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Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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==Problem 17==
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== Problem ==
 
The vertices of a quadrilateral lie on the graph of <math>y=\ln{x}</math>, and the <math>x</math>-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is <math>\ln{\frac{91}{90}}</math>. What is the <math>x</math>-coordinate of the leftmost vertex?
 
The vertices of a quadrilateral lie on the graph of <math>y=\ln{x}</math>, and the <math>x</math>-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is <math>\ln{\frac{91}{90}}</math>. What is the <math>x</math>-coordinate of the leftmost vertex?
  
 
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13</math>
 
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13</math>
  
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==Solution 1==
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== Solution 1 ==
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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
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<cmath>\begin{align*}
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&\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2} \\
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&=\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} \\
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&= \ln \left( \sqrt{\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \right)  \\
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&= \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) \\
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&= \ln \left( \frac{91}{90} \right).
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\end{align*}</cmath>
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We know 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>.
  
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Let the left-most <math>x</math>-coordinate be <math>n.</math>
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== Solution 2 ==
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Like above, use the shoelace formula to find that the area of the quadrilateral is equal to <math>\ln\frac{(n+1)(n+2)}{n(n+3)}</math>. Because the final area we are looking for is <math>\ln\frac{91}{90}</math>, the numerator factors into <math>13</math> and <math>7</math>, which one of <math>n+1</math> and <math>n+2</math> has to be a multiple of <math>13</math> and the other has to be a multiple of <math>7</math>. Clearly, the only choice for that is <math>\boxed{12}</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>
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~Solution by IronicNinja
  
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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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== Solution 3 ==
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How <math>f(x)=\ln(x)</math> is a concave function, then:
  
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<math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{91}{90}</math>
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[[File:Problema17NivelSuperior.png|frameless|border|400px|link=https://wiki-images.artofproblemsolving.com//7/7b/Problema17NivelSuperior.png]]
  
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<math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{182}{180}</math>
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Therefore <math>[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]</math>, all quadrilaterals of side right are trapezius
  
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<math>n^{2}+3n = 180</math>
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<cmath>[BCDE]=\frac{\ln(n+1)+\ln n}{2}+\frac{\ln(n+2)+\ln(n+1)}{2}+\frac{\ln(n+3)+\ln(n+2)}{2}-\frac{3(\ln(n+3)+\ln n)}{2}</cmath>
  
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<math>n^{2}+3n-180 = 0</math>
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<cmath>[BCDE]=\tfrac{2\ln(n+1)+2\ln(n+2)-2\ln(n+3)-2\ln n}{2}=\ln(n+1)+\ln(n+2)-\ln(n+3)-\ln n=\ln\tfrac{(n+1)(n+2)}{n(n+3)}</cmath>
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<cmath>\implies\ln\frac{(n+1)(n+2)}{n(n+3)}=\ln\frac{91}{90}\implies \frac{(n+1)(n+2)}{n(n+3)}=\frac{91}{90}</cmath>
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<cmath>\implies n^2+3n-180=0\implies (n+15)(n-12)=0\implies n=12</cmath>
  
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<math>(n-12)(n+15) = 0</math>
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~Solution by AsdrúbalBeltrán
  
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The <math>x</math>-coordinate is, therefore, <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm.
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==Video Solution by TheBeautyofMath==
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https://www.youtube.com/watch?v=Eq2A2TTahqU?t=583
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Another example of shoelace theorem included earlier in the video
  
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==Solution 2==
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~IceMatrix
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Like above, use the shoelace formula to find that the area of the triangle is equal to <math>\ln\frac{(n+1)(n+2)}{n(n+3)}</math>. Because the final area we are looking for is <math>\ln\frac{91}{90}</math>, the numerator factors into <math>13</math> and <math>7</math>, which one of <math>n+1</math> and <math>n+2</math> has to be a multiple of <math>13</math> and the other has to be a multiple of <math>7</math>. Clearly, the only choice for that is <math>\boxed{12}</math>
 
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~Solution by IronicNinja
 
  
 
==See Also==
 
==See Also==

Latest revision as of 20:43, 21 October 2023

Problem

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 \begin{align*} &\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(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+2)^2}{n^2(n+3)^2}} \right) \\ &= \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) \\ &= \ln \left( \frac{91}{90} \right). \end{align*} We know 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 $\boxed{12}$.

Solution 2

Like above, use the shoelace formula to find that the area of the quadrilateral 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

Solution 3

How $f(x)=\ln(x)$ is a concave function, then:

Problema17NivelSuperior.png

Therefore $[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]$, all quadrilaterals of side right are trapezius

\[[BCDE]=\frac{\ln(n+1)+\ln n}{2}+\frac{\ln(n+2)+\ln(n+1)}{2}+\frac{\ln(n+3)+\ln(n+2)}{2}-\frac{3(\ln(n+3)+\ln n)}{2}\]

\[[BCDE]=\tfrac{2\ln(n+1)+2\ln(n+2)-2\ln(n+3)-2\ln n}{2}=\ln(n+1)+\ln(n+2)-\ln(n+3)-\ln n=\ln\tfrac{(n+1)(n+2)}{n(n+3)}\] \[\implies\ln\frac{(n+1)(n+2)}{n(n+3)}=\ln\frac{91}{90}\implies \frac{(n+1)(n+2)}{n(n+3)}=\frac{91}{90}\] \[\implies n^2+3n-180=0\implies (n+15)(n-12)=0\implies n=12\]

~Solution by AsdrúbalBeltrán

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=Eq2A2TTahqU?t=583 Another example of shoelace theorem included earlier in the video

~IceMatrix

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AMC 12 Problems and Solutions

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