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?


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 = 22 + 22, giving d = radical8 approximately 2.83 cm. Therefore, the duct diameter will be radical8 + 2 approximately 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, 22 = h 2 + 12, therefore h = radical3 approximately 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 = 2radical3 / 3 approximately 1.15 cm.

So the duct radius will be 2radical3 / 3 + 1, hence the diameter will be 4radical3 / 3 + 2 approximately 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)

Problem ID: 28 (Dec 2000)     Difficulty: 3 Star

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