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A194940
The Square Peg in the Round Hole constant.
2
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
OFFSET
0,1
COMMENTS
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.
REFERENCES
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.
LINKS
Eric Weisstein's World of Mathematics, Circular Segment.
FORMULA
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
EXAMPLE
0.28445855040980187815920101812693174533005283078946269804587750...
MATHEMATICA
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 *)
PROG
(PARI) 4*acos(sqrt(Pi)/2) - sqrt(Pi*(4-Pi)) \\ G. C. Greubel, Mar 28 2017
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
STATUS
approved