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

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m (Undo revision 122235 by Srisainandan6 (talk))
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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>.
 
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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== Bashy Solution==
+
 
 +
== Bash ==
 +
Let <math>P(x) = x^2 + Bx + C</math> and <math>Q(x) = x^2 + Ex + F</math>.
 +
 
 +
Then <math>P(Q(x))</math> is <math>(x^2 + Ex + F)^2 + B(x^2 + Ex + F) + C</math>, which simplifies to:
 +
 
 +
<math>P(Q(x)) = x^4 + 2Ex^3 + (E^2 + 2F + B)x^2 + (2EF + BE)x + (F^2 + BF + C)</math>
 +
 
 +
We can find <math>Q(P(x))</math> by simply doing <math>B\Leftrightarrow E</math> and <math>C \Leftrightarrow F</math> to get:
 +
 
 +
<math>Q(P(x)) = x^4 + 2Bx^3 + (B^2 + 2C + E)x^2 + (2BC + BE)x + (C^2 + EC + F)</math>
 +
 
 +
The sum of the zeros of <math>P(Q(x))</math> is <math>-76</math>. From Vieta, the sum is <math>-2E</math>. Therefore, <math>E = 38</math>.
 +
 
 +
The sum of the zeros of <math>Q(P(x))</math> is <math>-216</math>. From Vieta, the sum is <math>-2B</math>. Therefore, <math>B = 108</math>.
 +
 
 +
Plugging in, we get:
 +
 
 +
<math>P(Q(x)) = x^4 + 76x^3 + (1552 + 2F)x^2 + (76F + 4104)x + (F^2 + 108F + C)</math>
 +
<math>Q(P(x)) = x^4 + 216x^3 + (11702 + 2C)x^2 + (216C + 4104)x + (C^2 + 38C + F)</math>
 +
 
 +
Let's tackle the <math>x^2</math> coefficients, which is the sum of the six double-products possible. Since <math>23 \cdot (21 + 17 + 15) + 21 \cdot (17 + 15) + 17 \cdot 15</math> gives the sum of these six double products of the roots of <math>P(Q(x))</math>, we have:
 +
 
 +
<math>1552 + 2F = 23 \cdot (21 + 17 + 15) + 21 \cdot (17 + 15) + 17 \cdot 15</math>
 +
 
 +
<math>1552 + 2F = 2146</math>
 +
 
 +
<math>F = 297</math>
 +
 
 +
Similarly with <math>Q(P(x))</math>, we get:
 +
 
 +
<math>11702 + 2C = 59(57 + 51 + 49) + 57(51 + 49) + 51(49)</math>
 +
 
 +
<math>11702 + 2C = 17462</math>
 +
 
 +
<math>C = 2880</math>
 +
 
 +
Thus, our polynomials are <math>P(x) = x^2 + 108x + 2880</math> and <math>Q(x) = x^2 + 38x + 297</math>.
 +
 
 +
The minimum value of <math>P(x)</math> happens at <math>x = -\frac{108}{2} = -54</math>, and is <math>54^2 - 108 \cdot 54 + 2880 = 2880 - 54^2</math>.
 +
 
 +
The minimum value of <math>Q(x)</math> happens at <math>x = -\frac{38}{2} = -19</math>, and is <math>19^2 - 38 \cdot 19 + 297 = 297 - 19^2</math>.
 +
 
 +
The sum of these minimums is <math>2880 +297 - 54^2 - 19^2 = \boxed{-100}</math>. -srisainandan6
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:42, 10 May 2020

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 $P(x) = (x + 54)^2 - b, Q(x) = (x + 19)^2 - d$. 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 $\boxed{\textbf{(A)}}$.


Bash

Let $P(x) = x^2 + Bx + C$ and $Q(x) = x^2 + Ex + F$.

Then $P(Q(x))$ is $(x^2 + Ex + F)^2 + B(x^2 + Ex + F) + C$, which simplifies to:

$P(Q(x)) = x^4 + 2Ex^3 + (E^2 + 2F + B)x^2 + (2EF + BE)x + (F^2 + BF + C)$

We can find $Q(P(x))$ by simply doing $B\Leftrightarrow E$ and $C \Leftrightarrow F$ to get:

$Q(P(x)) = x^4 + 2Bx^3 + (B^2 + 2C + E)x^2 + (2BC + BE)x + (C^2 + EC + F)$

The sum of the zeros of $P(Q(x))$ is $-76$. From Vieta, the sum is $-2E$. Therefore, $E = 38$.

The sum of the zeros of $Q(P(x))$ is $-216$. From Vieta, the sum is $-2B$. Therefore, $B = 108$.

Plugging in, we get:

$P(Q(x)) = x^4 + 76x^3 + (1552 + 2F)x^2 + (76F + 4104)x + (F^2 + 108F + C)$ $Q(P(x)) = x^4 + 216x^3 + (11702 + 2C)x^2 + (216C + 4104)x + (C^2 + 38C + F)$

Let's tackle the $x^2$ coefficients, which is the sum of the six double-products possible. Since $23 \cdot (21 + 17 + 15) + 21 \cdot (17 + 15) + 17 \cdot 15$ gives the sum of these six double products of the roots of $P(Q(x))$, we have:

$1552 + 2F = 23 \cdot (21 + 17 + 15) + 21 \cdot (17 + 15) + 17 \cdot 15$

$1552 + 2F = 2146$

$F = 297$

Similarly with $Q(P(x))$, we get:

$11702 + 2C = 59(57 + 51 + 49) + 57(51 + 49) + 51(49)$

$11702 + 2C = 17462$

$C = 2880$

Thus, our polynomials are $P(x) = x^2 + 108x + 2880$ and $Q(x) = x^2 + 38x + 297$.

The minimum value of $P(x)$ happens at $x = -\frac{108}{2} = -54$, and is $54^2 - 108 \cdot 54 + 2880 = 2880 - 54^2$.

The minimum value of $Q(x)$ happens at $x = -\frac{38}{2} = -19$, and is $19^2 - 38 \cdot 19 + 297 = 297 - 19^2$.

The sum of these minimums is $2880 +297 - 54^2 - 19^2 = \boxed{-100}$. -srisainandan6

See Also

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

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