Difference between revisions of "2013 AMC 12B Problems/Problem 17"
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==Solution== | ==Solution== | ||
<math>a+b= 2-c</math>. Now, by C-S, we have that <math>(a^2+b^2) \ge \frac{(2-c)^2}{2}</math>. Therefore, we have that <math>\frac{(2-c)^2}{2}+c^2 \le 12</math>. We then find the roots of <math>c</math> that satisfy equality and find the difference of the roots. This gives the answer, <math>\boxed{\textbf{(D)}\frac{16}{3}}</math>. | <math>a+b= 2-c</math>. Now, by C-S, we have that <math>(a^2+b^2) \ge \frac{(2-c)^2}{2}</math>. Therefore, we have that <math>\frac{(2-c)^2}{2}+c^2 \le 12</math>. We then find the roots of <math>c</math> that satisfy equality and find the difference of the roots. This gives the answer, <math>\boxed{\textbf{(D)}\frac{16}{3}}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=B|num-b=16|num-a=18}} | ||
Revision as of 17:11, 22 February 2013
Problem
Let 
and 
be real numbers such that
and
What is the difference between the maximum and minimum possible values of ?
Solution
. Now, by C-S, 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, 
.
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 | |
