A188858 Decimal expansion of the solution to a = cot(a/2).

1, 3, 0, 6, 5, 4, 2, 3, 7, 4, 1, 8, 8, 8, 0, 6, 2, 0, 2, 2, 2, 8, 7, 2, 7, 8, 3, 1, 9, 2, 3, 1, 1, 8, 2, 8, 4, 8, 4, 1, 2, 7, 9, 7, 5, 5, 6, 3, 5, 0, 6, 9, 1, 9, 3, 2, 2, 8, 0, 3, 2, 7, 2, 0, 0, 4, 6, 0, 5, 4, 3, 9, 5, 5, 6, 6, 2, 3, 2, 9, 3, 6, 1, 7, 7, 2, 5, 0, 2, 6, 7, 2, 3, 7, 8, 1, 3, 9, 3, 3

Offset

1,2

Comments

This angle is the maximal vertex angle that allows an isosceles triangle to be bisected the most efficient way by means of a circular arc.

The problem of "efficiently" [=least length of boundary] bisecting a triangle was stated and solved for an equilateral triangle by Paul Halmos in his book "Problems for Mathematicians, Young and Old". The angle 'a' given here is the limit vertex angle that separates two ways of bisecting an isosceles triangle: if the vertex angle is less than 'a', then the most efficient way is to bisect the triangle with a circular arc centered on the vertex; if the angle is greater than 'a', then the most efficient way is to use the altitude drawn from the vertex. It can be noticed that the equation cos(a/2) = sqrt(a)*sqrt(cos(a/2))*sqrt(sin(a/2)) equating altitude and arc length (if length of sides = 1), simplifies to a = cot(a/2) if 0<a<pi is assumed.

References

Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991.

Links

Table of n, a(n) for n=1..100.

Jean-François Alcover, Figure showing both ways of efficiently bisecting an isosceles triangle

Example

1.3065423741888... or in degrees : 74.859363796...

Mathematica

RealDigits[a /. FindRoot[a == Cot[a/2], {a, Pi/3}, WorkingPrecision -> 100]][[1]]

Crossrefs

Sequence in context: A062542 A360173 A109693 * A199610 A285871 A322579

Adjacent sequences: A188855 A188856 A188857 * A188859 A188860 A188861

Keyword

nonn,cons

Author

Jean-François Alcover, May 02 2011

Status

approved