String Patterns

Introduction Polygons and Stars Cyclic groups Snooker on a circular table Introduction

Next we look at these patterns in a slightly different way.

Imagine starting a snooker/pool ball on the circumference and hitting it so that it makes an angle θ with the tangent at that point.

It will hit the table again and rebound as if it hit a flat wall tangent to the circle.

When something hits a wall the angle at which it hits is equal to the angle at which it leaves.

Extending the tangents, and using the fact that tangents to a circle from the same point are equal, we see that the triangle in the diagram below is isosceles, and so all angles θ are equal as shown.

The ball starts at 0 degrees, and each time it hits it's angle increases by 2θ.

Similar to the nails, we can record its positions as {0, 2θ, 4θ, 8θ, 10θ,...}

Eventually have gone all around the circle, and so we must take these angles mod 360 degrees.

The ball gets back to the start and traces over its own path forever if it reaches zero again.

After 'k' bounces the snooker ball is at position 2kθ (mod 360)

If the ball has gone around the snooker table 'n' times and returned to the zero position, then

2kθ = 360n

or,

kθ = 180n

Unlike with our nails, θ can be any number, not just a whole number.

So since we're not dealing with whole numbers we can rearrange:

θ =  180 ×

n k

For example, if n = 2 and k = 9, θ = 40°

The ball goes around the circle twice and after hitting the side 9 times, it has got back to the start.

You can experiment below by entering θ = 40 and pressing 'play'.

What if we choose 75°?

In this case

75  =  180 ×

n k

Finding 'n' and 'k' as whole numbers:

75 180

=

5 12

=

n k

So n = 5 and k = 12.

That means the ball goes around the circle 5 times and hits the side of the table 12 times, repeating from then on.

(another possibility is n = 10, k = 24, but if you think about it that just means that the ball repeats what I said above twice over!).

What happens if we choose an angle for which there are no such 'n' and 'k'?

In that case the ball will never repeat its path.

This happens when

θ 180

is irrational, ie. cannot be written as a fraction.

1/v2 is such a number, it cannot be written as a fraction.

So    θ = 180/v2    gives a snooker ball that never repeats its path.

You can check this using θ = 127.27922061357855...

The computer can't handle the exact value, which has an infinite number of decimal digits, so we get an approximation, but you can still see how the ball never repeats its path.

The ball doesn't hit every single possible point on the outer circle.

The values it takes are {0, 2θ, 4θ, 6θ,...} mod 360   (where θ = 180/v2)

You might be able to prove that  2nθ   never equals 180/v3  mod 360.

That is, there is no value of 'n' for which  360n/v2 = 180/v3  +  360k   for some 'k'.

Equivalently, there is no value of 'n' for which  2n/v2 = 1/v3  +  2k  for some 'k'.

You might like to see if you can prove this, using the fact that v3 is irrational.

That means that  180/v3  is one of the numbers that the ball never hits.

However, for any θ for which θ/180 is irrational, the ball does hit eventually get as close as you please to any point on the boundary.

That is, if you give me any point α and say how close you want the ball to go (say within 0.0001 or 0.00000001, or...),

the ball will eventually get that close if you wait long enough!

We can prove this as follows:

Suppose it is not true.

Then there is an angle 'α' which the ball doesn't hit or get arbitrarily close to.

That means we can find two numbers, 'δ₁' and 'δ₂' for which the ball never enters the shaded part of the circle.

Actually, we can choose 'δ₁' and 'δ₂' as large as possible.

If we run the clock backwards, the ball goes back by an angle 2θ. So the ball cannot ever enter this second shaded region either.

Otherwise it would enter the first shaded region after the next bounce, but we assumed it doesn't!

In fact we can continue doing this:

These regions will go round and round the circle.

Eventually, after going round enough times, one of them will overlap the first region - since you can't have an infinite number of regions that don't overlap!

But then there is a larger region around α which the ball never enters - as shown above.

This isn't possible since we assumed we'd taken both δ₁ and δ₂ as large as possible to start with.

The only possibility is that those two regions must lie exactly on top of each other.

This means that after going backwards some number of times (say 'k'), we are back at α, mod 360.

ie.      α  -  2kθ  =  α   (mod 360)

or,     -  2kθ  =  0   (mod 360)

So 2kθ must be a multiple of 360

2kθ = 360n

Rearranging,

θ 180

=

n k

But we assumed that θ/180 was irrational, so this is impossible!

The proof is over, it's not possible for our statement to be false so it is true:

For any θ which is an irrational multiple of 180°, the ball gets as close as you please to any point on the boundary. That is, if you give me a point and say how close you want the ball to go (say within 0.0001 or 0.00000001, or...), the ball will eventually get that close!

I like this proof a lot, because of its simplicity, and also because of the way that the way the regions we defined spread around the circle in a similar way to how the ball bounces around!

Readers who want more precision, particularly in the definition of δ₁ and δ₂ being 'as large as possible', will probably also know how to make it more precise! So we will say no more on the subject.