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== | + | == Solutions == |
| − | + | === 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>\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}=</cmath><cmath>\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).</cmath>We now 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>. | 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>\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}=</cmath><cmath>\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).</cmath>We now 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== | + | === 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 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> | ||
~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> | ||
Revision as of 13:25, 19 January 2021
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?
Solutions
Solution 1
Let the coordinates of the quadrilateral be 
. We have by shoelace's theorem, that the area is![\[\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2}=\]](http://latex.artofproblemsolving.com/7/3/f/73f163caf2690b8d01e23b01661618202dd7c018.png)
![\[\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} = \ln \left( \sqrt{\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \right) = \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) = \ln \left( \frac{91}{90} \right).\]](http://latex.artofproblemsolving.com/5/0/6/50689c3733b3a590d9abbbc9ee58b4cba1a161da.png)
We now 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 triangle 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 ![$[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]$](http://latex.artofproblemsolving.com/e/3/9/e399a69e1911bd52b0cb18bd1d32edd4d8a658eb.png)
, all quadrilaterals of side right are trapezius
![\[[BCDE]=\frac{\ln(n+1)+\ln n}{2}+\frac{\ln(n+2)+\ln(n+1)}{2}+\frac{\ln(n+3)+\ln(n+2)}{2}-\frac{3(\ln(n+3)+\ln n)}{2}\]](http://latex.artofproblemsolving.com/6/b/4/6b49d20966d457b29fd23ca11cadd140ef0bb7a1.png)
![\[[BCDE]=\tfrac{2\ln(n+1)+2\ln(n+2)-2\ln(n+3)-2\ln n}{2}=\ln(n+1)+\ln(n+2)-\ln(n+3)-\ln n=\ln\tfrac{(n+1)(n+2)}{n(n+3)}\]](http://latex.artofproblemsolving.com/4/4/b/44b36bdf1379371b54537ebcccecb2d50a137c25.png)
![\[\implies\ln\frac{(n+1)(n+2)}{n(n+3)}=\ln\frac{91}{90}\implies \frac{(n+1)(n+2)}{n(n+3)}=\frac{91}{90}\]](http://latex.artofproblemsolving.com/b/5/4/b54d310717c1635b6e4b92a7cfb00d576864b245.png)
![\[\implies n^2+3n-180=0\implies (n+15)(n-12)=0\implies n=12\]](http://latex.artofproblemsolving.com/1/2/5/125a64661449db0308ff23335f0e46a41c0f00ab.png)
~Solution by AsdrúbalBeltrán
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. 
