Difference between revisions of "2013 AMC 12B Problems/Problem 17"
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− | This is similar to the first solution but is far more intuitive. From the given, we have <cmath> a + b = 2 - c</cmath> | + | This is similar to the first solution but is far more intuitive. From the given, we have <cmath> a + b = 2 - c</cmath> <cmath>a^2 + b^2 = 12 - c^2 </cmath> This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have <cmath> 2\,(a^2 + b^2) \geq (a + b)^2</cmath> Substitution of the above results and some algebra yields <cmath> 3c^2 - 4c - 20 \leq 0 </cmath> This quadratic inequality is easily solved, and it is seen that equality holds for <math>c = -2</math> and <math>c = \frac{10}{3}</math>. |
The difference between these two values is <math>\boxed{\textbf{(D)} \ \frac{16}{3}}</math>. | The difference between these two values is <math>\boxed{\textbf{(D)} \ \frac{16}{3}}</math>. |
Revision as of 17:30, 29 January 2016
Contents
Problem
Let and be real numbers such that
What is the difference between the maximum and minimum possible values of ?
Solution 1
. Now, by Cauchy-Schwarz, we have that . Therefore, we have that . We then find the roots of that satisfy equality and find the difference of the roots. This gives the answer, .
Solution 2
This is similar to the first solution but is far more intuitive. From the given, we have This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have Substitution of the above results and some algebra yields This quadratic inequality is easily solved, and it is seen that equality holds for and .
The difference between these two values is .
See also
2013 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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