Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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The <math>x</math>-coordinate is, therefore, <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm. | The <math>x</math>-coordinate is, therefore, <math>\boxed{\textbf{(D) } 12.}</math>~lopkiloinm. | ||
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+ | ==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> | ||
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+ | ~Solution by IronicNinja | ||
==See Also== | ==See Also== |
Revision as of 01:24, 2 February 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 left-most -coordinate be
Recall that, by the shoelace formula, the area of the triangle must be That equals to
The -coordinate is, therefore, ~lopkiloinm.
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
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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