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Difference between revisions of "2010 AIME I Problems/Problem 8"

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== Solution 2  ==
 
== Solution 2  ==
  
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When observing the equation
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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.  
−
The desired region consists of 12 boxes, whose lower-left corners are integers solutions of <math>x^2 + y^2 = 25</math>, namely <math>(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).</math> Since the points themselves are symmetric about <math>(0,0)</math>, the boxes are symmetric about <math>\left(\frac12,\frac12\right)</math>. The distance from <math>\left(\frac12,\frac12\right)</math> to the furthest point on a box that lays on an axis, for instance <math>(6,1)</math>, is <math>\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.</math> The distance from <math>\left(\frac12,\frac12\right)</math> to the furthest point on a box in the middle of a quadrant, for instance <math>(5,4)</math>, is <math>\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.</math> The latter is the larger, and is <math>\frac {\sqrt {130}}2</math>, giving an answer of <math>130 + 2 = \boxed{132}</math>.
 
  
−
<math>\lfloor a \rfloor</math>
+
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 21:49, 27 June 2024

Problem

For a real number $a$, let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$. Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$. The region $\mathcal{R}$ is completely contained in a disk of radius $r$ (a disk is the union of a circle and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$, where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$.

Solution

The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$, namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$, the boxes are symmetric about $\left(\frac12,\frac12\right)$. The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box that lays on an axis, for instance $(6,1)$, is $\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.$ The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box in the middle of a quadrant, for instance $(5,4)$, is $\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.$ The latter is the larger, and is $\frac {\sqrt {130}}2$, giving an answer of $130 + 2 = \boxed{132}$.

[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]

Solution 2

When observing the equation $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$, 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 $1$, $x < 6$ and $y < 6$. Note that $\lfloor 5.999...\rfloor = 5$, but if we add more $9's$ after the $5$, it will get infinitely close to $6$, so we can use $6$ as a bounding line. Also, with the same logic, when $x = 6$, $y = 1$ (the equal sign represents as $x$ approaches..., not actually equal to...) So, in quadrant one, we have points $(6,1)$ and $(1,6)$.

Moving to quadrant $2$, we must note that $\lfloor -4.999...\rfloor = -5$, so the circle will not be centered at $(0,0)$. In quadrant 2, $y$ is still positive, so we can have $y = 1$. When $y = 1$, $x = -5$, so we have our next point $(-5,1)$. With this method, other points can be found in quadrant $3$ and $4$.

Additionally, $3 ^2 + 4 ^2 = 5 ^2$, and with the same approaching limit, we know that quadrant $1$ also has lattice points $(4,5)$ and $(5,4)$. 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 $(5,4)$, the "opposite" point is $(-4,-3)$ located in quadrant 3.

Using the distance formula, we find that the distance between the two points is $\sqrt {130}$. 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 $\sqrt {130}$ by $2$ to get $\frac {\sqrt {130}}2$, and $m+n=130 + 2 = \boxed{132}$.

~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

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