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

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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2010 AMC 12A Problems/Problem 13

Problem 13

Solution