Difference between revisions of "2018 AMC 12B Problems/Problem 17"
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<cmath>\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},</cmath> | <cmath>\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},</cmath> | ||
− | which reduces to <math>q\geq b+d</math>. We can easily find that <math>p=a+c</math>, giving an answer of <math>\boxed{\textbf{(A)}\ 7}</math>. | + | which reduces to <math>q\geq b+d</math>. We can easily find that <math>p=a+c</math>, giving an answer of <math>\boxed{\textbf{(A)}\ 7}</math>. |
==Solution 2 (requires justification)== | ==Solution 2 (requires justification)== |
Revision as of 00:54, 19 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 .
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 |
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