2007 USAMO Problems/Problem 2
(Gregory Galperin) A square grid on the Euclidean plane consists of all points , where and are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least 5?
Lemma. Among 3 tangent circles with radius greater than or equal to 5, one can always fit a circle with radius greater than between those 3 circles.
Proof. Descartes' Circle Theorem states that if is the curvature of a circle (, positive for externally tangent, negative for internally tangent), then we have that Solving for , we get Take the positive root, as the negative root corresponds to internally tangent circle.
Now clearly, we have , and . Summing/square root/multiplying appropriately shows that . Incidently, , so , , as desired.
For sake of contradiction, assume that we have a satisfactory placement of circles. Consider 3 circles, where there are no circles in between. By Appolonius' problem, there exists a circle tangent to externally that is between those 3 circles. Clearly, if we move together, must decrease in radius. Hence it is sufficient to consider 3 tangent circles. By lemma 1, there is always a circle of radius greater than that lies between . However, any circle with must contain a lattice point. (Consider placing an unit square parallel to the gridlines in the circle.) That is a contradiction. Hence no such tiling exists.
It is not possible. The proof is by contradiction. Suppose that such a covering family exists. Let denote the disc with center and radius . Start with an arbitrary disc that does not overlap any member of . Then covers no grid point. Take the disc to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then is tangent to at least three discs in . Observe that there must be two of the three tangent discs, say and such that . By the Law of Cosines applied to triangle , which yields and thus Note that because covers no grid point, and because each disc in has radius at least 5. Hence , which gives and thus . Squaring both sides of this inequality yields . This contradiction completes the proof.
Remark: The above argument shows that no covering family exists where each disc has radius greater than . In the other direction, there exists a covering family in which each disc has radius . Take discs with this radius centered at points of the form , where and are integers. Then any grid point is with of one of the centers and the distance between any two centers is at least . The extremal radius of a covering family is unknown.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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