Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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~Solution by AsdrúbalBeltrán | ~Solution by AsdrúbalBeltrán | ||
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+ | ===Video Solution by TheBeautyofMath=== | ||
+ | https://www.youtube.com/watch?v=Eq2A2TTahqU?t=583 | ||
+ | Another example of shoelace theorem included earlier in the video | ||
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+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 00:32, 29 January 2021
Contents
[hide]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 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 .
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.