Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with ' == Problem 23 == Monic quadratic polynomial <math>P(x)</math> and <math>Q(x)</math> have the property that <math>P(Q(x))</math> has zeros at <math>x=-23, -21, -17,</math> and <m…')
 
m (→‎Solution 3 (Mild Bash))
 
(19 intermediate revisions by 12 users not shown)
Line 1: Line 1:
−
 
+
== Problem ==
−
== Problem 23 ==
 
 
Monic quadratic polynomial <math>P(x)</math> and <math>Q(x)</math> have the property that <math>P(Q(x))</math> has zeros at <math>x=-23, -21, -17,</math> and <math>-15</math>, and <math>Q(P(x))</math> has zeros at <math>x=-59,-57,-51</math> and <math>-49</math>. What is the sum of the minimum values of <math>P(x)</math> and <math>Q(x)</math>?  
 
Monic quadratic polynomial <math>P(x)</math> and <math>Q(x)</math> have the property that <math>P(Q(x))</math> has zeros at <math>x=-23, -21, -17,</math> and <math>-15</math>, and <math>Q(P(x))</math> has zeros at <math>x=-59,-57,-51</math> and <math>-49</math>. What is the sum of the minimum values of <math>P(x)</math> and <math>Q(x)</math>?  
  
 
<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 1==
 +
<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>.
 +
We now have <math> P(x) = (x + 54)^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>.
 +
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>.
 +
 +
 +
== Solution 2 (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
 +
 +
== Solution 3 (Mild Bash) ==
 +
Let <math>P(x) = x^2 - (a+b)x + ab</math> and <math>Q(x) = x^2 - (c+d)x + cd</math>. Notice that the roots of <math>P(x)</math> are <math>a,b</math> and the roots of <math>Q(x)</math> are <math>c,d.</math> Then we get:
 +
 +
<cmath>\begin{align*}
 +
P(Q(x)) &= a, b \\
 +
x^2 - (c+d)x + cd &= a, b
 +
\end{align*}</cmath>
 +
The two possible equations are then <math>x^2 - (c+d)x + cd-a=0</math> and <math>x^2 - (c+d)x + cd-b=0</math>. The solutions are <math>-23, -21, -17, -15</math>. From Vieta's we know that the total sum <math>2(c+d) = -76 \implies c+d = -38</math> so the roots are paired <math>-23, -15</math> and <math>-21, -17</math>. Then, <math>cd - a = 23*15</math> and <math>cd - b = 21*17</math>.
 +
 +
We can similarly get that <math>ab - c = 59*49</math> and <math>ab - d = 57*51</math>, and <math>a+b = -108</math>. Add the first two equations to get <cmath>2cd - (a+b) = 23*15 + 21*17 \implies cd = \frac{23*15+21*17 - 108}{2} = 297.</cmath> This means <math>Q(x) = x^2 + 38x + 297</math>.
 +
 +
Once more, we can similarly obtain <cmath>ab = \frac{59*49 + 57*51 - 38}{2} = 2880.</cmath> Therefore <math>P(x) = x^2 + 108x + 2880</math>.
 +
 +
Now we can find the minimums to be <cmath>19^2 - 19*38 + 297 = -64</cmath> and <cmath>54^2 - 54*108 + 2880 = -36.</cmath> Summing, the answer is <math>\boxed{\textbf{(A)} -100}.</math>
 +
 +
~Leonard_my_dude~
 +
 +
== Solution 4 ==
 +
 +
Let <math>P(x) = (x+a)(x+b)</math>, <math>Q(x) = (x+c)(x+d)</math>.
 +
 +
<math>P(Q(x)) = (x^2 + cx + dx + cd + a)(x^2 + cx + dx + cd + b)</math>
 +
 +
<math>Q(P(x)) = (x^2 + ax + bx + ab + c)(x^2 + ax + bx + ab + d)</math>
 +
 +
Notice how the coefficient for <math>x</math> has to be the same for the two quadratics that are multiplied to create <math>P(Q(x))</math>, and <math>Q(P(x))</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 + 2907)</math>
 +
 +
 +
<math>c + d = 38</math>, <math>cd + a = 345</math>, <math>cd + b = 357</math>, <math>a + b = 108</math>, <math>ab + c = 2891</math>, <math>ab + d = 2907</math>
 +
 +
<math>cd = \frac{345 + 357 - 108}{2} = 297</math>, <math>297 = 3^3 \cdot 11</math>, <math>c = 27</math>, <math>d = 11</math>
 +
 +
<math>a = 345 - 27*11 = 48</math>, <math>b = 357 - 27*11 = 60</math>
 +
 +
 +
<math>P(x) = (x+48)(x+60) = x^2 + 108x + 2880 = (x+54)^2 - 36</math>
 +
 +
<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]
 +
 +
==Video Solution by MOP 2024==
 +
https://youtu.be/FPhaUQoRtPs
 +
 +
~r00tsOfUnity
 +
 +
==See Also==
 +
{{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 16:29, 30 June 2024

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_23&oldid=221355"