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

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The <math>x</math>-coordinate is, therefore, <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm.
 
The <math>x</math>-coordinate is, therefore, <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm.
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==Solution 2==
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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==

Revision as of 02:24, 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.

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 (Problems • Answer Key • Resources)
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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