Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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~Solution by IronicNinja | ~Solution by IronicNinja | ||
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+ | ==Solution 3== | ||
+ | How <math>f(x)=\ln(x)</math> is a concave function, then: | ||
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+ | [[File:Problema17NivelSuperior.png|frameless|border|400px|link=https://wiki-images.artofproblemsolving.com//7/7b/Problema17NivelSuperior.png]] | ||
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+ | Therefore <math>[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]</math>, all quadrilaterals of side right are trapezius | ||
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+ | <math>[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}</math> | ||
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+ | <math>[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)}</math> | ||
+ | <cmath>\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}</cmath> | ||
+ | <cmath>n^2+3n-180=0\implies (n+15)(n-12)=0\implies n=12</cmath> | ||
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+ | ~Solution by AsdrúbalBeltrán | ||
==See Also== | ==See Also== |
Revision as of 12:54, 8 March 2020
Contents
[hide]Problem 17
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 isWe 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 .
~AopsUser101
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
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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