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? | ||
− | == | + | == Solutions == |
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− | + | === 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 <math>\frac{1}{\sqrt{2}}</math> between those 3 circles. | |
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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 | ||
+ | <cmath>(a+b+c+d)^2=2(a^2+b^2+c^2+d^2)</cmath> | ||
+ | Solving for <math>a</math>, we get | ||
+ | <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>. | ||
− | 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. | + | 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
Contents
[hide]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) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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