Mathematical Problems, 3A - The Value of Pi

Back when I was doing my A-levels, I remember learning about how it's possible to evaluate pi in various ways, one of which was through calculus.  I can only remember the basics, and I'm sure I can't recall how to do it now - at least not without some help!

The value of pi is the ratio of the circumference of a circle (the distance all the way around the outside of the circle) divided by the diameter (the straight-line distance from one side to the other, through the centre).  It's been known historically to be about 3, but I'm going to make some approximations from first principles.

Firstly, drawing a square around a circle, so that the square touches the circle.  See the diagram above, with the square ABCD around the circle.

If the diameter of the circle is d, then the perimeter of the square is 4d.  We can see by inspection that the square's perimeter is longer than the circle's circumference, and we've called the circumference pi d.  Therefore, we know that pi is less than 4.

It's not a dramatic result, I know, but it's a start!

Next, we need to look at setting a minimum value for pi, and we'll look at this next time.

Mathematical Problems, 2 - The Tetrahedron Path

In this article, I want to provide another puzzle, and the solution to it (you'll need to look closely at the diagram below to understand the solution).  This follows on from my previous post, which was a puzzle about a spider and a cube.  The spider was a mathematical spider, and she wanted to walk around all 12 edges of a cube, in both directions (i.e. from left to right and then right to left) in a single continuous path.  In the original puzzle (devised in a BBC Micro maths adventure game), you had to tell the spider to turn either left or right at each corner.  Now I'm a little older, I use a pen and paper and a diagram...

Following on from the cube, I set myself the challenge of doing the same for a tetrahedron.  I can't say if it was significantly harder or easier, perhaps a little easier with fewer edges to follow, but here's the solution.  The path through the letters goes along each edge twice, once in one direction and then later in the opposite direction, starting and finishing at the same point.

My next post will be a brief discussion on how to calculate a very approximate value of pi from first principles (by which I mean a few diagrams and some basic geometry).

Here's my solution to the tetrahedron path puzzle:

Mathematical Problems, 1 - The Cube Path

Now that the X-Factory has finished its on-screen cash-making routine, I'm returning to more interesting and productive matters - in this case, a mathematical problem that I first came across nearly 20 years ago.  Yes, I'm that old.

This puzzle was posed as part of a text-based adventure game (remember them?) with a strong mathematical bias - you had to solve certain problems in order to progress through the game.  If I remember rightly, for each part you solved, you were given an item or a shape that would come in useful later in the game.  The game was on the school's Maths Club's BBC Micro, and although I can't remember what it was called, there was one puzzle which I didn't solve at the time (although a friend of mine, James Leeson,did, after we started working on it together).  The puzzle recently resurfaced in my mind earlier this week, and I sat down to apply myself to solving it.

Anyways, enough preamble.  In the game, you met a spider, who had a challenge.  She wanted to walk along each of the 12 edges of a cube in a single continuous path that would take her along each edge twice - once in each direction.  You weren't allowed to go directly back on yourself (i.e. reverse) but at the end of each edge, you had to type in if you wanted to turn left or right.  Although the problem didn't specify that you had to start and finish at the same point, this becomes evident through the symmetry of the problem.  And you have to walk along each edge twice (once in one direction, and once in the reverse) otherwise you're guaranteed to hit a dead end.

I can't say much about the theory behind the solution - or if there's more than one solution - but here's mine (and I'm quite proud of it, nearly two decades later).  Start with A and follow each letter sequentially.  If you label each edge with an arrow as you go along, you'll see that it's a valid solution.

Next time... the same puzzle for a tetrahedron!