2010 AMC 12B Problems/Problem 23

Problem

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 1

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

Solution 2 (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

Solution 3 (Mild Bash)

Let $P(x) = x^2 - (a+b)x + ab$ and $Q(x) = x^2 - (c+d)x + cd$ . Notice that the roots of $P(x)$ are $a,b$ and the roots of $Q(x)$ are $c,d.$ Then we get:

$\begin{align*} P(Q(x)) &= a, b \\ x^2 - (c+d)x + cd &= a, b \end{align*}$ The two possible equations are then $x^2 - (c+d)x + cd-a=0$ and $x^2 - (c+d)x + cd-b=0$ . The solutions are $-23, -21, -17, -15$ . From Vieta's we know that the total sum $2(c+d) = -76 \implies c+d = -38$ so the roots are paired $-23, -15$ and $-21, -17$ . Then, $cd - a = 23*15$ and $cd - b = 21*17$ .

We can similarly get that $ab - c = 59*49$ and $ab - d = 57*51$ , and $a+b = -108$ . Add the first two equations to get $2cd - (a+b) = 23*15 + 21*17 \implies cd = \frac{23*15+21*17 - 108}{2} = 297.$ This means $Q(x) = x^2 + 38x + 297$ .

Once more, we can similarly obtain $ab = \frac{59*49 + 57*51 - 38}{2} = 2880.$ Therefore $P(x) = x^2 + 108x + 2880$ .

Now we can find the minimums to be $19^2 - 19*38 + 297 = -64$ and $54^2 - 54*108 + 2880 = -36.$ Summing, the answer is $\boxed{\textbf{(A)} -100}.$

~Leonard_my_dude~

Solution 4

Let $P(x) = (x+a)(x+b)$ , $Q(x) = (x+c)(x+d)$ .

$P(Q(x)) = (x^2 + cx + dx + cd + a)(x^2 + cx + dx + cd + b)$

$Q(P(x)) = (x^2 + ax + bx + ab + c)(x^2 + ax + bx + ab + d)$

Notice how the coefficient for $x$ has to be the same for the two quadratics that are multiplied to create $P(Q(x))$ , and $Q(P(x))$ .

$P(Q(x)) = (x+ 23)(x+ 21)(x+ 17)(x+ 15) = (x^2 + 38x + 345)(x^2 + 38x + 357)$

$Q(P(x)) = (x+ 59)(x+ 57)(x+ 51)(x+ 49) = (x^2 + 108x + 2891)(x^2 + 108x + 2907)$

$c + d = 38$ , $cd + a = 345$ , $cd + b = 357$ , $a + b = 108$ , $ab + c = 2891$ , $ab + d = 2907$

$cd = \frac{345 + 357 - 108}{2} = 297$ , $297 = 3^3 \cdot 11$ , $c = 27$ , $d = 11$

$a = 345 - 27*11 = 48$ , $b = 357 - 27*11 = 60$

$P(x) = (x+48)(x+60) = x^2 + 108x + 2880 = (x+54)^2 - 36$

$Q(x) = (x+27)(x+11) = x^2 + 38x + 297 = (x+19)^2 -64$

$-36 -64 = \boxed{\textbf{(A) } -100}$ .

~isabelchen

Video Solution by MOP 2024

https://youtu.be/FPhaUQoRtPs

~r00tsOfUnity

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