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

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== Problem 13 ==
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== Problem ==
 
For how many integer values of <math>k</math> do the graphs of <math>x^2+y^2=k^2</math> and <math>xy = k</math> not intersect?
 
For how many integer values of <math>k</math> do the graphs of <math>x^2+y^2=k^2</math> and <math>xy = k</math> not intersect?
  
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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>.  
 
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>.  
 
This is a total of <math>\boxed{2\ \textbf{(C)}}</math> values.
 
This is a total of <math>\boxed{2\ \textbf{(C)}}</math> values.
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== See also ==
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{{AMC12 box|year=2010|num-b=12|num-a=14|ab=A}}
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[[Category:Introductory Geometry Problems]]

Revision as of 23:31, 25 February 2010

Problem

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.

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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