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Triangle Inequality

ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm.

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How Far Does it Move?

Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage.

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Speeding Up, Slowing Down

Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its speed at each stage.

Rollin' Rollin' Rollin'

Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Simon sent us this solution :

The obvious answer is that the outer circle (B say) rotating around the stationary circle (A) revolves once but this is wrong. Looking at the interactivity I could see that the mark on the circumference started on the left of the outer circle and was on the left again when the outer circle had only gone half way around. This means that it must make two full turns as it rotates around the inner circle. Now I need to explain why.

Let's start with each of the circles having a radius r. I thought about the centre of the outer circle. When it makes a full circuit of the inner circle the centre of the outer circle will have drawn a circle of raduis 2r so it will have covered a distance of $2 \pi \times 2r = 4 \pi r$.

So how many rotations has this outer circle made?

I next imagined the circle rolling along a line. How many revolutions would be necessary for the centre to travel the distance of $4\pi r$

In one revolution the centre will travel the same distance as the circle (imagine a bicycle wheel) that is a distance of $2\pi r$. So to travel a distance $4\pi r$ the circle would need to revolve twice. This means the outer circle makes two full turns for every single circuit of the inner circle.

Kevin of Langley Grammer School explained what was happening for circles with different, as well as the same radii.
The centre of the moving circle moves round the circumference of a circle of radius $2r$, i.e. a distance $4\pi r$.

The centre of the moving circle moves a distance $2 \pi r$ when it makes one complete turn about its centre. Therefore when the moving circle returns to $P$ it has made 2 complete turns.

In this situation the ratio of the radii of the moving circle to the non-moving circle was 1.

Let the non-moving circle have a radius of $r$, and the moving circle have a radius of $nr$, so that the ratio of the radii is $n$.

Therefore the centre of the moving circle moves along the circumference of a circle with radius $(1+n)r$, i.e. a distance $2 \pi (1+n)r$. Therefore when the moving circle returns to $P$ it will have made $$\frac{2\pi(1+n)r}{2 \pi nr} = 1+ \frac{1}{n}$$ turns. Therefore if the moving circle has a greater radius than the non-moving circle then $n> 1$, and so $1/n$ would be less than 1, and so the number of turns would be less than 2. If the moving circle has a smaller radius than the non-moving circle then $n< 1$, and so $1/n$ would be greater than 1, meaning that the number of turns would be greater than 2.

It can also be seen that the moving coin makes at least 1 turn, regardless of the sizes of the circles, as 1/n is always positive.

So, if the moving circle is half the radius of the inner circle it will turn $1+2 = 3$ times

If the moving circle is one third the radius of the inner circle - it will turn $1+3 = 4$ times.

Surprised?