Efficient Packing

How efficiently can you pack together disks?

Perfect Eclipse

Use trigonometry to determine whether solar eclipses on earth can be perfect.

Construct the Solar System

Make an accurate diagram of the solar system and explore the concept of a grand conjunction.

Far Horizon

Stage: 4 Challenge Level:

The solution below is based on the one sent in by Barinder of Langley Grammar School. We had quite a number of correct, well laid out solutions to this problem this month including those from Roy of Allerton High School, Dan (no address given) and Calum of Wayland High School.
Although definitely not in proportion, this makes the problem seem a lot easier. The question is asking for the length of the arc I have coloured red. To get this, I decided to find the angle $\theta$ on the diagram, and use the equation

Arc Length $= \frac {\theta}{360} \times 2 \pi r$ where $\theta$ is measured in degrees and r is the radius.

$\angle OAB = 90^\circ$, since it is where a tangent and a radius of a circle meet - it is a circle theorem.

Thus, the triangle AOB can be drawn as follows:

We can now use trigonometry to find $\theta$:

$$\begin{align*} \cos \theta &= \frac {6367000}{6367025} \\ \cos \theta &= 0.99999607 \\ \theta &= \cos^{-1} (0.99999607) = 0.1606^\circ \mbox{(4 d.p.)}\end{align*}$$

Substitute this into the equation for the arc length of a circle earlier to obtain the length required:

Arc Length $= \frac {0.1606}{360} \times 2\pi r.$

Arc Length $= 0.000446 \times 2 \times \pi \times 6367000 = 17,842.3m = 17.8 km$

For this next part, we are given the arc length, since this corresponds to the distance between England and France. The diagram is therefore:

This is essentially the reverse of the previous question. We need to find the angle $\alpha$ first, and to do this, we consider the arc length of the sector OAD of the circle:

Arc Length $= \frac {\alpha}{360} \times 2 \pi r$.

So $32000 = \frac {\alpha}{360} \times 2 \times \pi \times 6367000$.

Then $32000 \times 360 = \alpha \times 2 \times \pi \times 6367000$.

So $\alpha = 0.288^\circ \mbox{(3 d.p.)}$

Since we now have the angle $\alpha$, we can consider the triangle AOB:

$\cos \alpha = \frac {6367000}{6367000 + h}$.

So $6367000 + h = \frac {6367000}{\cos (0.288)}$.

So $6367000 + h = 6367080.415$

$h = 6367080.415 - 6367000 = 80.415$ m high.

Thus, the cliffs of Dover are $80.4$ metres high.

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