# Difference between revisions of "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

Let the left-most $x$-coordinate be $n.$

Recall that, by the shoelace formula, the area of the quadrilateral must be $-\ln{(n)}+\ln{(n+1)}+\ln{(n+2)}-\ln{(n+3)}.$ That equals to $\ln\frac{(n+1)(n+2)}{n(n+3)}.$

$\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}$

$\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{91}{90}$

$\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{182}{180}$

$n^{2}+3n = 180$

$n^{2}+3n-180 = 0$

$(n-12)(n+15) = 0$

The $x$-coordinate is, therefore, $\boxed{\textbf{(D) } 12.}$~lopkiloinm.

## Solution 2

Like above, use the shoelace formula to find that the area of the triangle is equal to $\ln\frac{(n+1)(n+2)}{n(n+3)}$. Because the final area we are looking for is $\ln\frac{91}{90}$, the numerator factors into $13$ and $7$, which one of $n+1$ and $n+2$ has to be a multiple of $13$ and the other has to be a multiple of $7$. Clearly, the only choice for that is $\boxed{12}$

~Solution by IronicNinja