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

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==Solution 1==
 
==Solution 1==
  
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Realize that <math>\ln\frac{91}{90}</math> is extremely small, so the <math>x</math>-coordinates must be very close to each other. Because of how ridiculously small <math>\ln\frac{91}{90}</math> is, we must say that the <math>x</math>-coordinates must be consecutive. Also realize that <math>\frac{91}{90} = \frac{182}{180} = \frac{13*14}{12*15}.</math> <math>\boxed{\textbf{(D) } 12}</math> is the smallest number in that fraction, and therefore  must be the answer. ~lopkiloinm
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Let the <math>x</math>-intercept be <math>n.</math>
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Realize that by the shoelace formula the area of the triangle must be <math>-\lnn+\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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<math>\ln\frac{(n+1)(n+2)}{(n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}</math>  
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<math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{91}{90}</math>
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<math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{182}{180}</math>
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<math>n^{2}+3n = 180</math>
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<math>n^{2}+3n-180 = 0</math>
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<math>(n-12)(n+15) = 0</math>
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The <math>x</math>-intercept is <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm.

Revision as of 01:01, 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 $x$-intercept be $n.$

Realize that by the shoelace formula the area of the triangle must be $-\lnn+\ln{n+1}+\ln{n+2}-\ln{n+3}.$ (Error compiling LaTeX. Unknown error_msg) 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 $x$-intercept is $\boxed{\textbf{(D) } 12.}$~lopkiloinm.

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