Complex Ratio ProblemDate: 07/31/97 at 18:37:29 From: JdJ Subject: Complex ratio problem Given: If you randomly throw 3 points on a plane, you get a triangle..... What is the probability that the triangle will become obtuse.... The chance of two points being in the exact spot or exactly vertical or horizontal is so slim that it is ruled out. I tried using 2 of the points to make a line segment (line AB). Given that line segment, using pi, you can create a a circle where AB/2 is the radius. If the third point is inside the circle, you get an obtuse triangle. Having the 3rd point on or above or below the circle within pt. A and pt. B will result in a right or acute triangle; however, if the 3rd pt. is not within A and B, then you also get an obtuse triangle. I don't know how to get the ratio, so even if I'm on the right track... Thanks in advance. Date: 08/01/97 at 11:00:22 From: Doctor Anthony Subject: Re: Complex ratio problem These problems can give a variety of answers depending on how you define the way the random triangle is to be formed. You are probably aware of the classic example of the random chord drawn in a circle, and the probability that this chord is longer than the length of side of the inscribed equilateral triangle. You can get answers of 1/4, 1/3 and 1/2 depending on the rule you adopt for drawing the 'random' chord. Similarly, in the problem you quote above, a number of different answers can be expected depending on the rule you adopt for drawing the random triangle. A good example of the rule could be as follows: A circle can always be drawn through three points, so we could say that the three points are to be chosen at random on the circumference of a circle. We select the first point anywhere on the circumference and then draw a diameter and tangent to the circle at that point. For the other two points we take the angle between the tangent and the chord to one of the other points as x, and similarly the angle between the tangent and the chord to the third point as y. In both cases measure x and y anticlockwise from the tangent to the respective chords. Then both x and y are uniformly distributed between 0 and pi. The triangle will be obtuse if x and y are both less than pi/2 (probability of this (1/2)(1/2) = 1/4), or both greater than pi/2, again with probability 1/4. So altogether this probability is 1/4 + 1/4 = 1/2. We also get an obtuse angled triangle if x-y > pi/2, this gives y < x-pi/2, and also if y-x > pi/2 i.e., y > x+pi/2 To find the probability of this we set up a two-dimensional sample space, with the usual x and y axes, and with x varying uniformly from 0 to pi, y varying uniformly from 0 to pi. This gives a square of area pi^2. The regions cut off by the line y > x+pi/2 and the line y < x-pi/2 if put together will form a square of side pi/2, and an area pi^2/4. So the probability of this producing the obtuse angle is (pi^2/4)/pi^2 = 1/4. Now if we call first situation (probability = 1/2) event A and the second situation (with probability 1/4) event B then: P(A or B) = P(A) + P(B) - P(A and B) = 1/2 + 1/4 - P(A and B) However, P(A and B) is zero, since if both x and y < pi/2, or both x and y > pi/2, (event A), it is impossible for x-y or y-x to be greater than pi/2, (event B). We conclude that P(A or B) = 1/2 + 1/4 = 3/4 So a triangle drawn at random has a probability of 3/4 of being obtuse. This result depended on our definition of a 'random' triangle. Other definitions will produce a different result. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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