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Difference between revisions of "2010 AMC 12A Problems/Problem 13"

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== Solution ==
 
== Solution ==
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We can see that the function <math>xy = k</math> is symmetric to the line <math>y=x</math>, and the distance to the origin approaches infinity as the function approaches either the <math>x</math>-axis or the <math>y</math>-axis. Therefore, assuming that graphs don't intersect, the point at which the function <math>xy = k</math> is closest to the function <math>x^2+y^2=k^2</math> (which is clearly a circle) is when <math>|x|=|y|</math>. It follows that at these points, the magnitude of the <math>x</math> and <math>y</math> values for the function <math>xy = k</math> will be <math>\sqrt{|k|}</math>.
 
  
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All of these points are found at angles <math>\frac{\pi}{4}</math>, <math>\frac{3\pi}{4}</math>, <math>\frac{5\pi}{4}</math>, or <math>\frac{7\pi}{4}</math>, so the minimum distance from the origin to the function <math>xy = k</math> is <math>\sqrt{2}(\sqrt{|k|})</math>.
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The image below shows the two curves for <math>k=4</math>. The blue curve is <math>x^2+y^2=k^2</math>, which is clearly a circle with radius <math>k</math>, and the red curve is a part of the curve <math>xy=k</math>.
  
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The distance from the circle to the origin is always <math>k</math>. Therefore, we want to find all integer values such that
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<asy>
 +
import graph;
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size(200);
  
 +
real f(real x) {return 4/x;};
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real g1(real x) {return sqrt(4*4-x*x);};
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real g2(real x) {return -sqrt(4*4-x*x);};
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draw(graph(f,-20./3,-0.6),red);
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draw(graph(f,0.6,20./3),red);
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draw(graph(g1,-4,4),blue);
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draw(graph(g2,-4,4),blue);
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axes("$x$","$y$");
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</asy>
  
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<math>|k| < \sqrt{|2k|}}</math>
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In the special case <math>k=0</math> the blue curve is just the point <math>(0,0)</math>, and as <math>0\cdot 0=0</math>, this point is on the red curve as well, hence they intersect.
  
 +
The case <math>k<0</math> is symmetric to <math>k>0</math>: the blue curve remains the same and the red curve is flipped according to the <math>x</math> axis. Hence we just need to focus on <math>k>0</math>.
  
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It is then easy to see that the only values that satisfy the inequality are <math>-1</math> and <math>1</math>, a total of <math>\boxed{2\ \textbf{(C)}}</math> <math>k</math> values.
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Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as <math>x</math> approaches 0, <math>y</math> approaches <math>\infty</math>. Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most <math>k</math>.
  
  
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[Images of the graphs of these functions would really help to understand and visualize this solution.]
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At this point we can guess that on the red curve the point where <math>x=y</math> is always closest to the origin, and skip the rest of this solution.
 +
 
 +
 
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For an exact solution, fix <math>k</math> and consider any point <math>(x,y)</math> on the red curve. Its distance from the origin is <math>\sqrt{ x^2 + (k/x)^2 }</math>. To minimize this distance, it is enough to minimize <math>x^2 + (k/x)^2</math>. By the [[Arithmetic Mean-Geometric Mean Inequality]] we get that this value is at least <math>2k</math>, and that equality holds whenever <math>x^2 = (k/x)^2</math>, i.e., <math>x=\pm\sqrt k</math>.
 +
 
 +
 
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Now recall that the red curve intersects the blue one if and only if its closest point is at most <math>k</math> from the origin. We just computed that the distance between the origin and the closest point on the red curve is <math>\sqrt{2k}</math>. Therefore, we want to find all positive integers <math>k</math> such that <math>\sqrt{2k} > k</math>.
 +
 
 +
Clearly the only such integer is <math>k=1</math>, hence the two curves are only disjoint for <math>k=1</math> and <math>k=-1</math>.
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This is a total of <math>\boxed{2\ \textbf{(C)}}</math> values.

Revision as of 09:00, 16 February 2010

Problem 13

For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

Solution

The image below shows the two curves for $k=4$. The blue curve is $x^2+y^2=k^2$, which is clearly a circle with radius $k$, and the red curve is a part of the curve $xy=k$.

[asy] import graph; size(200); real f(real x) {return 4/x;}; real g1(real x) {return sqrt(4*4-x*x);}; real g2(real x) {return -sqrt(4*4-x*x);}; draw(graph(f,-20./3,-0.6),red); draw(graph(f,0.6,20./3),red); draw(graph(g1,-4,4),blue); draw(graph(g2,-4,4),blue); axes("$x$","$y$"); [/asy]

In the special case $k=0$ the blue curve is just the point $(0,0)$, and as $0\cdot 0=0$, this point is on the red curve as well, hence they intersect.

The case $k<0$ is symmetric to $k>0$: the blue curve remains the same and the red curve is flipped according to the $x$ axis. Hence we just need to focus on $k>0$.

Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as $x$ approaches 0, $y$ approaches $\infty$. Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most $k$.


At this point we can guess that on the red curve the point where $x=y$ is always closest to the origin, and skip the rest of this solution.


For an exact solution, fix $k$ and consider any point $(x,y)$ on the red curve. Its distance from the origin is $\sqrt{ x^2 + (k/x)^2 }$. To minimize this distance, it is enough to minimize $x^2 + (k/x)^2$. By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at least $2k$, and that equality holds whenever $x^2 = (k/x)^2$, i.e., $x=\pm\sqrt k$.


Now recall that the red curve intersects the blue one if and only if its closest point is at most $k$ from the origin. We just computed that the distance between the origin and the closest point on the red curve is $\sqrt{2k}$. Therefore, we want to find all positive integers $k$ such that $\sqrt{2k} > k$.

Clearly the only such integer is $k=1$, hence the two curves are only disjoint for $k=1$ and $k=-1$. This is a total of $\boxed{2\ \textbf{(C)}}$ values.

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