### Route to Root

A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?

### Rain or Shine

Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry.

### Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

# Ford Circles

##### Age 16 to 18Challenge Level
Navjot from Sherborne Qatar School sent us a lovely clear explanation:

Explore the value of ad−bc for the touching circles that you have found. What do you notice?

I noticed that $ad-bc$ was always equal to $\pm 1$. This happened when the pair
of fractions were Farey neighbours.
For example, $\frac01, \frac 11: (0 \times 1) - (1 \times 1) = -1$
Or, $\frac 1{11}, \frac 1{12}: (1 \times 12) - (1 \times 11) =1$

In both of these cases, the two circles were tangent to each other.

Can you prove that for any touching circles in the interactivity above, $|ad−bc|=1$?

So it is given to us that the centre of circle $A$ is  $\left(\frac ac , \frac1{2c^2}\right)$ with radius $\frac1{2c^2}$,
and the centre of circle B is $\left(\frac bd, \frac1{2d^2}\right)$ with radius
$\frac1{2d^2}$.

To show $|ad-bc| = 1$, I found the length between the centres of the two circles and equated it to the sum of the radii, which is also the length between the two points.

$|AB| =\sqrt{(\frac ac - \frac bd)^2 +(\frac1{2c^2} - \frac1{2d^2})^2)} = \frac1{2c^2} + \frac1{2d^2}$

$\Rightarrow \sqrt{\left(\frac{ad-bc}{cd}\right)^2 + \left(\frac{2d^2-2c^2}{4c^2d^2}\right)^2} = \frac{2d^2+2c^2}{4c^2d^2}$

$\Rightarrow \frac{(ad)^2 - 2abcd + (bc)^2}{(cd)^2}+ \frac{4d^4-8(dc)^2 + 4c^4}{16(cd)^4} = \frac{4d^4+8(dc)^2+4c^4}{16(cd)^4}$

$\Rightarrow \frac{(ad)^2 - 2abcd +(bc)^2}{(cd)^2}=\frac{(4d^4 + 8(dc)^2 + 4c^4) - (4d^4 - 8(dc)^2 + 4c^4)}{16(cd)^4}$

$\Rightarrow \frac{(ad-bc)^2}{(cd)^2} = \frac{16(cd)^2}{16(cd)^4}$
$\Rightarrow (ad-bc)^2=1$
$\Rightarrow |ad-bc|=1$.

We also received solutions from Sarith from Royal College in Sri Lanka, and Vignesh from Hymers College in the UK. You can read their solutions below:

Sarith's Solution
Vignesh's Solution