2020 AMC 12A Problems/Problem 17

Problem 17

The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?

$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$

Solution 1

Realize that $\ln\frac{91}{90}$ is extremely small, so the $x$ -coordinates must be very close to each other. Because of how small $\ln\frac{91}{90}$ is, we must say that the $x$ -coordinates must be consecutive. Also realize that $\frac{91}{90} = \frac{182}{180} = \frac{13*14}{12*15}.$ $\boxed{\textbf{(D) } 12}$ is the smallest number in that fraction, and therefore must be the answer. ~lopkiloinm

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2020 AMC 12A Problems/Problem 17

Problem 17

Solution 1