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

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==Solution 1==
 
==Solution 1==
  
Let the left-most <math>x</math>-coordinate be <math>n.</math>
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Let the coordinates of the quadrilateral be <math>(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))</math>. We have by shoelace's theorem, that the area is<cmath>\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)\ln(n) + \ln(n+2)\ln(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2}=</cmath><cmath>\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} = \ln \left( \sqrt{\frac{(n+1)^2(n+3)^2}{n^2(n+2)^2}} \right) = \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) = \ln \left( \frac{91}{90} \right).</cmath>We now that the numerator must have a factor of <math>13</math>, so given the answer choices, <math>n</math> is either <math>12</math> or <math>11</math>. If <math>n=11</math>, the expression <math>\frac{(n+1)(n+2)}{n(n+3)}</math> does not evaluate to <math>\frac{91}{90}</math>, but if <math>n=12</math>, the expression evaluates to <math>\frac{91}{90}</math>. Hence, our answer is <math>\boxed{12}</math>.
  
Recall that, by the shoelace formula, the area of the quadrilateral 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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~AopsUser101
 
 
<math>\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}</math>
 
 
 
<math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{91}{90}</math>
 
 
 
<math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{182}{180}</math>
 
 
 
<math>n^{2}+3n = 180</math>
 
 
 
<math>n^{2}+3n-180 = 0</math>
 
 
 
<math>(n-12)(n+15) = 0</math>
 
 
 
The <math>x</math>-coordinate is, therefore, <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm.
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 15:20, 19 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 coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$. We have by shoelace's theorem, that the area is\[\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)\ln(n) + \ln(n+2)\ln(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2}=\]\[\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} = \ln \left( \sqrt{\frac{(n+1)^2(n+3)^2}{n^2(n+2)^2}} \right) = \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) = \ln \left( \frac{91}{90} \right).\]We now that the numerator must have a factor of $13$, so given the answer choices, $n$ is either $12$ or $11$. If $n=11$, the expression $\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\frac{91}{90}$, but if $n=12$, the expression evaluates to $\frac{91}{90}$. Hence, our answer is $\boxed{12}$.

~AopsUser101

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

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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