Problem:
The set of points in -dimensional coordinate space that lie in the plane whose coordinates satisfy the inequalities
forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form , where and are positive integers and is not divisible by the square of any prime. Find .
Solution:
We can rewrite as i.e. . Similarly, if and only if .
The trick is to consider such that:
and . This is equivalent to .
We split our analysis by quadrant. Since , it is impossible for all three to be negative. If two are negative, then we must have and . Then, the only condition is , so the region is unbounded. Similarly, if one is negative, we must have and and the only restriction is , so the region is unbounded.
Therefore, the bounded region exists in the first quadrant, where we have an equilateral triangle formed by the points and and the only additional restriction is that . In the first quadrant, this is equivalent to , which covers of the triangle. Hence, the final area is:
and so the answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions