Difference between revisions of "2010 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
− | For a real number <math>a</math>, let <math>\lfloor a \rfloor</math> | + | For a real number <math>a</math>, let <math>\lfloor a \rfloor</math> denote the [[ceiling function|greatest integer]] less than or equal to <math>a</math>. Let <math>\mathcal{R}</math> denote the region in the [[coordinate plane]] consisting of points <math>(x,y)</math> such that <math>\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25</math>. The region <math>\mathcal{R}</math> is completely contained in a [[disk]] of [[radius]] <math>r</math> (a disk is the union of a [[circle]] and its interior). The minimum value of <math>r</math> can be written as <math>\frac {\sqrt {m}}{n}</math>, where <math>m</math> and <math>n</math> are integers and <math>m</math> is not divisible by the square of any prime. Find <math>m + n</math>. |
== Solution == | == Solution == | ||
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); </asy></center> | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); </asy></center> | ||
− | == | + | == Solution 2 == |
− | + | When observing the equation <math>\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25</math>, it is easy to see that it is the graph of a circle. So, we can draw a coordinate plane and find some points. | |
+ | |||
+ | In quadrant <math>1</math>, <math>x < 6 </math> and <math>y < 6 </math>. Note that <math>\lfloor 5.999...\rfloor = 5</math>, but if we add more <math>9's</math> after the <math>5</math>, it will get infinitely close to <math>6</math>, so we can use <math>6</math> as a bounding line. Also, with the same logic, when <math>x = 6</math>, <math>y = 1</math> (the equal sign represents as <math>x</math> approaches..., not actually equal to...) So, in quadrant one, we have points <math>(6,1)</math> and <math>(1,6)</math>. | ||
+ | |||
+ | Moving to quadrant <math>2</math>, we must note that <math>\lfloor -4.999...\rfloor = -5</math>, so the circle will not be centered at <math>(0,0)</math>. In quadrant 2, <math>y</math> is still positive, so we can have <math>y = 1</math>. When <math>y = 1</math>, <math>x = -5</math>, so we have our next point <math>(-5,1)</math>. With this method, other points can be found in quadrant <math>3</math> and <math>4</math>. | ||
+ | |||
+ | Additionally, <math>3 ^2 + 4 ^2 = 5 ^2</math>, and with the same approaching limit, we know that quadrant <math>1</math> also has lattice points <math>(4,5)</math> and <math>(5,4)</math>. We need a point that passes through the center of the circle (we don't actually need to find the center). If we focus on <math>(5,4)</math>, the "opposite" point is <math>(-4,-3)</math> located in quadrant 3. | ||
+ | |||
+ | Using the distance formula, we find that the distance between the two points is <math>\sqrt {130}</math>. Since the line connected from those two points passes through the center of the circle, it is the diameter. So, the radius can be found by diving <math>\sqrt {130}</math> by <math>2</math> to get <math>\frac {\sqrt {130}}2</math>, and <math>m+n=130 + 2 = \boxed{132}</math>. | ||
+ | |||
+ | ~hwan | ||
== See Also == | == See Also == |
Latest revision as of 22:49, 27 June 2024
Contents
Problem
For a real number , let
denote the greatest integer less than or equal to
. Let
denote the region in the coordinate plane consisting of points
such that
. The region
is completely contained in a disk of radius
(a disk is the union of a circle and its interior). The minimum value of
can be written as
, where
and
are integers and
is not divisible by the square of any prime. Find
.
Solution
The desired region consists of 12 boxes, whose lower-left corners are integers solutions of , namely
Since the points themselves are symmetric about
, the boxes are symmetric about
. The distance from
to the furthest point on a box that lays on an axis, for instance
, is
The distance from
to the furthest point on a box in the middle of a quadrant, for instance
, is
The latter is the larger, and is
, giving an answer of
.
![[asy]import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); filldraw((-3,4)--(-2,4)--(-2,5)--(-3,5)--cycle,evevff,blue); filldraw((3,4)--(4,4)--(4,5)--(3,5)--cycle,evevff,blue); filldraw((4,3)--(5,3)--(5,4)--(4,4)--cycle,evevff,blue); filldraw((5,0)--(6,0)--(6,1)--(5,1)--cycle,evevff,blue); filldraw((4,-3)--(5,-3)--(5,-2)--(4,-2)--cycle,evevff,blue); filldraw((3,-3)--(3,-4)--(4,-4)--(4,-3)--cycle,evevff,blue); filldraw((0,-5)--(1,-5)--(1,-4)--(0,-4)--cycle,evevff,blue); filldraw((-3,-4)--(-2,-4)--(-2,-3)--(-3,-3)--cycle,evevff,blue); filldraw((-4,-3)--(-3,-3)--(-3,-2)--(-4,-2)--cycle,evevff,blue); filldraw((-4,3)--(-3,3)--(-3,4)--(-4,4)--cycle,evevff,blue); filldraw((-5,0)--(-4,0)--(-4,1)--(-5,1)--cycle,evevff,blue); filldraw((0,6)--(0,5)--(1,5)--(1,6)--cycle,evevff,blue); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); draw(circle((0,0),5),linewidth(1.6)); draw(circle((0.5,0.5),5.7),linetype("2 2")); draw((-3,4)--(-2,4),zzttqq); draw((-2,4)--(-2,5),zzttqq); draw((-2,5)--(-3,5),zzttqq); draw((-3,5)--(-3,4),zzttqq); draw((3,4)--(4,4),zzttqq); draw((4,4)--(4,5),zzttqq); draw((4,5)--(3,5),zzttqq); draw((3,5)--(3,4),zzttqq); draw((4,3)--(5,3),zzttqq); draw((5,3)--(5,4),zzttqq); draw((5,4)--(4,4),zzttqq); draw((4,4)--(4,3),zzttqq); draw((5,0)--(6,0),zzttqq); draw((6,0)--(6,1),zzttqq); draw((6,1)--(5,1),zzttqq); draw((5,1)--(5,0),zzttqq); draw((4,-3)--(5,-3),zzttqq); draw((5,-3)--(5,-2),zzttqq); draw((5,-2)--(4,-2),zzttqq); draw((4,-2)--(4,-3),zzttqq); draw((3,-3)--(3,-4),zzttqq); draw((3,-4)--(4,-4),zzttqq); draw((4,-4)--(4,-3),zzttqq); draw((4,-3)--(3,-3),zzttqq); draw((0,-5)--(1,-5),zzttqq); draw((1,-5)--(1,-4),zzttqq); draw((1,-4)--(0,-4),zzttqq); draw((0,-4)--(0,-5),zzttqq); draw((-3,-4)--(-2,-4),zzttqq); draw((-2,-4)--(-2,-3),zzttqq); draw((-2,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-4),zzttqq); draw((-4,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-2),zzttqq); draw((-3,-2)--(-4,-2),zzttqq); draw((-4,-2)--(-4,-3),zzttqq); draw((-4,3)--(-3,3),zzttqq); draw((-3,3)--(-3,4),zzttqq); draw((-3,4)--(-4,4),zzttqq); draw((-4,4)--(-4,3),zzttqq); draw((-5,0)--(-4,0),zzttqq); draw((-4,0)--(-4,1),zzttqq); draw((-4,1)--(-5,1),zzttqq); draw((-5,1)--(-5,0),zzttqq); draw((0,6)--(0,5),zzttqq); draw((0,5)--(1,5),zzttqq); draw((1,5)--(1,6),zzttqq); draw((1,6)--(0,6),zzttqq); dot((0,5),ds); dot((3,4),ds); dot((4,3),ds); dot((5,0),ds); dot((4,-3),ds); dot((3,-4),ds); dot((0,-5),ds); dot((-3,-4),ds); dot((-4,-3),ds); dot((-4,3),ds); dot((-3,4),ds); dot((-2,4),ds); dot((-2,5),ds); dot((-3,5),ds); dot((4,4),ds); dot((4,5),ds); dot((3,5),ds); dot((5,3),ds); dot((5,4),ds); dot((4,4),ds); dot((6,0),ds); dot((6,1),ds); dot((5,1),ds); dot((5,-3),ds); dot((5,-2),ds); dot((4,-2),ds); dot((3,-3),ds); dot((4,-4),ds); dot((4,-3),ds); dot((1,-5),ds); dot((1,-4),ds); dot((0,-4),ds); dot((-2,-4),ds); dot((-2,-3),ds); dot((-3,-3),ds); dot((-3,-2),ds); dot((-4,-2),ds); dot((-3,3),ds); dot((-3,4),ds); dot((-4,4),ds); dot((-5,0),ds); dot((-4,0),ds); dot((-4,1),ds); dot((-5,1),ds); dot((0,6),ds); dot((1,5),ds); dot((1,6),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/f/f/4/ff43639961c43d52b0e31cf8627eaafbfbc3e27c.png)
Solution 2
When observing the equation , it is easy to see that it is the graph of a circle. So, we can draw a coordinate plane and find some points.
In quadrant ,
and
. Note that
, but if we add more
after the
, it will get infinitely close to
, so we can use
as a bounding line. Also, with the same logic, when
,
(the equal sign represents as
approaches..., not actually equal to...) So, in quadrant one, we have points
and
.
Moving to quadrant , we must note that
, so the circle will not be centered at
. In quadrant 2,
is still positive, so we can have
. When
,
, so we have our next point
. With this method, other points can be found in quadrant
and
.
Additionally, , and with the same approaching limit, we know that quadrant
also has lattice points
and
. We need a point that passes through the center of the circle (we don't actually need to find the center). If we focus on
, the "opposite" point is
located in quadrant 3.
Using the distance formula, we find that the distance between the two points is . Since the line connected from those two points passes through the center of the circle, it is the diameter. So, the radius can be found by diving
by
to get
, and
.
~hwan
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.