2010 AIME I Problems/Problem 8
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[hide]Problem
For a real number , let denote the greatest integer less than or equal to . Let denote the region in the coordinate plane consisting of points such that . The region is completely contained in a disk of radius (a disk is the union of a circle and its interior). The minimum value of can be written as , where and are integers and is not divisible by the square of any prime. Find .
Solution
The desired region consists of 12 boxes, whose lower-left corners are integers solutions of , namely Since the points themselves are symmetric about , the boxes are symmetric about . The distance from to the furthest point on a box that lays on an axis, for instance , is The distance from to the furthest point on a box in the middle of a quadrant, for instance , is The latter is the larger, and is , giving an answer of .
Solution 2
When observing the equation , it is easy to see that it is the graph of a circle. So, we can draw a coordinate plane and find some points.
In quadrant , and . Note that , but if we add more after the , it will get infinitely close to , so we can use as a bounding line. Also, with the same logic, when , (the equal sign represents as approaches..., not actually equal to...) So, in quadrant one, we have points and .
Moving to quadrant , we must note that , so the circle will not be centered at . In quadrant 2, is still positive, so we can have . When , , so we have our next point . With this method, other points can be found in quadrant and .
Additionally, , and with the same approaching limit, we know that quadrant also has lattice points and . We need a point that passes through the center of the circle (we don't actually need to find the center). If we focus on , the "opposite" point is located in quadrant 3.
Using the distance formula, we find that the distance between the two points is . Since the line connected from those two points passes through the center of the circle, it is the diameter. So, the radius can be found by diving by to get , and .
~hwan
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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