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Difference between revisions of "2007 USAMO Problems/Problem 2"

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'''Proof''': [[Descartes' Circle Theorem]] states that if a is the curvature of a circle (<math>a=\frac 1{r}</math>, positive for [[externally tangent]], negative for [[internally tangent]]), then we have that  
 
'''Proof''': [[Descartes' Circle Theorem]] states that if a is the curvature of a circle (<math>a=\frac 1{r}</math>, positive for [[externally tangent]], negative for [[internally tangent]]), then we have that  
  
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<div style='text-align:center;'><math>\displaystyle (a+b+c+d)^2=2(a^2+b^2+c^2+d^2)</math></div>
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<div style='text-align:center;'><math>(a+b+c+d)^2=2(a^2+b^2+c^2+d^2)</math></div>
  
 
Solving for a, we get  
 
Solving for a, we get  
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<div style='text-align:center;'><math>a=b+c+d+2 \sqrt{bc+cd+db}</math></div>
 
<div style='text-align:center;'><math>a=b+c+d+2 \sqrt{bc+cd+db}</math></div>
  
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Take the positive root, as the negative root corresponds to externally tangent circle.
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Take the positive root, as the negative root corresponds to internally tangent circle.
  
 
Now clearly, we have <math>b+c+d \le \frac 35</math>, and <math>bc+cd+db\le \frac 3{25}</math>.
 
Now clearly, we have <math>b+c+d \le \frac 35</math>, and <math>bc+cd+db\le \frac 3{25}</math>.

Revision as of 13:44, 8 July 2010

Problem

A square grid on the Euclidean plane consists of all points $(m,n)$, where $m$ and $n$ 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 $\frac{1}{\sqrt{2}}$ between those 3 circles.

Proof: Descartes' Circle Theorem states that if a is the curvature of a circle ($a=\frac 1{r}$, positive for externally tangent, negative for internally tangent), then we have that

$(a+b+c+d)^2=2(a^2+b^2+c^2+d^2)$

Solving for a, we get

$a=b+c+d+2 \sqrt{bc+cd+db}$

Take the positive root, as the negative root corresponds to internally tangent circle.

Now clearly, we have $b+c+d \le \frac 35$, and $bc+cd+db\le \frac 3{25}$. Summing/square root/multiplying appropriately shows that $a \le \frac{3 + 2 \sqrt{3}}5$. Incidently, $\frac{3 + 2\sqrt{3}}5 < \sqrt{2}$, so $a< \sqrt{2}$, $r > \frac 1{\sqrt{2}}$, as desired.

For sake of contradiction, assume that we have a satisfactory placement of circles. Consider 3 circles, $p,\ q,\ r$ where there are no circles in between. By Appolonius' problem, there exists a circle $t$ tangent to $p,\ q,\ r$ externally that is between those 3 circles. Clearly, if we move $p,\ q,\ r$ together, $t$ 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 $\frac{1}{\sqrt{2}}$ that lies between $p,\ q,\ r$. However, any circle with $r>\frac 1{\sqrt{2}}$ 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
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