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 = 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, 22 = h 2 + 12, 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)