Problem:
Three numbers, , are drawn randomly and without replacement from the set . Three other numbers, , are then drawn randomly and without replacement from the remaining set of 997 numbers. Let be the probability that, after a suitable rotation, a brick of dimensions can be enclosed in a box of dimensions , with the sides of the brick parallel to the sides of the box. If is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
Solution:
Since we may rotate the brick before we attempt to place it in the box, we may assume that and that . The brick will then fit in the box if and only if , and . Because each selection of dimensions is equally likely, there is no loss of generality in assuming the brick and box dimensions are selected from the set . There are ways to select the dimensions of a brick-box pair from .
If the brick does fit inside the box, then we must have and . In addition, we must have and , so .
If , then . Taking to be either or will result in a pair of dimensions for which the brick fits in the box.
If , then taking to be or will result in a box that can hold the brick. Thus there are ways to select the two sets of dimensions from so that the brick fits inside the box. It follows that the probability that the brick will fit inside the box is . The sum of the numerator and denominator is .
The problems on this page are the property of the MAA's American Mathematics Competitions