Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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Let the left-most <math>x</math>-coordinate be <math>n.</math> | Let the left-most <math>x</math>-coordinate be <math>n.</math> | ||
− | Recall that, by the shoelace formula, the area of the | + | Recall that, by the shoelace formula, the area of the quadrilateral must be <math>-\ln{(n)}+\ln{(n+1)}+\ln{(n+2)}-\ln{(n+3)}.</math> That equals to <math>\ln\frac{(n+1)(n+2)}{n(n+3)}.</math> |
<math>\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}</math> | <math>\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}</math> | ||
− | <math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{91}{90}</math> | + | <math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{91}{90}</math> |
− | <math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{182}{180}</math> | + | <math>\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{182}{180}</math> |
<math>n^{2}+3n = 180</math> | <math>n^{2}+3n = 180</math> |
Revision as of 00:05, 3 February 2020
Contents
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 quadrilateral 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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