2007 USAMO Problems/Problem 2
Contents
Problem
(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?
Solutions
Solution 1
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>viewtopic.php?t=145844 Discussion on AoPS/MathLinks</url>
2007 USAMO (Problems • Resources) | ||
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