## Cables

#### Problem

A telephone company places round cables in round ducts. Assuming the diameter of a cable is 2 cm, what would the diameter of the duct be for two cables, three cables and four cables?

#### Solution

Clearly the duct for two cables must have a diameter of 4 cm, however, the duct for three and four cables requires a little more effort. Let us first consider the duct for four cables:

Using the Pythagorean Theorem, `d` ^{2} = 2^{2} + 2^{2}, giving `d` = 8 2.83 cm. Therefore, the duct diameter will be 8 + 2 4.83 cm.

Now we consider the three cable duct,

First of all we use the Pythagorean Theorem to find the height of the triangle, 2^{2} = `h` ^{2} + 1^{2}, therefore `h` = 3 1.73 cm.

Using the geometric result that states the centre of an equilateral triangle is 1/3 the height of the triangle, we deduce that the distance from the centre of the triangle to its apex is (2/3)`h` = 23 / 3 1.15 cm.

So the duct radius will be 23 / 3 + 1, hence the diameter will be 43 / 3 + 2 4.31 cm.

Prove the geometric result that states the distance from the base of an equilateral triangle to the centre is 1/3 the distance from the base to the apex.

Find the diameter of a duct containing five cables.

(Think carefully about the way in which five cables would group.)

Can you generalise for `n` cables?

Note: This is an unsolved problem, but you never know... :o)