the radius of the small circles inscribed in the medium circles is \[\frac{1}{1+\sqrt{2}}\]

Now we’re left with the small circle in the middle of the picture. We already know that the line from its centre to the centre of a medium circle has length $\sqrt2$, so if we take away the radius of the middle circle we get the radius of the small one, which is \[\sqrt2-1\]

IMAG0456 - christianp - Flickr

So which small circle is bigger? The answer is annoying: \[ \frac{1}{1+\sqrt{2}} \cdot \left( \frac{\sqrt{2}-1}{\sqrt{2}-1} \right) = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1. \]

They’re the same!

I’m getting a bit tired of writing out solutions to puzzles now, so here are just a few statements of puzzles:

  1. The answer is $21$. A rumour emerged that somebody had found $22$ but it appears to be baseless.
  2. Why are all squares of primes one more than a multiple of 24? Andrew Lobb had a wonderful proof of this, which he might post in the comments.
  3. What is the biggest integer with distinct digits whose English name has all words starting with the same letter? (American naming system allowed and, just this once, encouraged)
  4. Matthew (who isn’t Matt) found very short solutions to the “Princess in a castle” puzzle from last month: $2332$ solves the four-door puzzle, and $23223$ solves the five-door puzzle.

Andrew Lobb came up with a good question:

“What is the narrowest part of the UK?”

Everybody pretty much immediately said something like “the tip of Cornwall”, or some other jaggy bit on the edge. This wasn’t a very satisfying answer because you can pick arbitrarily narrow bits of land using that logic.

So it was agreed that a definition of narrowness was needed, and it would have to involve a compromise between the length of the line of narrowness and the ratio of the areas of the land on either side. I don’t think we got any further than that.

IMAG0458 - christianp - Flickr

NP-hard news

NP-hardness came up twice! First of all, David asked if, given advance knowledge of the pieces that will fall in a game of Tetris, whether it’s possible to work out a strategy that means you don’t lose. My encyclopaedic knowledge of esoteric mathematics gave the answer: it’s NP-hard.

As Matt was leaving, he mentioned that flipping pancakes is NP-hard.

Talking of hard problems: I mentioned the world’s shortest explanation of Gödel’s theorem.

IMAG0459 - christianp - Flickr

Prisoners in hats

Many, many logic puzzles involving prisoners or hats or prisoners in hats were discussed. I can’t remember what they were, but I think we covered all of them.

Actually, I can remember one of them, which I think Tash posed, claiming that nobody manages to work out the answer before looking it up.

100 prisoners are locked in a room. In the room next-door are one-hundred boxes containing the names of the prisoners, one for each prisoner. The prisoners are taken one at a time into the room and allowed to open up to fifty boxes. If a prisoner opens the box with their name, they are set free, otherwise they are sent back to their cell. The room is put back how it was after each prisoner leaves. The prisoners are allowed to decide on a strategy before they begin. How do they maximise the chances of everyone being set free?

As promised, none of us worked it out. The solution’s quite clever! Maybe look it up!

Gray codes and the Towers of Hanoi

Gray codes have an interesting link to the Towers of Hanoi puzzle. I don’t think we added anything to what’s on the Wikipedia page, but it was very interesting.

Finally, a few card tricks were demonstrated but I wasn’t following along. Can anyone describe them?