Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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== 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>\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}= | + | 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 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 == |
Latest revision as of 10:51, 19 April 2021
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 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 , 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.