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Difference between revisions of "2020 AMC 12A Problems/Problem 17"

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Let the left-most <math>x</math>-coordinate be <math>n.</math>
 
Let the left-most <math>x</math>-coordinate be <math>n.</math>
  
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Realize that by the shoelace formula the area of the triangle must be <math>-\ln{n}+\ln{n+1}+\ln{n+2}-\ln{n+3}.</math> That equals to <math>\ln\frac{(n+1)(n+2)}{n(n+3)}.</math>  
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Recall that, by the shoelace formula, the area of the triangle must be <math>-\ln{n}+\ln{n+1}+\ln{n+2}-\ln{n+3}.</math> That equals to <math>\ln\frac{(n+1)(n+2)}{n(n+3)}.</math>  
  
 
<math>\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}</math>  
 
<math>\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}</math>  

Revision as of 01:12, 2 February 2020

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 triangle must be $-\ln{n}+\ln{n+1}+\ln{n+2}-\ln{n+3}.$ That equals to $\ln\frac{(n+1)(n+2)}{n(n+3)}.$

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

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

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

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

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

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

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

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