### Duplication of the cube

**Doubling the cube** (also known as the **Delian problem**) is one of the three most famous geometric problems unsolvable by compass and straightedge construction. It was known to the Egyptians, Greeks, and Indians.^{[1]}

To "double the cube" means to be given a cube of some side length *s* and volume *V*= *s*^{3}, and to construct the side of a new cube, larger than the first, with volume 2*V* and therefore side length $s\backslash cdot\backslash sqrt[3]\{2\}$. The problem is known to be impossible to solve with only compass and straightedge, because $\backslash sqrt[3]\{2\}$ ≈ 1.25992105... is not a constructible number.

## History

The problem owes its name to a story concerning the citizens of Delos, who consulted the oracle at Delphi in order to learn how to defeat a plague sent by Apollo.^{[2]} According to Plutarch^{[3]} it was the citizens of Delos who consulted the oracle at Delphi, seeking 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 consulted Plato, 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 of Delos to occupy themselves with the study of geometry and mathematics in order to calm down their passions.^{[4]}

According to Plutarch, Plato gave the problem to However another version of the story says that all three found solutions but they were too abstract to be of practical value**.
**

A significant development in finding a solution to the problem was the discovery by Hippocrates of Chios that it is equivalent to finding two mean proportionals between a line segment and another with twice the length.^{[6]} In modern notation, this means that given segments of lengths *a* and 2*a*, the duplication of the cube is equivalent to finding segments of lengths *r* and *s* so that

- $\backslash frac\{a\}\{r\}\; =\; \backslash frac\{r\}\{s\}\; =\; \backslash frac\{s\}\{2a\}\; .\backslash $

In turn, this means that

- $r=a\backslash cdot\backslash sqrt[3]\{2\}$

But Pierre Wantzel proved in 1837 that the cube root of 2 is not constructible; that is, it cannot be constructed with straightedge and compass.

## Solutions

Menaechmus' original solution involves the intersection of two conic curves. Other more complicated methods of doubling the cube involve the cissoid of Diocles, the conchoid of Nicomedes, or the Philo line. Archytas solved the problem in the fourth century B.C. using geometric construction in three dimensions, determining a certain point as the intersection of three surfaces of revolution.

False claims of doubling the cube with compass and straightedge abound in mathematical crank literature (pseudomathematics).

Origami may also be used to construct the cube root of two by folding paper.

### Using a marked ruler

There is a simple neusis construction using a marked ruler for a length which is the cube root of 2 times another length.^{[7]}

- Mark a ruler with the given length, this will eventually be GH.
- Construct an equilateral triangle with the given length as side.
- Extend AB an equal amount again to D.
- Extend the line BC forming the line CE.
- Extend the line DC forming the line CF
- Place the marked ruler so it goes through A and one end G of the marked length falls on CF and the other end of the marked length falls on ray CE. Thus GH is the given length.

The AG is the given length times the cube root of 2.

## References

## External links

- Template:Springer
- Doubling the cube. J. J. O'Connor and E. F. Robertson in the MacTutor History of Mathematics archive.
- To Double a Cube – The Solution of Archytas. Excerpted with permission from A History of Greek Mathematics by Sir Thomas Heath.
- cut-the-knot.