Date: 09/14/97 at 14:05:03 From: Joe Subject: IMP Math program, word problem I am totally stuck on this problem, which is a Problem Of The Week in IMP II. Suppose that you have a straight pipe cleaner of a given length. The two ends are points A and C. Now suppose that the pipe cleaner is bent into 2 portions, with one portion twice as long as the other. Label the place where it's bent as B, so that the segment from B to C is twice as long as from A to B. Then a fourth point, X, is chosen at random somewhere on the longer section, and the pipe cleaner is bent at that portion, too. Can the 3 segments of the pipe cleaner be made into a triangle by changing the angles at the two bends? As you might suppose, the answer depends on the location of point X. My question to you is: if point X is chosen at random along the section from B to C, what is the probability that the 3 segments of the pipe cleaner can be made into a triangle? Thanks.
Date: 09/14/97 at 18:48:00 From: Doctor Anthony Subject: Re: IMP Math program, word problem You use the property of a triangle that any two sides must be greater than the third side. |<--- 1 --->|<-------- 2 -------->| A-----------B----|------------------C x D 2-x If you look at the above figure, suppose the extra bend is made at D, a distance x from the point B, as shown. Then for a triangle to be possible in this configuration with D closer to B than C we require 1 + x > 2-x 2x > 1 and so x > 1/2 Similarly, if D is closer to C than B the condition for a triangle to be possible is: x < 1 + (2-x) 2x < 3 so x < 3/2 Putting these two inequalities together we get 1/2 < x < 3/2 Now x can vary with uniform probability between 0 and 2, and its permitted range, 1/2 to 3/2, is exactly half this range. So the probability that a triangle can be formed is 1/2. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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