Doubling Cube

Doubling a cube is a classical antique task of constructing a cube edge with a pair of compasses and a ruler , the volume of which is twice the volume of a given cube ^[1] .

Along with the trisection of the angle and the quadrature of the circle , it is one of the most famous unsolvable problems of building with the help of a compass and a ruler.

Content

History

According to ancient legend, once on the island of Delos, an epidemic of plague broke out. The inhabitants of the island turned to the Delphic oracle , and he said that it was necessary to double the altar of the sanctuary, which had the shape of a cube. The inhabitants of Delos built another similar cube and put it on the first, but the epidemic did not stop. After the second conversion, the oracle explained that the double altar should also be in the form of a cube.

Since then, the best mathematicians of the ancient world have been engaged in the Delhi problem , several solutions have been proposed, however, no one was able to complete such a construction using only a pair of compasses and a ruler, so gradually a general conviction was formed that such a problem could not be solved. More Aristotle in the IV century BC. e. wrote: “By means of geometry it is impossible to prove that ... two cubes make up one cube” ^[2] .

Attempted Solutions

Hippocrates of Chios (end of the 5th century BC) showed that the task boils down to finding two means proportional between one segment and another twice its length. In modern notation, to finding $x$ ${\ displaystyle x}$ and $y$ ${\ displaystyle y}$ such that
${\frac {a}{x}}={\frac {x}{y}}={\frac {y}{2a}}$ ${\ displaystyle {\ frac {a} {x}} = {\ frac {x} {y}} = {\ frac {y} {2a}}}$ . From here $x^{3}=2a^{3}$ ${\ displaystyle x ^ {3} = 2a ^ {3}}$ .

The architect Tarentsky (beginning of the 4th century BC) proposed a solution based on the intersection of the torus , cone, and circular cylinder .

Plato (first half of the 4th century BC) proposed a mechanical solution based on the construction of three rectangular triangles with the desired aspect ratio.

Menehm (mid IV century BC) found two solutions to this problem based on the use of conic sections. In the first solution, the intersection point of two parabolas is sought, and in the second, parabolas and hyperbolas.

Eratosthenes (III century BC) proposed another solution that uses a special mechanical tool - mesolabia , and also described the solutions of its predecessors.

Nycomed (II century BC) used to solve this problem the insertion method performed using a special curve - conchoid .

A group of similar solutions belonging to Apollonius , Philo of Byzantium and Heron also uses the insertion method.

In another group of similar solutions belonging to Diocles , Papp and Spore , the same idea is used as in the solution of Plato , while Diocles uses a special curve, a cisoid , to construct.

Their solutions were also proposed by Viet , Descartes , Gregoire de Saint-Vincent , Huygens , Newton .

Insolubility

In modern notation, the problem is reduced to solving the equation $x^{3}=2a^{3}$ ${\ displaystyle x ^ {3} = 2a ^ {3}}$ . The solution has the form $x=a{\sqrt[{3}]{2}}$ ${\ displaystyle x = a {\ sqrt [{3}] {2}}}$ . It all comes down to the problem of constructing a length segment ${\sqrt[{3}]{2}}$ ${\ displaystyle {\ sqrt [{3}] {2}}}$ . Pierre Vanzel proved in 1837 that this problem cannot be solved with the help of a compass and a ruler .

Doubling a cube is not resolvable using a compass and a ruler, but it can be done using some additional tools, for example, doubling a cube can be done by building with a flat origami .

Solution Using Extra Tools

Doubling a cube using nevisis

Fig. 1 Doubling a cube using nevsis

Take an equilateral triangle MPN with side a , extend the side PN, and build a point R at a distance a from point N (Fig. 1). Let us extend the segments NM and RM to the left. We take a Nevisis ruler with a diastema a and using the straight line NM as a guide, the point P as a pole and the straight line RM as a target line, we construct the segment AB . The length of the segment BP corresponds to the side of the cube of doubled volume in comparison with the cube with side a .

Literature

Belozerov S. E. Five famous tasks of antiquity. History and modern theory. - Rostov: publishing house of the Rostov University, 1975. - 320 p.
Gleizer G.I. History of mathematics at school . - M .: Enlightenment, 1964 .-- S. 324-325.
Prasolov V.V. Three classic construction problems. Doubling a cube, trisection of an angle, squaring a circle . - M .: Nauka, 1992 .-- 80 p. - ( Popular Lectures in Mathematics , Issue 62).
Chistyakov V. D. Three famous tasks of antiquity. - M .: State. student-ped. Publishing House of the Ministry of Education of the RSFSR, 1963. - S. 8-28. - 96 p. .
Shchetnikov A. I. How were some solutions to the three classical problems of antiquity found? // Mathematical education. - 2008. - No. 4 (48) . - S. 3-15 .
- Shchetnikov A. I. How were some solutions to the problem of doubling a cube found? Historical and mathematical research , No. 15 (50), 2014, S. 65–78.

Notes

↑ Doubling the cube // Great Soviet Encyclopedia / V. A. Vvedensky. - 2nd edition. - The Great Soviet Encyclopedia, 1956. - T. 43. - S. 648. - 300,000 copies.
↑ Aristotle . Second Analytics, Part I, Ch. 7. M.: Gospolitizdat, 1952.

[bse1956-1] Doubling the cube // Great Soviet Encyclopedia / V. A. Vvedensky. - 2nd edition. - The Great Soviet Encyclopedia, 1956. - T. 43. - S. 648. - 300,000 copies.

[2] Aristotle . Second Analytics, Part I, Ch. 7. M.: Gospolitizdat, 1952.