Congratulations to four students from Madras College, Gordon and Alan from S6, Sue Liu from S4 and David from S3. They all sent excellent solutions to this problem including complete proofs of the general case. Answers also arrived from St Peter's College in Adelaide, Australia. This is David's proof of the first part: Lines CH and CI can be drawn in. Both have length R or 2 cm.

ICH is an isosceles triangle which can be split into two congruent right angled triangles by drawing line CJ, where J is the midpoint of chord IH.

Triangle AJC is similar to triangle AGD, with a ratio of 6 cm to 10 cm or 3:5.

Line GD = R = 2 cm,

Line CJ = 3/5 GD = 1.2 cm.

Both right angled triangles CJH and CJI have lengths as below and the length JH can be worked out using Pythagoras theorem. ${\mathrm{JH}}^2=2^2-1.2^2=4-1.44=2.56=1.6^2$ So JH = 1.6 cm and the chord JH = 2(1.6)= 3.2 cm.

This is Gordon's proof of the general case of n circles where AG cuts the m ^th circle at I and H.

$\angle \mathrm{ACJ}$ is similar to $\angle \mathrm{ADG}$

$\Rightarrow \mathrm{AD}=2x2(n-1)+2=4n-2$ units

$\Rightarrow \mathrm{AC}=4m-2$ units.

$\begin{array}{cc}\multicolumn{1}{c}{\frac{\mathrm{JC}}{\mathrm{AC}}}& =\frac{\mathrm{DG}}{\mathrm{AD}}\\ \multicolumn{1}{c}{\frac{\mathrm{JC}}{4m-2}}& =\frac{2}{4n-2}\\ \multicolumn{1}{c}{\mathrm{JC}}& =\frac{4m-2}{2n-1}\end{array}$

$\angle \mathrm{IHC}$ is isosceles and $\angle \mathrm{CJI}=\angle \mathrm{CJH}={90}^{\circ }$.

Hence, by Pythagoras theorem,

${\displaystyle \mathrm{IH}=2\mathrm{JH}=2\sqrt{2^2-{\left(\frac{4m-2}{2n-1}\right)}^2}}$

Simplifying this expression gives

${\displaystyle \mathrm{IH}=\frac{8}{2n-1}\sqrt{n(n-1)-m(m-1)}}$

Checking this where $n=3$ and $m=2$

${\displaystyle \mathrm{IH}=\frac{8}{5}\sqrt{6-2}=\frac{16}{5}=3.2}$