2013 AMC 12B Problems/Problem 17

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 1

$a+b= 2-c$ . Now, by Cauchy-Schwarz, 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, $\boxed{\textbf{(D)} \ \frac{16}{3}}$ .

Solution 2

This is similar to the first solution but is far more intuitive. From the given, we have $a + b = 2 - c \\a^2 + b^2 = 12 - c^2$ This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have $2\,(a^2 + b^2) \geq (a + b)^2$ Substitution of the above results and some algebra yields $3c^2 - 4c - 20 \leq 0$ This quadratic inequality is easily solved, and it is seen that equality holds for $c = -2$ and $c = \frac{10}{3}$ .

The difference between these two values is $\boxed{\textbf{(D)} \ \frac{16}{3}}$ .

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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2013 AMC 12B Problems/Problem 17

Contents

Problem

Solution 1

Solution 2

See also