### Two Trees

Two trees 20 metres and 30 metres long, lean across a passageway between two vertical walls. They cross at a point 8 metres above the ground. What is the distance between the foot of the trees?

### Golden Triangle

Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.

### Strange Rectangle

ABCD is a rectangle and P, Q, R and S are moveable points on the edges dividing the edges in certain ratios. Strangely PQRS is always a cyclic quadrilateral and you can find the angles.

# Chord

##### Stage: 5 Challenge Level:

 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. $JH^2 = 2^2 - 1.2^2 = 4 - 1.44 = 2.56 = 1.6^2$ 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 $H$.

$\angle ACJ$ is similar to $\angle ADG$

$\Rightarrow AD = 2 \times2(n - 1) + 2 = 4n - 2$ units
$\Rightarrow AC = 4m - 2$ units.

\eqalign{ \frac{JC}{AC} &= \frac{DG}{AD} \\ \frac{JC}{4m - 2} &= \frac{2}{4n - 2} \\ JC &= \frac{4m - 2}{2n - 1}}

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

Hence, by Pythagoras' theorem, $$IH = 2JH = 2\sqrt{2^2 - \left(\frac{4m - 2}{2n - 1}\right)^2}$$ Simplifying this expression gives $$IH = \frac{8}{2n - 1}\sqrt{n(n - 1) - m(m - 1)}$$ Checking this where $n=3$ and $m=2$ $$IH = \frac{8}{5}\sqrt{6 - 2} = \frac{16}{5} = 3.2$$