Duplicating the Cube

Steven Dutch, Natural and Applied Sciences, Universityof Wisconsin - Green Bay
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Duplicating the Cube isone of the three classic unsolved problems of antiquity.  They are allknown to be unsolvable under the rules used by theGreeks. The problemis to find a cube equal in volume to twice a given cube, or in other words, ifthe edge of the given cube is one, find a line equal in length to the cube rootof two. The cube root of two is1.2599210498948732........ This problem amountsto 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, andsince it deals with solid geometry, this problem is the least famous of thethree classic problems and gets the least attention from amateurs. Plus, it'sperhaps the easiest problem to approximate with high precision.

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

A slightly better approximation is 10*SQRT(7)/21 =1.25988157...; the differenceis  -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, butthis approach was not permitted under ancient Greek rules. Here's a niftyconstruction using a marked ruler (Robin Hartshorne, Geometry: Euclid andBeyond, Springer, 2000, p. 270; by the way, Hartshorne has a pretty readableexplanation of why this and other constructions are impossible, but it will takeserious study to understand it.)

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

Slide the ruler so it passes through A, and points D and E rest on OB and OCas 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 20 January 2020

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