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

Solution 3

(no Cauchy-Schwarz)

From the first equation, we know that $c=2-a-b$ . We substitute this into the second equation to find that $a^2+b^2+(2-a-b)^2=12.$ This simplifies to $2a^2+2b^2-4a-4b+2ab=8$ , which we can write as the quadratic $a^2+(b-2)a+(b^2-2b-4)=0$ . We wish to find real values for $a$ and $b$ that satisfy this equation. Therefore, the discriminant is nonnegative. Hence, $(b-2)^2-4(b^2-2b-4)\ge0,$ or $-3b^2+4b+20\ge 0$ . This factors as $-(3b-10)(b+2)\ge 0$ . Therefore, $-2\le b\le \frac{10}{3}$ , and by symmetry this must be true for $a$ and $c$ as well.

Now $a=b=2$ and $c=-2$ satisfy both equations, so we see that $c=-2$ must be the minimum possible value of $c$ . Also, $c=\frac{10}{3}$ and $a=b=-\frac{2}{3}$ satisfy both equations, so we see that $c=\frac{10}{3}$ is the maximum possible value of $c$ . The difference between these is $\frac{10}{3}-(-2)=\frac{16}{3}$ , or $\boxed{\textbf{(D)}}$ .

Solution 4

We take a geometrical approach.

From the given, we have $a + b = 2 - c$ and $a^2 + b^2 = 12 - c^2$ . The first equation is a line with x and y intercepts of $2-c$ and the second equation is a circle centered at the origin with radius $\sqrt{12-c^2}$ . Intuitively, if we want to find the minimum / maximum $c$ such that there still exist real solutions, the two graphs of the equations should be tangent.

Thus, we have that $\sqrt{2} \cdot \sqrt{12-c^2} = 2-c$ , which simplifies to $3c^2-4c-20=0$ . Solving the quadratic, we get that the values of $c$ for which the two graphs are tangent are $c=-2$ and $c=\frac{10}{3}$ . Thus, our answer is $\boxed{\frac{16}{3}}$ .

Solution 5

Draw the sphere and the plane represented by the two equations in Cartesian space, with the $z$ -axis representing $c$ . The intersection between the sphere and plane are a circle. We wish to find the point on the circle where $z$ is minimized and the point where $z$ is maximized. Looking at the graph, it is clear by symmetry that $x = y$ when $z$ is maximized or minimized. Thus, we can set $a = b$ . This gives us the following system of equations:

$2a - c = 2$

$2a^2 - c^2 = 12$

Solving gives $c = \frac{10}{3}, -2$ , which are the maximum and minimum values of $c$ respectively. The answer follows from here.

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

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