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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 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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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

Revision as of 20:18, 1 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

Realize that $\ln\frac{91}{90}$ is extremely small, so the $x$-coordinates must be very close to each other. Because of how ridiculously 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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