1965 AHSME Problems/Problem 13

Problem

Let $n$ be the number of number-pairs $(x,y)$ which satisfy $5y - 3x = 15$ and $x^2 + y^2 \le 16$. Then $n$ is:

$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ \text{more than two, but finite}\qquad \textbf{(E) }\ \text{greater than any finite number}$

Solution

[asy]  draw((-16,0)--(16,0), arrow=Arrows); label("$x$",(17.5,0)); draw((0,-16)--(0,16), arrow=Arrows); label("$y$", (0,17.5));  draw(circle((0,0),8),red); draw((-14,-12/5)--(14, 72/5), blue, arrow=Arrows);  [/asy]

For an ordered pair $(x,y)$ to satisfy the restrictions, the point $(x,y)$ must lie on the graph of the line $5y-3x=15$ (shown in blue in the diagram) and on the closed disk given by $x^2+y^2 \leq 16$ (whose boundary is red in the diagram). Because the $y$-intercept of the line, $(0,3)$, is within the disk (and not on the boundary of the disk), the line cannot be tangent to the disk, and so the disk must contain infinitely many of the points on the line. Thus, our answer is $\fbox{\textbf{(E)} greater than any finite number}$

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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