2011 AMC 10B Problem 24

At an integer $x$, the point $(x, mx + 2)$ is a lattice point exactly when $mx$ is an integer, i.e. when $m = \frac{k}{x}$ for some integer $k$. So the line avoids lattice points for $0 < x \le 100$ precisely when $m$ is not a fraction with denominator (before or after reduction) at most $100$.

Therefore the interval $\left(\frac12, a\right)$ is safe as long as it contains no fraction $\frac{k}{x}$ with $1 \le x \le 100$, and the largest possible $a$ is the smallest such fraction exceeding $\frac12$.

For $\frac{k}{x} > \frac12$ we need $2k - x \ge 1$, so
$$\frac{k}{x} - \frac12 = \frac{2k - x}{2x} \ge \frac{1}{2x}.$$
To make this gap as small as possible we want $2k - x = 1$ (so $x$ must be odd) and $x$ as large as possible: $x = 99$, $k = 50$. The candidate is $\frac{50}{99}$, with gap $\frac{1}{198}$.

Could an even denominator do better? With $x = 100$ the best is $2k - x = 2$, giving $\frac{51}{100}$ and gap $\frac{1}{100}$, which is larger. Cross-check: $50 \cdot 100 = 5000 < 51 \cdot 99 = 5049$, so indeed $\frac{50}{99} < \frac{51}{100}$.

Hence every $m$ in $\left(\frac12, \frac{50}{99}\right)$ is safe, while $m = \frac{50}{99}$ itself hits the lattice point $(99, 52)$. The maximum $a$ is $\frac{50}{99}$.

The answer is $\boxed{\textbf{(B)}\ \frac{50}{99}}$.

The lattice-point condition is really a statement about rational slopes with bounded denominators, so the geometry converts into "find the fraction closest to $\frac12$ from above with denominator at most $100$." The gap formula $\frac{2k - x}{2x}$ shows that the winning fraction has numerator-over-denominator difference $1$ and the largest odd denominator available.

Check the choices directly. $\frac{51}{101}$ and $\frac{52}{101}$ have denominator $101 > 100$, so they are never hit and cannot be the boundary; but they are also smaller or larger than the true boundary, so test them: $\frac{51}{101} < \frac{50}{99}$? $51 \cdot 99 = 5049 < 50 \cdot 101 = 5050$, yes, so choosing $a = \frac{51}{101}$ is safe but not maximal. Among the rest, $\frac{50}{99}$ is the smallest fraction that is actually hit, so it is the answer.

• Picking $\frac{51}{100}$ because $100$ is the largest allowed denominator, without checking that the odd denominator $99$ gives the closer fraction $\frac{50}{99}$.
• Choosing $\frac{51}{101}$, which is safe but smaller than $\frac{50}{99}$, so it is not the maximum $a$.
• Forgetting that the intercept $2$ is an integer, so the condition is simply "$mx$ is an integer"; an irrational or large-denominator intercept would change the problem entirely.