2010 AMC 12B Problems/Problem 23

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

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2010 AMC 12B Problems/Problem 23

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Problem 23

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