One of the great problems of Greek geometry is the *doubling of the cube*. When the sides are doubled, the area becomes four times as great, and the volume is tripled. So how to double the volume of the cube? Or in general, how to double the volume of a given solid without changing its form (obtaining a similar figure)? This feels like a simple thing to do, yet it presented a phenomenal difficulty.
Today (since 1837), we know this is an impossible task if we only use a compass and a straight-hedge. This impossibility is related to the fact that ∛2 is not constructible by compass and straight-hedge.
Hippocrates of Chios was the first to observe that the problem is equivalent to find two *mean proportionals* in *continued proportion* between a line of length A and another line of length 2A.
Mean proportional = geometric mean. The geometric mean of 2 and 3 is sqrt(2·3).
Continued proportion: a, b and c are in continued proportion of a/b = b/c.
Then, we want two mean proportionals, let's call them l and m. And we want them to be in continued proportion with A and 2A. This means
A/l = l/m = m/(2A).
This is equivalent to say that l is the geometric mean of A and m and that m is the geometric mean of l and 2A.
l=sqrt(A·m) => l² = A·m => A/l = l/m m=sqrt(l·2A) => m² = l·2A => l/m = m/(2A).
So, we can deduce from these equations that
m² = 2Al m = 2A²/l => m² = 4A⁴/l².
As m²=m², we have 2Al = 4A⁴/l², so we arrive to l = ∛2·A. In conclusion, if we would be able to build l from A, we could obtain a new solid with a volume enhanced by a factor l³ / A³ = 2.
There is a story of some Delians (from Delos) trying to double one of the altars, and falling into the same difficulty. The oracle, when the Delians asked how to defeat a plague, responded that they must double the altar to Apollo, which was a cube.
They went to the Academy of Plato to ask for help. Plato gave the problem to Eudoxus, Archytas and Menaechmus. It is said that Menaechmus found a solution using mechanical means, and not by compass and straight-edge.
Menaechmus' solution required the intersection of two conic curves. This suggests that Menaechmus could have been the discoverer of conic sections. It also seems that he gave two solutions!
Let's use again those two segments, but now let's say the smaller has length a and the larger length b. We call now x to the first mean (what we called l before) and the second one y (what was m). Then, a/x = x/y = y/b. Notice how from here we get
x² = ay ---> parabola y² = bx ---> parabola xy = ab ---> hyperbola.
which are two parabolas and a hyperbola. The solutions correspond to intersect any pair of these equations.
For example, take the hyperbola y=ab/x and the parabola y=x²/a. The intersection gives y=y => ab/x = x²/a => x³ = ba², but as b=2a => x³ = 2a³.
We can define a parabola as the points (loci) that fulfill the condition y=x²/a. Similarly, we can define a hyperbola as the loci for which y=ab/x. It is possible that Menaechmus looked for positions at which these conditions were met, and came up with those curves. If this is true, conic curves did not appear as an investigation of how to cut a cone.
It also seems that Archytas of Tarentum also found a solution, involving the intersection of three surfaces of revolution.