Duplicating the Cube

Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
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Duplicating the Cube is one of the three classic unsolved problems of antiquity. They are all known to be unsolvable under the rules used by the Greeks. The problem is to find a cube equal in volume to twice a given cube, or in other words, if the edge of the given cube is one, find a line equal in length to the cube root of two. The cube root of two is1.2599210498948732........ This problem amounts to solving a cubic equation using geometry and is thus the same problem, algebraically, as trisecting the angle.

Since this problem is algebraically the same as trisecting the angle, and since it deals with solid geometry, this problem is the least famous of the three classic problems and gets the least attention from amateurs. Plus, it's perhaps the easiest problem to approximate with high precision.

There are a couple of rational approximations that come close. 5/4 is accurate to within less than one per cent. 63/50 is accurate to within 0.00008. That means if you use 63/50 to duplicate a cube a meter on a side, the duplicate will be accurate to within 0.008 cm or 0.08 millimeters, or 0.003 inches, about the thickness of a sheet of paper.

A slightly better approximation is 10*SQRT(7)/21 = 1.25988157...; the difference is -0.00003947... or about twice as good as 63/50.

An Exact Construction With Marked Ruler

If you mark a ruler you can trisect the angle and duplicate the cube, but this approach was not permitted under ancient Greek rules. Here's a nifty construction using a marked ruler (Robin Hartshorne, Geometry: Euclid and Beyond, Springer, 2000, p. 270; by the way, Hartshorne has a pretty readable explanation of why this and other constructions are impossible, but it will take serious study to understand it.)

Let AO be the edge of the cube to be duplicated, call its length 1. Construct OB perpendicular to AB and OC at a 30 degree angle to OB as shown. Mark off DE on a ruler, with DE = AO

Slide the ruler so it passes through A, and points D and E rest on OB and OC as shown. AD is the edge of the doubled cube (AD = AO times cube root of two).

Proof:

Using similarity, we have x/1 = (x+1)/(m+1) or xm + x = x + 1 or xm = 1
Also, h/1 = sqrt(3)m/(m+1)
Since xm = 1, h = sqrt(3)(1/x)/(1/x+1) = sqrt(3)/(1+x)
By the Pythagorean Theorem, we have x² = 1 + h²
x² = 1 + h² = 1 + (sqrt(3)/(1+x))² = 1 + 3/(1+x)²
Clearing fractions, we get x² (1+x)² = (1 + x)² + 3
Expanding, we have x² + 2x³ + x⁴ = 1 + 2x + x² + 3
Thus 2x³ + x⁴ = 4 + 2x
And x³(2 + x) = 2(2 + x)
Thus x³ = 2

Two Carpenter's Squares

This construction is attributed to Plato (450 B.C.). Construct axes as shown and mark off OA = 1 and OD = 2. Arrange two carpenter's squares so that they pass through A and D and the corners lie on the axes at B and C. Note that the outer edge of the square passes through C and D and the inner edge passes through A and B.

The proof is simple proportion: BO/AO = CO/BO = DO/CO. AO = 1 and DO = 2, and solve for BO.

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Created 6 September 2001, Last Update 21 May 2004

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