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>.
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<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 ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2013|ab=B|num-b=16|num-a=18}}

Revision as of 22:08, 23 February 2013

Problem

Let $a,b,$ and $c$ be real numbers such that

\[a+b+c=2, \text{ and}\] \[a^2+b^2+c^2=12\]

What is the difference between the maximum and minimum possible values of $c$?

$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$

Solution

$a+b= 2-c$. Now, by C-S, we have that $(a^2+b^2) \ge \frac{(2-c)^2}{2}$. Therefore, we have that $\frac{(2-c)^2}{2}+c^2 \le 12$. We then find the roots of $c$ that satisfy equality and find the difference of the roots. This gives the answer, $\boxed{\textbf{(D)} \ \frac{16}{3}}$.

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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