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

Revision as of 21:08, 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, $\boxed{\textbf{(D)} \ \frac{16}{3}}$ .

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

@@ Line 12: / Line 12: @@
 ==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}}

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

Revision as of 21:08, 23 February 2013

Problem

Solution

See also