Obtuse Triangle

An obtuse triangle is a Triangle in which one of the Angles is an Obtuse Angle. (Obviously, only a single Angle in a Triangle can be Obtuse or it wouldn't be a Triangle.) A triangle must be either obtuse, Acute, or Right.

A famous problem is to find the chance that three points picked randomly in a Plane are the Vertices of an obtuse triangle (Eisenberg and Sullivan 1996). Unfortunately, the solution of the problem depends on the procedure used to pick the ``random'' points (Portnoy 1994). In fact, it is impossible to pick random variables which are uniformly distributed in the plane (Eisenberg and Sullivan 1996). Guy (1993) gives a variety of solutions to the problem. Woolhouse (1886) solved the problem by picking uniformly distributed points in the unit Disk, and obtained

$$P_2=1-\left({{4\over\pi^2}-{1\over 8}}\right)= {9\over 8}-{4\over\pi^2}=0.719715\ldots.$$ (1)

The problem was generalized by Hall (1982) to $n$-D Ball Triangle Picking, and Buchta (1986) gave closed form evaluations for Hall's integrals.

Lewis Carroll (1893) posed and gave another solution to the problem as follows. Call the longest side of a Triangle $AB$, and call the Diameter $2r$. Draw arcs from $A$ and $B$ of Radius $2r$. Because the longest side of the Triangle is defined to be $AB$, the third Vertex of the Triangle must lie within the region $ABCA$. If the third Vertex lies within the Semicircle, the Triangle is an obtuse triangle. If the Vertex lies on the Semicircle (which will happen with probability 0), the Triangle is a Right Triangle. Otherwise, it is an Acute Triangle. The chance of obtaining an obtuse triangle is then the ratio of the Area of the Semicircle to that of $ABCA$. The Area of $ABCA$ is then twice the Area of a Sector minus the Area of the Triangle.

$$A_{\rm whole\ figure} = 2\left({4\pi r^2\over 6}\right)-\sqrt{3}\,r^2 = r^2({\textstyle{4\over 3}}\pi-\sqrt{3}\,).$$ (2)

Therefore,

$$P={{\textstyle{1\over 2}}\pi r^2\over r^2({\textstyle{4\over 3}}\pi-\sqrt{3}\,)} = {3\pi\over 8\pi-6\sqrt{3}}=0.63938\ldots.$$ (3)

Let the Vertices of a triangle in $n$-D be Normal (Gaussian) variates. The probability that a Gaussian triangle in $n$-D is obtuse is

$$P_n = {3\Gamma(n)\over\Gamma^2({\textstyle{1\over 2}}n)} \int_0^{1/3} {x^{(n-2)/2}\over(1+x)^n}\,dx$$

$$= {3\Gamma(n)\over\Gamma^2({\textstyle{1\over 2}}n)2^{n-1}}\int_0^{\pi/3} \sin^{n-1}\theta\,d\theta$$

$$= {6\Gamma(n)\,{}_2F_1({\textstyle{1\over 2}}n, n; 1+{\textstyle{1\... ...2}}n; -{\textstyle{1\over 3}})\over 3^{n/2}n\Gamma^2({\textstyle{1\over 2}}n)},$$ (4)

where $\Gamma(n)$ is the Gamma Function and ${}_2F_1(a,b;c;x)$ is the Hypergeometric Function. For Even $n\equiv 2k$,

$$P_{2k}=3\sum_{j=k}^{2k-1}{2k-1\choose j}\left({1\over 4}\right)^j\left({3\over 4}\right)^{2k-1-j}$$ (5)

(Eisenberg and Sullivan 1996). The first few cases are explicitly

$$P_2 = {\textstyle{3\over 4}}=0.75$$ (6)

$$P_3 = 1-{3\sqrt{3}\over 4\pi}=0.586503\ldots$$ (7)

$$P_4 = {\textstyle{15\over 32}}=0.46875$$ (8)

$$P_5 = 1-{9\sqrt{3}\over 8\pi}=0.37975499\ldots.$$ (9)

See also Acute Angle, Acute Triangle, Ball Triangle Picking, Obtuse Angle, Right Triangle, Triangle

References

Buchta, C. ``A Note on the Volume of a Random Polytope in a Tetrahedron.'' Ill. J. Math. 30, 653-659, 1986.

Carroll, L. Pillow Problems & A Tangled Tale. New York: Dover, 1976.

Eisenberg, B. and Sullivan, R. ``Random Triangles $n$ Dimensions.'' Amer. Math. Monthly 103, 308-318, 1996.

Guy, R. K. ``There are Three Times as Many Obtuse-Angled Triangles as There are Acute-Angled Ones.'' Math. Mag. 66, 175-178, 1993.

Hall, G. R. ``Acute Triangles in the $n$-Ball.'' J. Appl. Prob. 19, 712-715, 1982.

Portnoy, S. ``A Lewis Carroll Pillow Problem: Probability on at Obtuse Triangle.'' Statist. Sci. 9, 279-284, 1994.

Wells, D. G. The Penguin Book of Interesting Puzzles. London: Penguin Books, pp. 67 and 248-249, 1992.

Woolhouse, W. S. B. Solution to Problem 1350. Mathematical Questions, with Their Solutions, from the Educational Times, 1. London: F. Hodgson and Son, 49-51, 1886.