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Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

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Doodles

A 'doodle' is a closed intersecting curve drawn without taking pencil from paper. Only two lines cross at each intersection or vertex (never 3), that is the vertex points must be 'double points' not 'triple points'. Number the vertex points in any order. Starting at any point on the doodle, trace it until you get back to where you started. Write down the numbers of the vertices as you pass through them. So you have a [not necessarily unique] list of numbers for each doodle. Prove that 1)each vertex number in a list occurs twice. [easy!] 2)between each pair of vertex numbers in a list there are an even number of other numbers [hard!]

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Russian Cubes

How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted in order 'blue then red then blue then red then blue then red'. Having finished one cube, they begin to paint the next one. Prove that the girl can choose the faces she paints so as to make the second cube the same as the first.

Archimedes and Numerical Roots

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

This problem builds on Approximating Pi. This brilliant man Archimedes managed to establish that $3 1/10 < \pi < 3 1/7$.

The problem is how did he calculate the lengths of the sides of the polygons, which he needed to be able to calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.

How might he have calculated $\sqrt{3}$?

This must be somewhere between 1 and 2. How do I know this?

Now calculate the average of $3/2$ and $2$ (which is 1.75) - this is a second approximation to $\sqrt{3}$. i.e. we are saying that a better approximation to $\sqrt{3}$ is $(3/n + n)/2$ where $n$ is an approximation to $\sqrt{3}$.

We then repeat the process to find the new (third) approximation to $\sqrt{3}$. $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214\dots$$ to find a fourth approximation repeat this process using 1.73214 and so on...

How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places?

Why do you think it works?

Will it always work no matter what I take as my first approximation and does the same apply to finding other roots?