Doubling the cube

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Doubling the cube is one of the three most famous geometric problems unsolvable by compass and straightedge construction. It was known to the Greeks, and earlier to the mathematicians of India.

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

[edit] History

According to legend, the citizens of Athens consulted the oracle of Apollo at Delos in 430 BC, in order to learn how to defeat a plague which was ravaging their lands. The oracle responded that to stop the plague, they must double the size of their altar. The Athenians dutifully doubled each side of the altar, and the plague increased. The correct interpretation was that they must double the volume of their altar, not merely its side length; this proved to be a most difficult problem indeed. It is due to this legend that the problem is often known as the Delian problem.

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

Although doubling the cube is impossible in the plane with the compass and straightedge, ancient Greek mathematician Archytas did solve this problem in 4th century B.C. using geometric construction in three dimensions, determining a certain point as the intersection of three surfaces of revolution...[1][2][3].

[edit] Solutions

An illustration of the ruler-and-compass method

There are many ways to construct $\sqrt[3]{2}$ which involve tools other than compass and straightedge. In fact, some of these tools can themselves be constructed using compass and straightedge, but must be cut out of a sheet of paper before they can be used. For example, construct a ruler with a single unit distance marked on it. Construct an equilateral triangle ABC with side length 1, and extend side $\overline{AB}$ by one unit to form the line segment $\overline{ABD}$ . Extend side $\overline{BC}$ to form the ray $\overrightarrow{BCE}$ , and draw the ray $\overrightarrow{DCF}$ . Now take the ruler and place it such that it passes through vertex A and intersects $\overline{DCF}$ at G and $\overline{BCE}$ at H, such that the distance GH is exactly 1. The distance AG will then be precisely $\sqrt[3]{2}$ .