Difference between revisions of "2020 AMC 12A Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
| − | Realize that <math>\ln\frac{91}{90}</math> is extremely small, so the <math>x</math>-coordinates must be very close to each other. Also realize that <math>\frac{91}{90} = \frac{182}{180} = \frac{13*14}{12*15}.</math> <math>\boxed{\textbf{(D) } 12}</math> is the smallest number in that fraction, and therefore must be the answer. ~lopkiloinm | + | Realize that <math>\ln\frac{91}{90}</math> is extremely small, so the <math>x</math>-coordinates must be very close to each other. Because of how small <math>\ln\frac{91}{90}</math> is, we must say that the <math>x</math>-coordinates must be consecutive. Also realize that <math>\frac{91}{90} = \frac{182}{180} = \frac{13*14}{12*15}.</math> <math>\boxed{\textbf{(D) } 12}</math> is the smallest number in that fraction, and therefore must be the answer. ~lopkiloinm |
Revision as of 20:15, 1 February 2020
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
Realize that 
is extremely small, so the -coordinates must be very close to each other. Because of how small is, we must say that the -coordinates must be consecutive. Also realize that 
is the smallest number in that fraction, and therefore must be the answer. ~lopkiloinm
