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

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:

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

@@ Line 16: / Line 16: @@
 == Solution 2 ==
-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>
+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>
 ~Solution by IronicNinja

2020 AMC 12A (Problems • Answer Key • Resources)
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