Fix a line segment L in the plane.
Among all triangles which have L as longest side one is chosen at random.
What is the probability it is obtuse?
Like other problems of this type (choosing a random compound object) there may be different answers depending on how one chooses it at random. One way might be to choose a random, with uniform distribution, side length shorter than the given length, and then choose a random angle with the first line, between zero and the angle that would make the triangle isoscelese. Another would be to choose random numbers lower than the length of L and keep doing so until a set is chosen to that the two new lengths add up to more than the length of L.
But I will assume that the random triangle is produced by selecting a point for the third vertex at random from all those that would make L the longest side.
To make L the longest side, the third vertex must be within the intersection of two circles: one centered at each of the endpoints of L, both with radius that is the length of L. The angle is obtuse if the point falls within a circle with L as the diameter centered on the midpoint of L; this circle is completely within the first area given.
As we are going for a ratio, it's easier to consider only one side of the setup (i.e., one side of line L), as the other side has the same ratio of included to excluded area.
The arc defined by one endpoint of L with its bounding radii encloses an area of pi*L^2 / 6, using L as the length of segment L. If we add these two equal areas, the sum is pi*L^2 / 3. But it includes the area of the equilateral triangle of their intersection twice; that area is L^2*sqrt(3) / 4. So the total area where the third vertex might be located so that there is a triangle whose largest side is L is L^2*(pi/3  sqrt(3)/4)
The area where the angle is obtuse is the semicircle with radius L/2. It has area pi*L^2 / 4 / 2 = pi*L^2 / 8.
The ratio, and therefore probability, is (pi/8) / (pi/3  sqrt(3)/4). You might want to simplify this as 3*pi/(8*pi  6*sqrt(3)). Each evaluates to .63938256071....
Another method of choosing the "random triangle" would be to choose an angle for the left hand endpoint and another for the right hand endpoint. Each should be under 90° so that the given line is indeed the longest, and further, 180AB (the size of the angle opposite the given line) should not be both larger than A and larger than B. That seems a daunting challenge in setting up bounds for a double definite integral. But I have no doubt the answer would be different from that obtained for the result based on uniform distribution choice of third vertex.
Another randomization:
WLOG, let L be of unit length.
Let the x and y axes represent the lengths of the other two sides.
The sample space is the triangle with vertices (0,1), 1=(1,0), (1,1), in which the other two sides add up to at least L, but neither exceeds 1.
to be obtuse, x^2+y^2 must be less than 1. This is a quarter circle of radius 1, but the portion outside the sample space, with area 1/2, must be excluded, to the portion within the sample space has area pi/4  1/2. That constitutes (pi/4  1/2)/(1/2) of the sample space, or .57079632679..., which is then the probability assuming the other side lengths are what are randomized.

Posted by Charlie
on 20171025 15:02:17 