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

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

@@ Line 1: / Line 1: @@
-{{solution}}
+== 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>?
 <math>\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0</math>
-==Solution==
+==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}}

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

Page

Toolbox

Search