# 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 $\boxed{\textbf{(D)} \ \frac{16}{3}}$.