Difference between revisions of "2007 USAMO Problems/Problem 2"
5849206328x (talk | contribs) (→Solutions: official solution) |
|||
(9 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | (''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 == | |
− | == Solution == | + | === 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. | ||
+ | |||
+ | ''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. | ||
+ | |||
+ | 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. <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. | ||
+ | |||
+ | === Solution 2 === | ||
+ | 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>, | ||
+ | <cmath>(a + b)^2\leq (a + r)^2 + (b + r)^2 + (a + r)(b + r),</cmath> | ||
+ | which yields | ||
+ | <cmath>ab\leq 3(a + b)r + 3r^2,</cmath> | ||
+ | and thus | ||
+ | <cmath>12r^2\geq (a - 3r)(b - 3r).</cmath> | ||
+ | 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. | ||
+ | |||
+ | '''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. | ||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | == See also == | ||
+ | * <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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 03:41, 7 August 2014
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.
Solution 2
It is not possible. The proof is by contradiction. Suppose that such a covering family exists. Let
denote the disc with center
and radius
. Start with an arbitrary disc
that does not overlap any member of
. Then
covers no grid point. Take the disc
to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then
is tangent to at least three discs in
. Observe that there must be two of the three tangent discs, say
and
such that
. By the Law of Cosines applied to triangle
,
which yields
and thus
Note that
because
covers no grid point, and
because each disc in
has radius at least 5. Hence
, which gives
and thus
. Squaring both sides of this inequality yields
. This contradiction completes the proof.
Remark: The above argument shows that no covering family exists where each disc has radius greater than . In the other direction, there exists a covering family in which each disc has radius
. Take discs with this radius centered at points of the form
, where
and
are integers. Then any grid point is with
of one of the centers and the distance between any two centers is at least
. 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) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.