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A194940
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The Square Peg in the Round Hole constant.
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2
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2, 8, 4, 4, 5, 8, 5, 5, 0, 4, 0, 9, 8, 0, 1, 8, 7, 8, 1, 5, 9, 2, 0, 1, 0, 1, 8, 1, 2, 6, 9, 3, 1, 7, 4, 5, 3, 3, 0, 0, 5, 2, 8, 3, 0, 7, 8, 9, 4, 6, 2, 6, 9, 8, 0, 4, 5, 8, 7, 7, 5, 0, 0, 3, 0, 1, 1, 8, 9, 8, 9, 5, 8, 4, 8, 2, 9, 2, 3, 9, 7, 5, 3, 8, 6, 9, 4, 7, 2, 3, 6, 0, 6, 2, 2, 7, 2, 2, 1, 4, 6, 7, 6, 4, 6, 1, 7, 2, 4, 4, 7
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OFFSET
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0,1
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COMMENTS
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Given a unit circle and a square of equal area, what is the amount of the square peg shavings (or filings) which would allow the peg to be inserted into the circle? It turns out to be not quite two sevenths.
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REFERENCES
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Daniel Zwillinger, Editor, CRC Standard Mathematical Tables and Formulae, 31st Edition, Chapman & Hall/CRC, Boca Raton, Section 4.6.6 Circles, page 334 & figure 4.18, 2003.
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LINKS
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FORMULA
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Area = 4*arccos(sqrt(Pi)/2) - sqrt(Pi*(4-Pi)).
Area = Pi + sqrt(2*Pi(2 - sqrt(Pi*(4 - Pi)))) - 4*arcsin(sqrt(Pi/4)). - Robert G. Wilson v, Mar 19 2014
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EXAMPLE
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0.28445855040980187815920101812693174533005283078946269804587750...
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MATHEMATICA
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RealDigits[ 4*ArcCos[ Sqrt[Pi]/2] - Sqrt[ Pi(4 - Pi)], 10, 111][[1]]
RealDigits[Pi + Sqrt[ 2Pi(2 - Sqrt[Pi (4 - Pi)])] - 4 ArcSin[ Sqrt[Pi/4]], 10, 111][[1]] (* Robert G. Wilson v, Sep 20 2011 *)
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PROG
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(PARI) 4*acos(sqrt(Pi)/2) - sqrt(Pi*(4-Pi)) \\ G. C. Greubel, Mar 28 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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