2007 USAMO Problems/Problem 2
Problem
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?
Solution
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 a is the curvature of a circle (, positive for externally tangent, negative for internally tangent), then we have that
![$\displaystyle (a+b+c+d)^2=2(a^2+b^2+c^2+d^2)$](http://latex.artofproblemsolving.com/c/a/5/ca52c5d7434091d8f03105806dede7e239f36891.png)
Solving for a, we get
![$a=b+c+d+2 \sqrt{bc+cd+db}$](http://latex.artofproblemsolving.com/1/8/3/183b327e7219cc585ddb9f8737fa738816070cc3.png)
Take the positive root, as the negative root corresponds to externally 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.
See also
2007 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |