Difference between revisions of "2003 AMC 12B Problems/Problem 24"
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# As we are looking for a <math>c<1002</math>, we have <math>b+c\leq 2001</math>, hence <math>2003+a-b-c > a</math>. To make sure that the value falls outside <math>\left(a,b\right]</math>, we need to make it larger than <math>b</math>, thus <math>2003+a-b-c > b</math>, or equivalently <math>2003+a > 2b+c</math>. | # As we are looking for a <math>c<1002</math>, we have <math>b+c\leq 2001</math>, hence <math>2003+a-b-c > a</math>. To make sure that the value falls outside <math>\left(a,b\right]</math>, we need to make it larger than <math>b</math>, thus <math>2003+a-b-c > b</math>, or equivalently <math>2003+a > 2b+c</math>. | ||
− | # The condition we just derived, <math>2003+a > 2b+c</math>, can be rewritten as <math>2003+a+b > 3b+c</math>, then as <math>2003+a+b-c > 3b</math>, which becomes <math>\frac{2003+a+b-c}3 > b</math>. | + | # The condition we just derived, <math>2003+a > 2b+c</math>, can be rewritten as <math>2003+a+b > 3b+c</math>, then as <math>2003+a+b-c > 3b</math>, which becomes <math>\frac{2003+a+b-c}3 > b</math>. Thus to make sure that the second value falls outside <math>\left(b,c\right]</math>, we need to make it larger than <math>c</math>. The inequality <math>\frac{2003+a+b-c}3 > c</math> simplifies to <math>2003+a+b > 4c</math>. |
− | Thus to make sure that the second value falls outside <math>\left(b,c\right]</math>, we need to make it larger than <math>c</math>. | ||
− | The inequality <math>\frac{2003+a+b-c}3 > c</math> simplifies to <math>2003+a+b > 4c</math>. | ||
# To avoid the last solution, we must have <math>\frac{2003+a+b+c}5\leq c</math>, which simplifies to <math>2003+a+b \leq 4c</math>. | # To avoid the last solution, we must have <math>\frac{2003+a+b+c}5\leq c</math>, which simplifies to <math>2003+a+b \leq 4c</math>. | ||
Revision as of 00:08, 29 January 2009
Contents
Problem
Positive integers and are chosen so that , and the system of equations
has exactly one solution. What is the minimum value of ?
Solution
Step 1: Finding some promising bound
Does the system have a solution where ?
For such a solution we would have , hence , which solves to . If we want to avoid this solution, we need to have , hence , hence . In other words, if , there will always be one solution such that .
Step 2: Showing one solution
We will now find out whether there is a for which (and some ) the system has only one solution. We already know of one such solution, so we need to make sure that no other solution appears.
Obviously, there are three more theoretically possible solutions: one in , one in , and one in . The first case solves to , the second to , and the third to . We need to make sure that the following three conditions hold:
- .
Let and . We then have:
Hence for , and any valid the system has exactly one solution .
Step 3: Proving the optimality of our solution
We will now show that for the system always has a solution such that . This will mean that the system has at least two solutions, and thus the solution with is optimal.
- As we are looking for a , we have , hence . To make sure that the value falls outside , we need to make it larger than , thus , or equivalently .
- The condition we just derived, , can be rewritten as , then as , which becomes . Thus to make sure that the second value falls outside , we need to make it larger than . The inequality simplifies to .
- To avoid the last solution, we must have , which simplifies to .
The last two inequalities contradict each other, thus there are no that would satisfy both of them.
Conclusion
We just showed that whenever , the system has at least two different solutions: one with and one with .
We also showed that for there are some for which the system has exactly one solution.
Hence the optimal value of is .
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |