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

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(''Gregory Galperin'') A [[square]] grid on the [[Cartesian plane|Euclidean plane]] consists of all [[point]]s <math>(m,n)</math>, where <math>m</math> and <math>n</math> are [[integer]]s.  Is it possible to cover all grid points by an infinite family of [[circle|discs]] with non-overlapping interiors if each disc in the family has [[radius]] at least 5?
 
(''Gregory Galperin'') A [[square]] grid on the [[Cartesian plane|Euclidean plane]] consists of all [[point]]s <math>(m,n)</math>, where <math>m</math> and <math>n</math> are [[integer]]s.  Is it possible to cover all grid points by an infinite family of [[circle|discs]] with non-overlapping interiors if each disc in the family has [[radius]] at least 5?
  
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== Solution ==
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== Solutions ==
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===Solution 1===
 
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====Lemma====
 
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Among 3 [[tangent]] circles with radius greater than or equal to 5, one can always fit a circle with radius greater than <math>\frac{1}{\sqrt{2}}</math> between those 3 circles.
 
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====Proof====
 
  
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[[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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=== Solution 1 ===
  
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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>
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'''Lemma.''' Among 3 [[tangent]] circles with radius greater than or equal to 5, one can always fit a circle with radius greater than <math>\frac{1}{\sqrt{2}}</math> between those 3 circles.
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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>
 
  
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''Proof.'' [[Descartes' Circle Theorem]] states that if <math>a</math> 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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<cmath>(a+b+c+d)^2=2(a^2+b^2+c^2+d^2)</cmath>
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Solving for <math>a</math>, we get
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<cmath>a=b+c+d+2 \sqrt{bc+cd+db}</cmath>
 
Take the positive root, as the negative root corresponds to internally tangent circle.
 
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>.
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Summing/[[square root]]/multiplying appropriately shows that <math>a \le \frac{3 + 2 \sqrt{3}}5</math>. Incidently, <math>\frac{3 + 2\sqrt{3}}5 < \sqrt{2}</math>, so <math>a< \sqrt{2}</math>, <math>r > \frac 1{\sqrt{2}}</math>, as desired.
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Summing/[[square root]]/multiplying appropriately shows that <math>a \le \frac{3 + 2 \sqrt{3}}5</math>. Incidently, <math>\frac{3 + 2\sqrt{3}}5 < \sqrt{2}</math>, so <math>a< \sqrt{2}</math>, <math>r > \frac 1{\sqrt{2}}</math>, as desired. <math>\blacksquare</math>
  
 
For sake of [[contradiction]], assume that we have a satisfactory placement of circles. Consider 3 circles, <math>p,\ q,\ r</math> where there are no circles in between. By [[Appolonius' problem]], there exists a circle <math>t</math> tangent to <math>p,\ q,\ r</math> externally that is between those 3 circles. Clearly, if we move <math>p,\ q,\ r</math> together, <math>t</math> 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 <math>\frac{1}{\sqrt{2}}</math> that lies between <math>p,\ q,\ r</math>. However, any circle with <math>r>\frac 1{\sqrt{2}}</math> 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.
 
For sake of [[contradiction]], assume that we have a satisfactory placement of circles. Consider 3 circles, <math>p,\ q,\ r</math> where there are no circles in between. By [[Appolonius' problem]], there exists a circle <math>t</math> tangent to <math>p,\ q,\ r</math> externally that is between those 3 circles. Clearly, if we move <math>p,\ q,\ r</math> together, <math>t</math> 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 <math>\frac{1}{\sqrt{2}}</math> that lies between <math>p,\ q,\ r</math>. However, any circle with <math>r>\frac 1{\sqrt{2}}</math> 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.
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{{alternate solutions}}
  
 
== See also ==
 
== See also ==

Revision as of 01:49, 7 August 2014

Problem

(Gregory Galperin) 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?

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 $\frac{1}{\sqrt{2}}$ 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. $\blacksquare$

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.

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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Problem 3
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All USAMO Problems and Solutions

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