Doubling the Cube

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Doubling the cube by Greek Mathematicians

The idea of two mean proportionals

Hippocrates of Chios (circa 470 BC-410 BC) discovered that the duplication of the cube is equivalent to finding two mean proportionals in continued proportion between two given straight lines.
That is, if a/x = x/y = y/b where a and b are given, then (a/x)³ = a/b
Therefore if a = 1 and b = 2, x = (2)^1/3
So we have cubic root of 2.
After this discovery, the Delian problem became a problem of finding these two mean proportionals geometrically.

1. Archytas- 3D solids intersection solution

Archytas of Tarentum (circa 428 BC-350 BC) discovered that a solution is finding the intersection of three surfaces of revolution.
The three surfaces of revolution used are
(1) the right cone : x² + y² + z² = (a/b)²x²
(2) the cylinder : x² + y² = a x
(3) the torus : x² + y² + z² = a {x² + y²}^1/2
where AC = a, and AB = b
Then in the drawing, AC/AP = AP/AM = AM/AB.
This gives AC/AB = (AM/AB)³ and AM is the solution.

For detail, go to the section Archytas - 3D solids intersection solution.

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2. Solution by Eudoxus

Eudoxus of Cnidus (408 BC - 355 BC) was a student of Archytas. He solved the problem by projecting the intersection of the cone and torus onto xy plane. The solution is the the point where such a curve crosses the circle ABC.

AB, and AC is the line segment between which two mean proportionals have to be found.

The steps go as follows:
1. Drop a line BF perpendiculat to AC.
2. Take any point G on BF, and with A as center and AG as radius, draw an arc cutting AC at M.
3. Erect a line normal to AC at M. This line intersects line AG at L.
4. When L is on the circle, then AL is the solution.

You can see the process in animation.
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For detail, go to the section Eudoxus_Delian.

3. Menaechmus- Conic Solution

Menaechmus (circa 380 BC - 320 BC), who is said to have discovered the conic sections, is also famous by his answer to Alexander the great. "In geometry there is no royal road .".
He used parabolas and rectangular hyperbola to solve the Delian problem.

If x, y are two mean proportional between straight lines a, b, a / x = x / y = y / b then x² = ay, y² = bx , and xy = ab Therefore any two pair of these three curves, will give the solution to Delian problem.

Animation to be added here.

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For detail, go to the section Menaechmus_Delian.

4. Plato's Cube doubling Tool

Many historians agree that this solution is wrongly attributed to Plato (427 BC - 347 BC) .
Generally accepted conclusion is that the solution was invented in the Plato's Academy by someone contemporary with or later than Menaechmus .

The basic concept is to find points M & N

on y and x -axis ,respectively

while keeping angle AMN and MNB

being equal to 90 degrees.

Then, 3 triangles AOM, MON, and NOB are similar.
Therefore, AO:OM = OM:ON = ON:BO
and ON³ = AO.BO²
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For detail, go to the section Plato_Delian.

5. Eratosthenes' Cubic Root Tool

Eratosthenes of Cyrene (276 BC - 194 BC) is also famous for estimating the size of the Earth.

You can see the process in animation.

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For detail, go to the section Eratosthenes_Delian

6. Conchoid by Nicomedes

Nicomedes (circa 280 BC - 210 BC) ,the inventor of Conchoid curve, used his own invention to solve Delian problem.

The length of AB and BC are given.
D , E are the bisectors of AB , BC ,respectively.
CG = AB, CF = AD, and GF is parallel to CZ.
Objective is to find length equal to: {AB.BC²}^1/3 and {AB².BC}^1/3

Result : AB / AM = AM / CK = CK / BC

Key : To find point K such that CF = HK

Conchoid is used F as "pole", CZ as "ruler",
and "distance" equal to CF(= AD).

You can see the process in animation.
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For detail, go to the section Nicomedes_Delian

7. Philon

Philon of Byzantine (circa 280 BC - 220 BC) solved the "Delian"problem as follows:

AB and AC are the two given straight lines.
draw a rectangle ABDC
point E is the intersection of diagonals,AD and BC.
draw a semi-circle BDC with diameter,BC .
from D ,draw a line.This intersects AB and AC at F and G, respectively.
This line also cuts the semi-circle at H.
When this line is chosen such that GH = DF,
Then, BF and CG are the solutions.
Note: Precision bar shows the value of |GH - DF| .

You can see the process in animation.

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For detail, go to the section Philon_Delian

8. Apollonius

The solution by Apollonius of Perga (circa 262 BC - 190 BC) is very similar to that of Philon.

Again AB and AC are given straight lines,and semi-circle BDC is also drawn.
Draw a circle with center E, cutting the extended lines of AC & AB at F and G, respectively.
Connect FD and DG.

When 3 points F,D, and G are colinear, CG & BF are solutions.

You can see the process in animation.

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For detail, go to the section Apollonius_Delian

9. Heron

The Delian solution by Heron of Alexandria (circa 10 AD - 75 AD) is also very similar to that of Philon.
The difference is that a line passing through D is chosen such that EF = EG
Then, CG and BF are the solutions.