2011 AMC 10B Problem 24

Shift the graph down by $2$. The problem is equivalent to finding the smallest slope $m\gt\dfrac12$ for which $y=mx$ passes through a lattice point with $0\lt x\le100$. For a fixed integer $x$, the smallest integer $y$ with $\frac{y}{x}\gt\frac{1}{2}$ is $\frac{x}{2}+1$ when $x$ is even, and $\frac{x+1}{2}$ when $x$ is odd. Thus the candidate slopes are $\dfrac12+\dfrac1x$ for even $x$, minimized at $x=100$ as $\dfrac{51}{100}$, and $\dfrac12+\dfrac1{2x}$ for odd $x$, minimized at $x=99$ as $\dfrac{50}{99}$. The smaller of these is $\dfrac{50}{99}$, so every $m$ with $\dfrac12\lt m\lt\dfrac{50}{99}$ avoids such lattice points, and this upper endpoint is best possible. Thus, B is the correct answer.