### All in the Mind

Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface of the water make around the cube?

### Triangle Incircle Iteration

Start with any triangle T1 and its inscribed circle. Draw the triangle T2 which has its vertices at the points of contact between the triangle T1 and its incircle. Now keep repeating this process starting with T2 to form a sequence of nested triangles and circles. What happens to the triangles? You may like to investigate this interactively on the computer or by drawing with ruler and compasses. If the angles in the first triangle are a, b and c prove that the angles in the second triangle are given (in degrees) by f(x) = (90 - x/2) where x takes the values a, b and c. Choose some triangles, investigate this iteration numerically and try to give reasons for what happens. Investigate what happens if you reverse this process (triangle to circumcircle to triangle...)

### Circumspection

M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.

# Arclets Explained

##### Stage: 3 and 4

Published March 2003,February 2011.

The Arclets problem set in September 2002 produced some very interesting and inspiring work from Madras College. This short article gives a flavour of the way that Sheila, Shona, Alison Colvin, Sarah, Kathryn and Gordan tackled the problem.

Each of the following shapes is made from arcs of a circle of radius r.

What is the perimeter of a shape with $3$, $4$, $5$ and $n$ "nodes".

What happens when $n$ is very large?

Explain the relationship between your answers and the perimeter of the original circle

Here are arclets with $3$, $4$ and $5$ nodes:

### 3-node solution

The angles at the centre of the inner circle are $60^{\circ}$

So the angles at the centre of the outer circles are $120^{\circ}$ and $240^{\circ}$ ($120^{\circ}+240^{\circ}= 360^{\circ}$.

We can therefore divide each circle into a $1/3$ ($120^{\circ}$ out of $360^{\circ}$) part and a $2/3$ ($240^{\circ}$ out of $360^{\circ}$) part.

The perimeter of the arclet is made up of 3 "inward" arcs of $1/3$ of the circumference and 3 "outward" arcs of $2/3$ of the circumference.

\begin{eqnarray} \mbox{Perimeter} &=&((3 \times \frac{1}{3}) + (3 \times \frac{2}{3}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

### 4-node solution

For the 4 node arclets the angles at the centre are $90^{\circ}$

The inward arcs are $1/4$ of the circumference ($90^{\circ}$ - there are 4 inward arcs.

The outward arcs are $1/2$ of the circumference (4 lots of $45^{\circ}$ - there are 4 outward arcs.

\begin{eqnarray} \mbox{Perimeter} &=&((4 \times \frac{1}{4}) + (4 \times \frac{2}{4}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

### 5-node solution

Because the inner circle is surrounded by five outer circles there are 5 angles - all of $72^{\circ}$ at the centre.

Using the properties of isoseles triangles the outward arcs are $2/5$ of the circumference and the inward arcs are $1/5$ of the circumference.

\begin{eqnarray} \mbox{Perimeter} &=&((5 \times \frac{1}{5}) + (5 \times \frac{2}{5}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

An image of part of the work Sarah, Kathryn and Gordon did to find the perimeter of the 5-node arclet is shown below

### 6-node solution

Scanned diagrams showing the work of Sheila, Shona and Alison to find the perimeter of a 6-node arclet:

\begin{eqnarray} \mbox{Perimeter for 6 nodes} &=&((6 \times \frac{1}{6}) + (6 \times \frac{2}{6}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

### Finding the perimeter of the N - node arclet

We know that these are the equations for the perimeter of 3, 4, 5 and 6 node arclets:

 Number of nodes 3 4 5 6 Perimeter $\quad 3\times 2 \pi r = 6 \pi r \quad$ $\quad 3\times 2 \pi r = 6 \pi r \quad$ $\quad 3\times 2 \pi r = 6 \pi r \quad$ $\quad 3\times 2 \pi r = 6 \pi r \quad$

If we substitute $N$ for the node number we get (in every case):

\begin{eqnarray} \mbox{Perimeter for N nodes} &=&((N \times \frac {1}{N}) + (N \times \frac{2}{N}))\pi d \\ &= & (1 + 2)\pi d \\ &= & 3 \pi d \\ & = & 6 \pi r \end{eqnarray}

This is based on the fact that the angles at the centres of the circles will be $1/N$ of a full turn.

If $N$ is very large the node shape begins to look like a circle:

In other words. No matter how many nodes the perimeter will always be 3 circumferences.