Difference between revisions of "2011 AMC 12B Problems/Problem 19"
Xhypotenuse (talk | contribs) |
|||
(2 intermediate revisions by 2 users not shown) | |||
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 | + | 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>. 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{n+1}{2n+1}</math>, <math>\frac{n+1}{2n}</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> (In other words we are trying to find the smallest <math>m>\frac{1}{2}</math> such that <math>b\le 100</math>.) | ||
==Solution 2== | ==Solution 2== | ||
Line 14: | Line 13: | ||
A line with a slope of <math>\frac{1}{2}</math> passes through (among other lattice points) the lattice point <math>(100,50)</math>. As the slope of the line increases from <math>\frac{1}{2}</math>, the first lattice point it hits is at <math>(99, 50)</math>, the slope of that line being <math>\frac{50}{99}</math> . So the answer is <math>\boxed{\textbf{(B)}}</math> | A line with a slope of <math>\frac{1}{2}</math> passes through (among other lattice points) the lattice point <math>(100,50)</math>. As the slope of the line increases from <math>\frac{1}{2}</math>, the first lattice point it hits is at <math>(99, 50)</math>, the slope of that line being <math>\frac{50}{99}</math> . So the answer is <math>\boxed{\textbf{(B)}}</math> | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | Like in the first solution, we note that the <math>+2</math> does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line <math>y = mx</math>. | ||
+ | |||
+ | We will start with the first answer choice. Notice how it is impossible for a line with slope <math>\frac{51}{101}</math> to pass through a lattice point when <math>0 < x \leq 100</math>. | ||
+ | |||
+ | However, if the slope is <math>\frac{50}{99}</math>, then the line will pass through the point <math>(99, 50)</math>. This sets the upper bound for <math>a</math> as <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | ~xHypotenuse | ||
== See also == | == See also == |
Latest revision as of 20:02, 13 July 2024
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
(In other words we are trying to find the smallest
such that
.)
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
Solution 3
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
.
We will start with the first answer choice. Notice how it is impossible for a line with slope to pass through a lattice point when
.
However, if the slope is , then the line will pass through the point
. This sets the upper bound for
as
.
~xHypotenuse
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.