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Difference between revisions of "2010 AMC 12B Problems/Problem 23"

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<math>\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0</math>
 
<math>\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0</math>
 
==Solution==
 
==Solution==
 +
<math> P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d</math>. Notice that <math> P(x)</math> has roots <math> a\pm \sqrt {b}</math>, so that the roots of <math> P(Q(x))</math> are the roots of <math> Q(x) = a + \sqrt {b}, a - \sqrt {b}</math>. For each individual equation, the sum of the roots will be <math> 2c</math> (symmetry or Vieta's). Thus, we have <math> 4c = - 23 - 21 - 17 - 15</math>, or <math> c = - 19</math>. Doing something similar for <math> Q(P(x))</math> gives us <math> a = - 54</math>.
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We now have <math> P(x) = (x + 35)^2 - b, Q(x) = (x + 19)^2 - d</math>. Since <math> Q</math> is monic, the roots of <math> Q(x) = a + \sqrt {b}</math> are "farther" from the axis of symmetry than the roots of <math> Q(x) = a - \sqrt {b}</math>. Thus, we have <math> Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}</math>, or <math> 16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}</math>. Adding these gives us <math> 20 - 2d = - 108</math>, or <math> d = 64</math>. Plugging this into <math> 16 - d = - 54 + \sqrt {b}</math>, we get <math> b = 36</math>.
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The minimum value of <math> P(x)</math> is <math> - b</math>, and the minimum value of <math> Q(x)</math> is <math> - d</math>. Thus, our answer is <math> - (b + d) = - 100</math>, or answer <math> \boxed{\textbf{(A)}}</math>.
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==See Also==
 
==See Also==
 
{{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}}
 
{{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}}

Revision as of 18:19, 31 May 2011

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Problem 23

Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$, and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$. What is the sum of the minimum values of $P(x)$ and $Q(x)$?

$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$

Solution

$P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d$. Notice that $P(x)$ has roots $a\pm \sqrt {b}$, so that the roots of $P(Q(x))$ are the roots of $Q(x) = a + \sqrt {b}, a - \sqrt {b}$. For each individual equation, the sum of the roots will be $2c$ (symmetry or Vieta's). Thus, we have $4c = - 23 - 21 - 17 - 15$, or $c = - 19$. Doing something similar for $Q(P(x))$ gives us $a = - 54$. We now have $P(x) = (x + 35)^2 - b, Q(x) = (x + 19)^2 - d$. Since $Q$ is monic, the roots of $Q(x) = a + \sqrt {b}$ are "farther" from the axis of symmetry than the roots of $Q(x) = a - \sqrt {b}$. Thus, we have $Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}$, or $16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}$. Adding these gives us $20 - 2d = - 108$, or $d = 64$. Plugging this into $16 - d = - 54 + \sqrt {b}$, we get $b = 36$. The minimum value of $P(x)$ is $- b$, and the minimum value of $Q(x)$ is $- d$. Thus, our answer is $- (b + d) = - 100$, or answer $\boxed{\textbf{(A)}}$.

See Also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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