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Difference between revisions of "2013 AMC 12B Problems/Problem 17"

(Created page with "==Problem== Let <math>a,b,</math> and <math>c</math> be real numbers such that <math>a+b+c=2,</math> and <math> a^2+b^2+c^2=12 </math> What is the difference between the max...")
 
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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>
 +
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==Solution==
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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>.

Revision as of 15:47, 22 February 2013

Problem

Let $a,b,$ and $c$ be real numbers such that $a+b+c=2,$ 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, $\boxed{\textbf{(D)}\frac{16}{3}}$.

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