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Difference between revisions of "2010 AMC 12B Problems/Problem 23"
(→‎Solution 4)
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== Solution 4 ==
 
== Solution 4 ==
  
−
Let <math>P(x) = (x+a)(x+b)</math>, and Q(x) = (x+c)(x+d)
+
Let <math>P(x) = (x+a)(x+b)</math>, and <math>Q(x) = (x+c)(x+d)</math>, WLOG, <math>a > b </math><math>c > d</math>,
  
 
<math>P(Q(x)) = (x^2 + cx + dx + cd + a)(x^2 + cx + dx + cd + b)</math>
 
<math>P(Q(x)) = (x^2 + cx + dx + cd + a)(x^2 + cx + dx + cd + b)</math>
Line 81: Line 81:
 
Notice how the coefficient for <math>x</math> has to be the same for the two quadratics that are multiplied together above.
 
Notice how the coefficient for <math>x</math> has to be the same for the two quadratics that are multiplied together above.
  
−
<math>P(Q(x)) = (x^+ 23)(x^+ 21)(x^+ 17)(x^+ 15) = (x^2 + 38x + 345)(x^2 + 38x + 357)</math>
+
<math>P(Q(x)) = (x+ 23)(x+ 21)(x+ 17)(x+ 15) = (x^2 + 38x + 345)(x^2 + 38x + 357)</math>
  
−
<math>Q(P(x)) = (x^+ 59)(x^+ 57)(x^+ 51)(x^+ 49) = (x^2 + 108x + 2891)(x^2 + 108x + 2897)</math>
+
<math>Q(P(x)) = (x+ 59)(x+ 57)(x+ 51)(x+ 49) = (x^2 + 108x + 2891)(x^2 + 108x + 2897)</math>
  
 +
<math>c + d = 38</math>, <math>cd + a = 345</math>, <math>cd + 5 = 357</math>, <math>a + b = 108</math>, <math>ab + c = 2891</math>, <math>ab + d = 2907</math>
  
 +
By solving the above equations <math>a = 60</math>, <math>b = 48</math>, <math>c = 27</math> and <math>d = 11</math>
 +
 +
<math>P(x) = (x+60)(x+48) = x^2 + 108x + 2880 = (x+54)^2 - 36</math>, and <math>Q(x) = (x+27)(x+11) = x^2 + 38x + 297 = (x+19)^2 -64</math>
 +
 +
<math>-36 -64 = \boxed{\textbf{(A) } -100}</math>.
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==See Also==
 
==See Also==

Revision as of 04:45, 24 December 2022

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 $-21, -17$. Let $cd - a = 23*15$ 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 $P(x) = (x+a)(x+b)$, and $Q(x) = (x+c)(x+d)$, WLOG, $a > b$$c > 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 together above.

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

$Q(P(x)) = (x+ 59)(x+ 57)(x+ 51)(x+ 49) = (x^2 + 108x + 2891)(x^2 + 108x + 2897)$

$c + d = 38$, $cd + a = 345$, $cd + 5 = 357$, $a + b = 108$, $ab + c = 2891$, $ab + d = 2907$

By solving the above equations $a = 60$, $b = 48$, $c = 27$ and $d = 11$

$P(x) = (x+60)(x+48) = x^2 + 108x + 2880 = (x+54)^2 - 36$, and $Q(x) = (x+27)(x+11) = x^2 + 38x + 297 = (x+19)^2 -64$

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

~isabelchen

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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