Since the days of ancient Greece known three geometric problems - trisection of the angle, squaring the circle and the doubling cube, which are considered intractable using ruler and compass (the main instruments of the ancient geometers).

For the doubling cube invented already: mezolyabiya Eratosthenes, cylinders Archytas, the triad Menehma and others (see, for example, http://hijos.ru/2011/04/10/udvoenie-kuba/), this solution is different.

Perhaps someone found it before, but I came to it alone, and until proven otherwise, I consider myself the author of this solution.

According to legend, - oracle said (in response to a request to save from the plague the island Delos): "double the altar" (which was a cube-shaped).

In the oracle formulation is no restriction - using only a compass and a ruler, or rather it would solve the problem using only the tools available to the ancient Greeks.

Scratching the main thinking place, it can be concluded that the easiest way to solve the problem - the direct measurement of the volume, splitting along one coordinate desired value (this can be achieved by special construction of the meter).

Note - the volume of the cube can be found by placing a cube in a liquid and measuring the volume of of fluid displaced(Archimedes used this method for measuring the volume of the Hieron’s crown). So, this method of volume measurement was known to ancient Greeks, and it can be used to solve this problem.

Let us take as a measure of volume of fluid - an inverted pyramid with a regular square base. We take the length of the base edge of L, the length of the edge going to the top - kL. The problem of doubling cube (=to reduce the search from the cube root - to the construction of several square roots) can be solved by varying the only parameter k.

Double the amount of liquid (obtain the desired volume of a cube), and place it in a vertically mounted top down pyramid (without lid - so you can fill it). Express the volume of liquid poured into the pyramid through L, kL, and equate it to the desired volume of a cube:

It is seen that L is equal to the desired edge of the cube, if k - the square root of 9 1/2. This length can be obtained as the hypotenuse of a triangle with sides

the length of the second leg can be obtained as the half of the diagonal of a square with side of L.

We produce a pyramid with a square base of 4 triangles with aspect ratio, as indicated above, set it vertically, point downwards. Now, filling it with water, equal 2 volumes of the original cube - any edge of the base gives the desired length of the edge of the cube, doubling original - the task of doubling the cube is solved.

To check the verticality of the pyramid can be installed to measure the length of all edges of the base - vertical mounting edge lengths should be equal.

If you want to find the cube root of three - you need to pour volume of three original cube into the measuring pyramid, etc.

I wish this decision was found on Delos 2 000 years ago - as an oracle said, it could save this island from plague. But it was found 2000 years later in Russia - maybe, it saves Russia from horrors of multiculturalism, liberalism and free trade. Not sure, but i hope...

If you find a mistake in the argument - please let me know.

According to legend, - oracle said (in response to a request to save from the plague the island Delos): "double the altar" (which was a cube-shaped).

In the oracle formulation is no restriction - using only a compass and a ruler, or rather it would solve the problem using only the tools available to the ancient Greeks.

Scratching the main thinking place, it can be concluded that the easiest way to solve the problem - the direct measurement of the volume, splitting along one coordinate desired value (this can be achieved by special construction of the meter).

Note - the volume of the cube can be found by placing a cube in a liquid and measuring the volume of of fluid displaced(Archimedes used this method for measuring the volume of the Hieron’s crown). So, this method of volume measurement was known to ancient Greeks, and it can be used to solve this problem.

Let us take as a measure of volume of fluid - an inverted pyramid with a regular square base. We take the length of the base edge of L, the length of the edge going to the top - kL. The problem of doubling cube (=to reduce the search from the cube root - to the construction of several square roots) can be solved by varying the only parameter k.

Double the amount of liquid (obtain the desired volume of a cube), and place it in a vertically mounted top down pyramid (without lid - so you can fill it). Express the volume of liquid poured into the pyramid through L, kL, and equate it to the desired volume of a cube:

It is seen that L is equal to the desired edge of the cube, if k - the square root of 9 1/2. This length can be obtained as the hypotenuse of a triangle with sides

the length of the second leg can be obtained as the half of the diagonal of a square with side of L.

We produce a pyramid with a square base of 4 triangles with aspect ratio, as indicated above, set it vertically, point downwards. Now, filling it with water, equal 2 volumes of the original cube - any edge of the base gives the desired length of the edge of the cube, doubling original - the task of doubling the cube is solved.

To check the verticality of the pyramid can be installed to measure the length of all edges of the base - vertical mounting edge lengths should be equal.

If you want to find the cube root of three - you need to pour volume of three original cube into the measuring pyramid, etc.

I wish this decision was found on Delos 2 000 years ago - as an oracle said, it could save this island from plague. But it was found 2000 years later in Russia - maybe, it saves Russia from horrors of multiculturalism, liberalism and free trade. Not sure, but i hope...

If you find a mistake in the argument - please let me know.

## Error

Your reply will be screened

Your IP address will be recorded

You must follow the Privacy Policy and Google Terms of use.