Mock AIME 1 2010 Problems/Problem 3
Problem
Let be a line segment of length , and let be the set of all points such that . Find the last three digits of the largest integer less than the area of .
Solution
Let point be on the perpendicular bisector of such that . We know that (because is on the perpendicular bisector of ), so we can draw circle . Do the same on the other side of with point and draw circle . The central angle of in both circles is . Thus, by the properties of inscribed angles, we know that the locus of all points such that is the border of the area enclosed by the two circles. Going out of this area would force , so the inequality is not satified there. Conversely, going inside the area would force , so the inequality is satisfied. Thus, we look for the area enclosed by the two circles in the diagram below.
We know that and are isosceles right triangles, so their leg lengths are . Thus, the radius of the circles is , and so the area of the enclosed region is minus the region enclosed by both circles. To find the area of this doubly enclosed region, we consider the square in the diagram below.
The entire square has an area of , and the circle has area . Thus, each of the sections of the circle not inside the square has an area of . Thus, looking back at the first diagram, we see that the doubly enclosed area (which we would count twice if we just added the areas of the two circles) is composed of two of these regions. Thus, it has area . Subtracting this from the area of the two circles, we see that the area of the enclosed region is . The greatest integer less than this amount is , so our answer is .
See Also
Mock AIME 1 2010 (Problems, Source) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |