Difference between revisions of "2011 AMC 12B Problems/Problem 19"
Megaboy6679 (talk | contribs) m |
|||
| Line 4: | Line 4: | ||
<math>\textbf{(A)}\ \frac{51}{101} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{51}{100} \qquad \textbf{(D)}\ \frac{52}{101} \qquad \textbf{(E)}\ \frac{13}{25}</math> | <math>\textbf{(A)}\ \frac{51}{101} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{51}{100} \qquad \textbf{(D)}\ \frac{52}{101} \qquad \textbf{(E)}\ \frac{13}{25}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | It is very easy to see that the <math>+2 </math> in the graph does not impact whether it passes through lattice. | + | It is very easy to see that the <math>+2 </math> in the graph does not impact whether it passes through the lattice. |
| − | We need to make sure that <math>m</math> cannot be in the form of <math>\frac{a}{b}</math> for <math>1\le b\le 100</math>, | + | We need to make sure that <math>m</math> cannot be in the form of <math>\frac{a}{b}</math> for <math>1\le b\le 100</math>. Otherwise, the graph <math>y= mx</math> passes through the lattice point at <math>x = b</math>. We only need to worry about <math>\frac{a}{b}</math> very close to <math>\frac{1}{2}</math>, <math>\frac{m+1}{2m+1}</math>, <math>\frac{m+1}{2m}</math> will be the only case we need to worry about and we want the minimum of those, clearly for <math>1\le b\le 100</math>, the smallest is <math>\frac{50}{99} </math>, so answer is <math>\boxed{\frac{50}{99} \textbf{(B)}}</math> |
Revision as of 18:29, 7 August 2023
Contents
Problem
A lattice point in an 
-coordinate system is any point 
where both 
and 
are integers. The graph of 
passes through no lattice point with 
for all 
such that 
. What is the maximum possible value of 
?
Solution 1
It is very easy to see that the 
in the graph does not impact whether it passes through the lattice.
We need to make sure that cannot be in the form of 
for 
. Otherwise, the graph 
passes through the lattice point at 
. We only need to worry about very close to 
, 
, 
will be the only case we need to worry about and we want the minimum of those, clearly for , the smallest is 
, so answer is 
Solution 2
Like in the first solution, we note that the does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line 
.
A line with a slope of passes through (among other lattice points) the lattice point 
. As the slope of the line increases from , the first lattice point it hits is at 
, the slope of that line being . So the answer is 
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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. 
