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2020 AMC 12A Problems/Problem 17

Revision as of 18:11, 25 April 2020 by Aopsuser101 (talk | contribs) (→‎Solution 1)

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\[\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)\ln(n) + \ln(n+2)\ln(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2}=\]\[\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} = \ln \left( \sqrt{\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \right) = \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) = \ln \left( \frac{91}{90} \right).\]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 $\boxed{12}$.

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

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

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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