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

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== Problem ==
 
== Problem ==
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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?
  
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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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== Solutions ==
  
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== Solution ==
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=== Solution 1 ===
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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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'''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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'''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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<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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''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>
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Take the positive root, as the negative root corresponds to internally tangent circle.
  
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Solving for a, we get
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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. <math>\blacksquare</math>
  
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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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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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Take the positive root, as the negative root corresponds to internally tangent circle.
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=== Solution 2 ===
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It is not possible. The proof is by contradiction. Suppose that such a covering family <math>\mathcal{F}</math> exists. Let <math>D(P,\rho)</math> denote the disc with center <math>P</math> and radius <math>\rho</math>. Start with an arbitrary disc <math>D(O,r)</math> that does not overlap any member of <math>\mathcal{F}</math>. Then <math>D(O,r)</math> covers no grid point. Take the disc <math>D(O,r)</math> to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then <math>D(O,r)</math> is tangent to at least three discs in <math>\mathcal{F}</math>. Observe that there must be two of the three tangent discs, say <math>D(A,a)</math> and <math>D(B,b)</math> such that <math>\angle AOB\leq 120^\circ</math>. By the Law of Cosines applied to triangle <math>ABO</math>,
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<cmath>(a + b)^2\leq (a + r)^2 + (b + r)^2 + (a + r)(b + r),</cmath>
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which yields
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<cmath>ab\leq 3(a + b)r + 3r^2,</cmath>
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and thus
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<cmath>12r^2\geq (a - 3r)(b - 3r).</cmath>
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Note that <math>r < 1/\sqrt{2}</math> because <math>D(O,r)</math> covers no grid point, and <math>(a - 3r)(b - 3r)\geq (5 - 3r)^2</math> because each disc in <math>\mathcal{F}</math> has radius at least 5. Hence <math>2\sqrt{3}r\geq 5 - 3r</math>, which gives <math>5\leq (3 + 2\sqrt{3})r < (3 + 2\sqrt{3})/\sqrt{2}</math> and thus <math>5\sqrt{2} < 3 + 2\sqrt{3}</math>. Squaring both sides of this inequality yields <math>50 < 21 + 12\sqrt{3} < 21 + 12\cdot 2 = 45</math>. This contradiction completes the proof.
  
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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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'''Remark:''' The above argument shows that no covering family exists where each disc has radius greater than <math>(3 + 2\sqrt{3})/\sqrt{2}\approx 4.571</math>. In the other direction, there exists a covering family in which each disc has radius <math>\sqrt{13}/2\approx 1.802</math>. Take discs with this radius centered at points of the form <math>\left(2m + 4n + \frac{1}{2}, 3m + \frac{1}{2}\right)</math>, where <math>m</math> and <math>n</math> are integers. Then any grid point is with <math>\sqrt{13}/2</math> of one of the centers and the distance between any two centers is at least <math>\sqrt{13}</math>. The extremal radius of a covering family is unknown.
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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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----
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{{alternate solutions}}
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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.
 
  
 
== See also ==
 
== See also ==
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* <url>viewtopic.php?t=145844 Discussion on AoPS/MathLinks</url>
  
 
{{USAMO newbox|year=2007|num-b=1|num-a=3}}
 
{{USAMO newbox|year=2007|num-b=1|num-a=3}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 02:41, 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.

Solution 2

It is not possible. The proof is by contradiction. Suppose that such a covering family $\mathcal{F}$ exists. Let $D(P,\rho)$ denote the disc with center $P$ and radius $\rho$. Start with an arbitrary disc $D(O,r)$ that does not overlap any member of $\mathcal{F}$. Then $D(O,r)$ covers no grid point. Take the disc $D(O,r)$ to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then $D(O,r)$ is tangent to at least three discs in $\mathcal{F}$. Observe that there must be two of the three tangent discs, say $D(A,a)$ and $D(B,b)$ such that $\angle AOB\leq 120^\circ$. By the Law of Cosines applied to triangle $ABO$, \[(a + b)^2\leq (a + r)^2 + (b + r)^2 + (a + r)(b + r),\] which yields \[ab\leq 3(a + b)r + 3r^2,\] and thus \[12r^2\geq (a - 3r)(b - 3r).\] Note that $r < 1/\sqrt{2}$ because $D(O,r)$ covers no grid point, and $(a - 3r)(b - 3r)\geq (5 - 3r)^2$ because each disc in $\mathcal{F}$ has radius at least 5. Hence $2\sqrt{3}r\geq 5 - 3r$, which gives $5\leq (3 + 2\sqrt{3})r < (3 + 2\sqrt{3})/\sqrt{2}$ and thus $5\sqrt{2} < 3 + 2\sqrt{3}$. Squaring both sides of this inequality yields $50 < 21 + 12\sqrt{3} < 21 + 12\cdot 2 = 45$. This contradiction completes the proof.

Remark: The above argument shows that no covering family exists where each disc has radius greater than $(3 + 2\sqrt{3})/\sqrt{2}\approx 4.571$. In the other direction, there exists a covering family in which each disc has radius $\sqrt{13}/2\approx 1.802$. Take discs with this radius centered at points of the form $\left(2m + 4n + \frac{1}{2}, 3m + \frac{1}{2}\right)$, where $m$ and $n$ are integers. Then any grid point is with $\sqrt{13}/2$ of one of the centers and the distance between any two centers is at least $\sqrt{13}$. 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.

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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