2010 AMC 12A Problems/Problem 13

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

We can see that the function $xy = k$ is symmetric to the line $y=x$ , and the distance to the origin approaches infinity as the function approaches either the $x$ -axis or the $y$ -axis. Therefore, assuming that graphs don't intersect, the point at which the function $xy = k$ is closest to the function $x^2+y^2=k^2$ (which is clearly a circle) is when $|x|=|y|$ . It follows that at these points, the magnitude of the $x$ and $y$ values for the function $xy = k$ will be $\sqrt{|k|}$ .

All of these points are found at angles $\frac{\pi}{4}$ , $\frac{3\pi}{4}$ , $\frac{5\pi}{4}$ , or $\frac{7\pi}{4}$ , so the minimum distance from the origin to the function $xy = k$ is $\sqrt{2}(\sqrt{|k|})$ .

The distance from the circle to the origin is always $k$ . Therefore, we want to find all integer values such that

$|k| < \sqrt{|2k|}}$ (Error compiling LaTeX. Unknown error_msg)

It is then easy to see that the only values that satisfy the inequality are $-1$ and $1$ , a total of $\boxed{2\ \textbf{(C)}}$ $k$ values.

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2010 AMC 12A Problems/Problem 13

Problem 13

Solution