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

Revision as of 04:37, 8 November 2016

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 $\boxed{\textbf{(D)}}$ .

@@ Line 18: / Line 18: @@
 The difference between these two values is  <math>\boxed{\textbf{(D)} \ \frac{16}{3}}</math>.
+==Solution 3==
+(no Cauchy-Schwarz)
+From the first equation, we know that <math>c=2-a-b</math>. We substitute this into the second equation to find that
+<cmath>a^2+b^2+(2-a-b)^2=12.</cmath>
+This simplifies to <math>2a^2+2b^2-4a-4b+2ab=8</math>, which we can write as the quadratic <math>a^2+(b-2)a+(b^2-2b-4)=0</math>. We wish to find real values for <math>a</math> and <math>b</math> that satisfy this equation. Therefore, the discriminant is nonnegative. Hence,
+<cmath>(b-2)^2-4(b^2-2b-4)\ge0,</cmath>
+or <math>-3b^2+4b+20\ge 0</math>. This factors as <math>-(3b-10)(b+2)\ge 0</math>. Therefore, <math>-2\le b\le \frac{10}{3}</math>, and by symmetry this must be true for <math>a</math> and <math>c</math> as well.
+Now <math>a=b=2</math> and <math>c=-2</math> satisfy both equations, so we see that <math>c=-2</math> must be the minimum possible value of <math>c</math>. Also, <math>c=\frac{10}{3}</math> and <math>a=b=-\frac{2}{3}</math> satisfy both equations, so we see that <math>c=\frac{10}{3}</math> is the maximum possible value of <math>c</math>. The difference between these is <math>\frac{10}{3}-(-2)=\frac{16}{3}</math>, or <math>\boxed{\textbf{(D)}}</math>.
 == See also ==

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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