2013 AMC 12B Problems/Problem 19

Problem

A lattice point in an $xy$ -coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y = mx + 2$ passes through no lattice point with $0 < x \leq 100$ for all $m$ such that $\frac{1}{2} < m < a$ . What is the maximum possible value of $a$ ?

$\textbf{(A)}\ \frac{51}{101} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{51}{100} \qquad \textbf{(D)}\ \frac{52}{101} \qquad \textbf{(E)}\ \frac{13}{25}$

Solution

Since $\angle{AFB}=\angle{ADB}=90$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ . In addition, since $\angle{AFB}=\angle{AED}=90$ , triangles $ABF$ and $ADE$ are similar. It follows that $AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})$ . By Ptolemy, we have $13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})$ . Cancelling $13$ , the rest is easy. We obtain $DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{\textbf{(B)} 21}$ (Error compiling LaTeX. Unknown error_msg)

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2013 AMC 12B Problems/Problem 19

Problem

Solution