Difference between revisions of "2013 AMC 12B Problems/Problem 17"
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<math> \text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3} </math> | <math> \text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3} </math> | ||
| − | ==Solution== | + | ==Solution 1== |
<math>a+b= 2-c</math>. Now, by Cauchy-Schwarz, 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 Cauchy-Schwarz, 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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | This is similar to the first solution but is far more intuitive. From the given, we have <cmath> a + b = 2 - c \\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>. | ||
== See also == | == See also == | ||
Revision as of 20:05, 24 January 2014
Contents
Problem
Let 
and 
be real numbers such that
![\[a+b+c=2, \text{ and}\]](http://latex.artofproblemsolving.com/0/e/4/0e45d0bbc5f0b07071a5820e33584f5e2a829350.png)
![\[a^2+b^2+c^2=12\]](http://latex.artofproblemsolving.com/0/9/2/0920fd752e68f85248ce35b7acebfdaecf68ac0a.png)
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 ![\[a + b = 2 - c \\a^2 + b^2 = 12 - c^2\]](http://latex.artofproblemsolving.com/b/2/b/b2bdba1c73e7683bf7c07023d376c40ae416992d.png)
This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have ![\[2\,(a^2 + b^2) \geq (a + b)^2\]](http://latex.artofproblemsolving.com/0/6/6/066668e2daf96b383b80abc2b108d69081b6745b.png)
Substitution of the above results and some algebra yields ![\[3c^2 - 4c - 20 \leq 0\]](http://latex.artofproblemsolving.com/8/5/c/85cc6e74d1a62145c732fc51d42e220167bae028.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
