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2013 AMC 12B Problems/Problem 17: Difference between revisions

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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>.
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>.
==Solution 4==
We take a geometrical approach.
From the given, we have <math> a + b = 2 - c</math> and <math>a^2 + b^2 = 12 - c^2 </math>. The first equation is a line with x and y intercepts of <math>2-c</math> and the second equation is a circle centered at the origin with radius <math>\sqrt{12-c^2}</math>. Intuitively, if we want to find the minimum / maximum <math>c</math> such that there still exist real solutions, the two graphs of the equations should be tangent.
Thus, we have that <math>\sqrt{2} \cdot \sqrt{12-c^2} = 2-c</math>, which simplifies to <math>3c^2-4c-20=0</math>. Solving the quadratic, we get that the values of <math>c</math> for which the two graphs are tangent are <math>c=-2</math> and <math>c=\frac{10}{3}</math>. Thus, our answer is <math>\boxed{\frac{16}{3}}</math>.


== See also ==
== See also ==

Revision as of 01:52, 14 January 2018

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 $\boxed{\frac{16}{3}}$.


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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