Doubling the cube, also known as theDelian problem, is an ancient[a][1]: 9 geometric problem. Given theedge of acube, the problem requires the construction of the edge of a second cube whosevolume is double that of the first. As with the related problems ofsquaring the circle andtrisecting the angle, doubling the cube is now known to be impossible to construct by using only acompass and straightedge, but even in ancient times solutions were known that employed other methods.
According to Eutocius, Archytas was the first to solve the problem of doubling the cube (the so-called Delian problem) with an ingenious geometric construction.[2][3][4] The nonexistence of a compass-and-straightedge solution was finally proven byPierre Wantzel in 1837.
In algebraic terms, doubling aunit cube requires the construction of aline segment of lengthx, wherex3 = 2; in other words,x =, thecube root of two. This is because a cube of side length 1 has a volume of13 = 1, and a cube of twice that volume (a volume of 2) has a side length of thecube root of 2. The impossibility of doubling the cube is thereforeequivalent to the statement that is not aconstructible number. This is a consequence of the fact that the coordinates of a new point constructed by a compass and straightedge are roots ofpolynomials over the field generated by the coordinates of previous points, of no greaterdegree than aquadratic. This implies that thedegree of thefield extension generated by a constructible point must be a power of 2. The field extension generated by, however, is of degree 3.
We begin with the unit line segment defined bypoints (0,0) and (1,0) in theplane. We are required to construct a line segment defined by two points separated by a distance of. It is easily shown that compass and straightedge constructions would allow such a line segment to be freely moved to touch theorigin,parallel with the unit line segment - so equivalently we may consider the task of constructing a line segment from (0,0) to (, 0), which entails constructing the point (, 0).
Respectively, the tools of a compass and straightedge allow us to createcirclescentred on one previously defined point and passing through another, and to create lines passing through two previously defined points. Any newly defined point either arises as the result of theintersection of two such circles, as the intersection of a circle and a line, or as the intersection of two lines. An exercise of elementaryanalytic geometry shows that in all three cases, both thex- andy-coordinates of the newly defined point satisfy a polynomial of degree no higher than a quadratic, withcoefficients that are additions, subtractions, multiplications, and divisions involving the coordinates of the previously defined points (and rational numbers). Restated in more abstract terminology, the newx- andy-coordinates haveminimal polynomials of degree at most 2 over thesubfield of generated by the previous coordinates. Therefore, thedegree of thefield extension corresponding to each new coordinate is 2 or 1.
So, given a coordinate of any constructed point, we may proceedinductively backwards through thex- andy-coordinates of the points in the order that they were defined until we reach the original pair of points (0,0) and (1,0). As every field extension has degree 2 or 1, and as the field extension over of the coordinates of the original pair of points is clearly of degree 1, it follows from thetower rule that the degree of the field extension over of any coordinate of a constructed point is apower of 2.
Now,p(x) =x3 − 2 = 0 is easily seen to beirreducible over – anyfactorisation would involve alinear factor(x −k) for some, and sok must be aroot ofp(x); but alsok must divide 2 (by therational root theorem); that is,k = 1, 2, −1 or−2, and none of these are roots ofp(x). ByGauss's Lemma,p(x) is also irreducible over, and is thus a minimal polynomial over for. The field extension is therefore of degree 3. But this is not a power of 2, so by the above:
The problem owes its name to a story concerning the citizens ofDelos, who consulted the oracle atDelphi in order to learn how to defeat a plague sent byApollo.[5][1]: 9 According toPlutarch,[6] however, the citizens ofDelos consulted theoracle atDelphi to find a solution for their internal political problems at the time, which had intensified relationships among the citizens. The oracle responded that they must double the size of the altar to Apollo, which was a regular cube. The answer seemed strange to the Delians, and they consultedPlato, who was able to interpret the oracle as the mathematical problem of doubling the volume of a given cube, thus explaining the oracle as the advice of Apollo for the citizens ofDelos to occupy themselves with the study of geometry and mathematics in order to calm down their passions.[7]
According toPlutarch, Plato gave the problem toEudoxus andArchytas andMenaechmus, who solved the problem using mechanical means, earning a rebuke from Plato for not solving the problem usingpure geometry.[8] This may be why the problem is referred to in the 350s BC by the author of the pseudo-PlatonicSisyphus (388e) as still unsolved.[9] However another version of the story (attributed toEratosthenes byEutocius of Ascalon) says that all three found solutions but they were too abstract to be of practical value.[10]
A significant development in finding a solution to the problem was the discovery byHippocrates of Chios that it is equivalent to finding twogeometric mean proportionals between a line segment and another with twice the length.[11] In modern notation, this means that given segments of lengthsa and2a, the duplication of the cube is equivalent to finding segments of lengthsr ands so that
In turn, this means that
ButPierre Wantzel proved in 1837 that thecube root of 2 is notconstructible; that is, it cannot be constructed withstraightedge and compass.[12]
Menaechmus' original solution involves the intersection of twoconic curves. Other more complicated methods of doubling the cube involveneusis, thecissoid of Diocles, theconchoid of Nicomedes, or thePhilo line.Pandrosion, a probably female mathematician of ancient Greece, found a numerically accurate approximate solution using planes in three dimensions, but was heavily criticized byPappus of Alexandria for not providing a propermathematical proof.[13]Archytas solved the problem in the 4th century BC using geometric construction in three dimensions, determining a certain point as the intersection of three surfaces of revolution.
Descartes' theory of geometric solution of equations uses a parabola to introduce cubic equations, in this way it is possible to set up an equation whose solution is a cube root of two. Note that the parabola itself is not constructible except by three dimensional methods.
False claims of doubling the cube with compass and straightedge abound in mathematicalcrank literature (pseudomathematics).
Origami may also be used to construct thecube root of two by folding paper.
There is a simpleneusis construction using a marked ruler for a length which is the cube root of 2 times another length.[14]
Then AG is the given length times.
Inmusic theory, a natural analogue of doubling is theoctave (a musical interval caused by doubling the frequency of a tone), and a natural analogue of a cube is dividing the octave into three parts, each the sameinterval. In this sense, the problem of doubling the cube is solved by themajor third inequal temperament. This is a musical interval that is exactly one third of an octave. It multiplies the frequency of a tone by, the side length of the Delian cube.[15]
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