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

Revision as of 01:14, 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 left-most $x$ -coordinate be $n.$

Recall that, by the shoelace formula, the area of the triangle must be $-\ln{n}+\ln{n+1}+\ln{n+2}-\ln{n+3}.$ 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 $x$ -coordinate is, therefore, $\boxed{\textbf{(D) } 12.}$ ~lopkiloinm.

2020 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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 The <math>x</math>-coordinate is, therefore, <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm.
+==See Also==
+{{AMC12 box|year=2020|ab=A|num-b=16|num-a=18}}
+{{MAA Notice}}

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

Revision as of 01:14, 2 February 2020

Problem 17

Solution 1

See Also