A075549 Decimal expansion of 9 - 12*log(2).

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

Offset: 0

Author

Zak Seidov, Oct 11 2002

Keywords

• nonn
• cons

Comments

Choose two numbers at random from the interval [0,1] (using a uniform distribution). This will give three subintervals of lengths a, b and c. What is the probability that there is a triangle with sides a, b and c? Given that a such a triangle exists, what is the probability that it is obtuse? Answer: Probability that there is a triangle is 1/4. Probability for this triangle to be obtuse is = 9 - 12 * log(2) = 0.68223... .

The problem proposed by Singmaster (1973) is to the calculate the probability that the three segments can form an obtuse triangle. Its solution is 1/4 of this constant, i.e., 9/4 - 3*log(2) = 0.170558... . - Amiram Eldar, May 20 2023

Examples

			0.682233833280656286993214542501881183...
		

References

• Paul J. Nahin, Digital Dice: Computational Solutions to Practical Probability Problems, Princeton University Press, 2008, pp. 31-32.

Links

• Zak Levi, The Problem #28.
• Robert Nelson, Pictures, probability, and paradox, The Two-Year College Mathematics Journal, Vol. 10, No. 3 (1979), pp. 182-190; alternative link. See Problem 3, p. 184.
• David Singmaster, Problem 858, Mathematics Magazine, Vol. 46, No. 1 (1973), p. 43; Probability of an Obtuse Triangle, Solution to Problem 85 by Major G. C. Holterman, ibid., Vol. 46, No. 5 (1973), pp. 294-295.
• Index entries for transcendental numbers.

Crossrefs

Cf. A002162 (log(2)).

Programs

Mathematica

RealDigits[9 - 12Log[2], 10, 100][[1]] (* Alonso del Arte, Nov 02 2013 *)

PARI

9 - 12*log(2) \\ Michel Marcus, Oct 14 2020

Extensions

Offset corrected by R. J. Mathar, Feb 05 2009