Problem:
A triangle has vertices , and . The probability that a randomly chosen point inside the triangle is closer to vertex than to either vertex or vertex can be written as , where and are relatively prime positive integers. Find .
Solution:
Because the perpendicular bisector of a line segment is the set of points equidistant from and , a point inside the triangle is closer to than to either or if and only if it is to the right of the perpendicular bisector of and below the perpendicular bisector of . The equation of the first perpendicular bisector is . The second perpendicular bisector passes through and has slope , and thus it has the equation . The two perpendicular bisectors intersect at . Let be . The area of is the sum of the area of , which is , and the area of , which is . The probability that the randomly chosen point is closer to than to either or is the area of quadrilateral divided by the area of , which is . Thus the required probability is . The requested sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions