Saturday, July 30, 2005
It's math puzzle time again.
Good Lord, it seems like forever since we've had a math-type puzzle so, without further ado: Given two points selected entirely at random on a stick, if you break the stick at those two points, what are the odds that the resulting three pieces can form a triangle?
Solutions accepted starting Saturday at noon. Until then, keep it to yourself.
OPEN FOR SOLUTIONS. Once you solve this one, we'll move on to the next one.
THE ANSWER: The correct answer is 1/4, and the best way to figure this out is to solve the problem in two stages.
Say the stick is 1 "unit" long. First, it should be obvious that you can build a triangle iff (if and only if) none of the resulting three pieces is longer than .5 units. And the only way that can happen is if the two randomly-selected points lie on opposite sides of centre, and the chance of that is 1/2. So far, so good? But, of course, that's not enough. (This first observation means that you can at least claim that the probability is no better than 1/2.)
In addition to both points lying on opposite sides of centre, it should be obvious that they can't be further apart than .5 units as well. If your randomly-selected points were, say, .1 and .9, it's clear you'd be hosed. So this means you have to further calculate the odds that the two points are within .5 of each other.
Effectively, this is solving a new, second problem. The first problem was, what are the odds that the points are on opposite sides of centre? And we've already solved that: 1/2.
The second problem: Given that the points are already on opposite sides, what are the odds that they're within .5 of each other? Turns out, that's also 1/2. (That second part is left as an exercise for the reader, but Jeff in the comments section does a pretty good job of explaining it.)
Ergo, the final probability is the product of those two values: 1/2 x 1/2 = 1/4.
(The ambitious might want to read more here.)