Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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− | ==Problem | + | == 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> | ||
− | ==Solution 1== | + | == Solution 1 == |
+ | 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 | ||
+ | <cmath>\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*}</cmath> | ||
+ | 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>. | ||
− | + | == Solution 2 == | |
− | + | 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 2== | ||
− | Like above, use the shoelace formula to find that the area of the | ||
~Solution by IronicNinja | ~Solution by IronicNinja | ||
− | ==Solution 3== | + | == Solution 3 == |
How <math>f(x)=\ln(x)</math> is a concave function, then: | How <math>f(x)=\ln(x)</math> is a concave function, then: | ||
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Therefore <math>[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]</math>, all quadrilaterals of side right are trapezius | Therefore <math>[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]</math>, all quadrilaterals of side right are trapezius | ||
− | |||
<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> | <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> | ||
− | |||
<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> | <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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~Solution by AsdrúbalBeltrán | ~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== | ==See Also== |
Latest revision as of 21:43, 21 October 2023
Contents
Problem
The vertices of a quadrilateral lie on the graph of , and the
-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is
. What is the
-coordinate of the leftmost vertex?
Solution 1
Let the coordinates of the quadrilateral be . We have by shoelace's theorem, that the area is
We know that the numerator must have a factor of
, so given the answer choices,
is either
or
. If
, the expression
does not evaluate to
, but if
, the expression evaluates to
. Hence, our answer is
.
Solution 2
Like above, use the shoelace formula to find that the area of the quadrilateral is equal to . Because the final area we are looking for is
, the numerator factors into
and
, which one of
and
has to be a multiple of
and the other has to be a multiple of
. Clearly, the only choice for that is
~Solution by IronicNinja
Solution 3
How is a concave function, then:
Therefore , all quadrilaterals of side right are trapezius
~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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.