Squaring the Circle
Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
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Squaring the circle, constructing a square equal in area to a given circle, is one of the three classic unsolved problems of antiquity. They are all known to be unsolvable under the rules used by the Greeks. If the circle has unit radius, the problem amounts to constructing a line with length square root of pi. The square root of pi is 1.7724538509055159.......
| You can also get there by constructing a line of length pi, then taking its square root. Here's how you get a square root graphically. |  |
In the 19th century mathematicians determined that all three problems were unsolvable. Pi is a transcendental number, one that cannot be calculated by any finite sequence of algebraic operations. Geometric constructions are just graphic equivalents of algebraic operations, and a limited set of them at that, so it follows that no finite series of geometric steps can ever calculate pi. That doesn't stop people from trying. There are approximate constructions, however, some very good.
Some Approximations
Not elegant, but accurate: √(3) + 4/99 = 1.7724548479729176, which differs from the square root of pi by only 0.000001. Square root of 3 is easy. If you construct a hexagon in a unit circle, it's the distance between any two alternate vertices. The 4/99 is the clunky part, but dividing a line segment into an arbitrary number of parts is a standard geometric construction.
If you can do this (or any other) approximation accurately enough to show the difference between the approximate value and the true value, you can lament the lack of exact constructions.
Another close approximation is 9 √(14)/19 = 1.772364.... a difference of 0.00009 from the true value.
Still another, shown below, is √(30)/10 + √(6)/2 = 1.772467..., differing from the true value of 1.772454... by 0.000013.
Some Graphical Approximations
| Draw a circle with radius OA=1. Construct AB tangent to the circle with AB=3. Construct right triangle COD with angle O=30 degrees. BD is approximately equal to pi. |  |
If OC = 1 then CD = tan 30 = √(3)/3. So by the Pythagorean Theorem, BD2 = 22 + (3-√(3)/3)2 = 4 + 9 - 2√(3) + 1/3 = 40/3 - 2√(3) and BD = 3.14153, differing from pi by 0.00006.
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Draw a circle with radius 1. Construct AB = 3 times diameter AOX. Construct OU at a 30 degree angle to OX, then draw UV perpendicular to OX. BV is approximately the circumference of the circle. |
Proof: OV = cos 30 = √(3)/2, so AOV = 1 + √(3)/2. BV = √(62 + (1 + √(3)/2))2) = √(36 + 1 + 2√(3)/2) + 3/4) = √(151/4 + √(3)) = 6.28347. 2 pi = 6.28318.
| This approximation is based on the fact that 355/113 = 3.1415929, differing from pi by only 0.0000003. Also, 355/113 = 3+16/113 = 3+42/(72+82).
AH=16/113, the fractional part of the approximation. |
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Draw a circle with radius AC=1 and draw diameter DC perpendicular to AB. Locate E so that EC = 7/8. Thus AE2=(72+82)/82 and AE=√(72+82)/8.
Locate F so that AF = 1/2. Draw FG parallel to CD. Draw EG. By similarity we have FG/aF = EC/aE, so FG = AFxEC/aE = (1/2)(7/8)√(82/(72+82)) = (7/2)/√(72+82).
Draw FH parallel to EG. Again by similarity we have AH/aF = AG/aE, and AH=AFxAG/aE. Also AG/fG = AC/cE = 8/7, so AG = (8/7)FG. Thus we have AH = AFx(8/7)FG/aE = (1/2)(8/7)[(7/2)/√(72+82)][8/√(72+82)] = 42/(72+82).
Somebody with way too much free time figured this one out.
| Construct a circle with radius OA=1 and draw a line through AE perpendicular to radius OA.
Locate B, where AB=11/5, and C, where BC=1/5. Extend OA and locate D so that AD=OB. Draw DE parallel to OC. DE is approximately 2pi. |
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By the Pythagorean Theorem, OB2=1+121/25=146/25. OC=12/5 and OA=1=5/5, so triangle COA has the well known Pythagorean proportions 5:12:13 and OC=13/5. By similarity, ED/aD=OC/oA=13/5. Also AD=OB=√(146)/5. So ED = ADxOC/oA = (13/5)(√(146)/5) = (13/25)√(146) = 6.2831839, differing from the true value of 6.2831853 by 0.0000014.
| With O as center, construct semicircle AB and call the radius AO=1.
Locate points D, E and F such that OD = 3/5, OE = 1/2 and OF = 3/2. Draw line GOH perpendicular to AOF. GH is very nearly the square root of pi. (GH looks longer than AB, but that is an optical illusion, as you can check easily.) |
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The radius of semicircle DE is (1/2+3/5)/2 = 11/20, and the center of the semicircle lies 1/20 left of O. Hence OG2 = (11/20)2 - (1/20)2. Thus OG = √(120)/20 = √(30)/10
The radius of semicircle AF is (1+3/2)/2 = 5/4, and and the center of the semicircle lies 1/4 right of O. Hence OH2 = (5/4)2 - (1/4)2. Thus OH = √(24)/4 = √(6)/2
Thus GH = √(30)/10 + √(6)/2 = 1.772467..., differing from the true value of 1.772454... by 0.000013.
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Created 8 September 2001, Last Update 20 January 2020
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