Difference between revisions of "2018 AMC 12B Problems/Problem 17"
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− | Graph the regions <math>y > \frac{5}{9}x</math> and <math>y < \frac{4}{7}x</math>. Note that the lattice point (9,16) is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is <math>\frac{9}{16}</math> and the answer is <math>16-9= \boxed{\textbf{(A)}\ 7}</math>. | + | Graph the regions <math>y > \frac{5}{9}x</math> and <math>y < \frac{4}{7}x</math>. Note that the lattice point (9,16) is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is <math>\frac{9}{16}</math> and the answer is <math>16-9= \boxed{\textbf{(A)}\ 7}</math>. // |
Remark: This also gives an intuitive geometric proof of the mediant. | Remark: This also gives an intuitive geometric proof of the mediant. | ||
Revision as of 22:09, 16 February 2018
Contents
Problem
Let and
be positive integers such that
and
is as small as possible. What is
?
Solution 1
We claim that, between any two fractions and
, if
, the fraction with smallest denominator between them is
. To prove this, we see that
which reduces to
. We can easily find that
, giving an answer of
. (pieater314159)
Solution 2 (requires justification)
Assume that the difference results in a fraction of the form
. Then,
Also assume that the difference results in a fraction of the form
. Then,
Solving the system of equations yields and
. Therefore, the answer is
Solution 3
Cross-multiply the inequality to get
Then,
Since ,
are integers,
is an integer. To minimize
, start from
, which gives
. This limits
to be greater than
, so test values of
starting from
. However,
to
do not give integer values of
.
Once , it is possible for
to be equal to
, so
could also be equal to
The next value,
, is not a solution, but
gives
. Thus, the smallest possible value of
is
, and the answer is
.
Solution 4
Graph the regions and
. Note that the lattice point (9,16) is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is
and the answer is
. //
Remark: This also gives an intuitive geometric proof of the mediant.
See Also
2018 AMC 12B (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.