The Delian Problem.

Around 430 BCE it is reported that a typhoid epidemic carried off about a quarter of the population of Athens in ancient Greece. As usual the gods were blamed for this disaster, but could be approached for help through the high priests in the Temple of Apollo in Delos. When the gods were asked what could be done to halt this raging epidemic, they apparently replied that the altar in the temple would have to be doubled in size.

Now the altar was in the shape of a cube. First, another cube identical to the altar was placed alongside. Although it doubled the size, the gods were apparently unhappy because the altar was no longer a cube. The next solution was to build a cubic altar with its edges double the length of the original edges. This increased the volume by a factor of eight, so the gods were still not happy. Suppose that the length of an edge of the old altar was one unit. Then the length of the edge of the new altar would have to be the cube root of 2 units; hence, the mathematicians were called in.

This is an opportunity for your students to put down their calculators, and to try and solve the Delian problem using logic. Clearly, the solution lies somewhere between 1 and 2 units.

Investigating the problem using logic

Firstly, if we try to use 1.5 or 3/2 this produces a result of:

(3/2) x (3/2) x (3/2) = 27/8 = 3.375

All done without calculators, I hope.

Next try using 1.25 or 5/4. The result produced is,

(5/4) x (5/4) x (5/4) = 125/64 = 1.953

by using long division, but it could be checked by calculators. This is very close to 2, so can it be improved without resorting to a calculator or using the 19th century method of finding the cube root of a number (Hutton, 1846, p.62). This gives your students the chance to discover the method of approximation.

The answer for the cube root of 2 is going to be 1.25+[epsilon], where [epsilon] is a very small number. Now:

2 = [(1.25 + [epsilon]).sup.3] = (125/64) + (75/16)[epsilon] + (15/4)[[epsilon].sup.2] + [[epsilon].sup.3]

Since e is small, its square and cube will be very much smaller and, therefore, a good approximation is given by neglecting the last two terms. Thus,

2 [approximately equal to] (125/64) + (75/16)[epsilon]

The solution is [epsilon] = (1/100) = 0.01 and hence [(1.26).sup.3] is very close to 2, and evaluates to 2.000376. The cube root of 2 is 1.25992104989 ... and therefore, the approximate value 1.26 is within 0.02%.

See if your students can use the method of approximation to evaluate [square root of 7] and then [square root of 5]. Next try the cube root of 4.

Investigating the problem using geometry

Now the Greek mathematicians were great geometers, and so they sought a practical 2-dimensional geometrical solution involving only a straight edge and compass. It was not proved until the 19th century that this was impossible. Many attempts to find a geometrical solution were made in the intervening years involving parabolas, hyperbolas and even 3-dimensional geometry. Along the way the curves known as the conchoid and the cissoid were discovered.

Here is one simple geometric method to obtain the cube root of 2 as a length.

1. Start with an equilateral triangle ABC of side length 1, which is easily obtained by your students using a circle of radius 1 with a chord AB of length 1 joined to the centre C.

2. Extend the radius AC to the opposite side of the circle creating the diameter AD.

3. Join DB, and extend it as shown in Figure 1.

4. Extend CB similarly.

5. Mark off 1 unit on a straight-edged ruler, and fit it exactly between these two extensions at points E and F while adjusting the ruler to pass through A. The length AE is the cube root of 2.

Figure 1 shows a sketch of the proof (after using steps 1 to 5), which some of your senior school students should be able to obtain.

Proving the cube root of 2 as a length

AC = CD = CB = EF = 1

Denote AE by x and BF by p.

Using the Cosine Rule on the triangle ACF produces:

(1 + [(p + 1).sup.2] - [(x + 1).sup.2])/2(p + 1) = cos(60[degrees]) = 1/2

Simplifying and rearranging yields:

[x.sup.2] + 2x - [p.sup.2] - p = 0 Equation (1)

Applying the Sine Rule to triangle ADE yields:

x/sin(30[degrees]) = 2/sin(< AED)

Applying the Sine Rule to triangle BEF yields:

x/sin (30[degrees]) = 22/sin (< BEF)

Since sin(< AED) = sin(< BEF) then px = 2. Hence, from equation (1) it is soon seen that x satisfies the quartic equation

[x.sup.4] + 2[x.sup.3] - 2x - 4 = 0

which factorises to:

(x + 2) ([x.sup.3] - 2) = 0

The result follows that AE is the cube root of 2. The method also presents BF = p as the cube root of 4. Astute readers will also have noted that BD = [square root of 3].

Now the ancient Greeks did not have algebra or trigonometry to be able to prove this construction, and they would not have accepted this method of straight edge and compass because it involved juggling a marked length until it was in the correct position. Nevertheless, there is a lot of useful mathematics in this problem.

Author's note

The idea for this Discovery was based on an article in the 1979 New Zealand Mathematical Digest: A Journal for Schools.

Reference

Hutton, C. (1846). A Course of Mathematics. London: Thomas Tegg.

Caption: Figure 1. Sketch of finding the cube root of 2 as a length using Steps 1 to 5.
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