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…') |
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− | + | == Problem == | |
− | == Problem | ||
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 17:29, 30 June 2024
Contents
Problem
Monic quadratic polynomial and
have the property that
has zeros at
and
, and
has zeros at
and
. What is the sum of the minimum values of
and
?
Solution 1
. Notice that
has roots
, so that the roots of
are the roots of
. For each individual equation, the sum of the roots will be
(symmetry or Vieta's). Thus, we have
, or
. Doing something similar for
gives us
.
We now have
. Since
is monic, the roots of
are "farther" from the axis of symmetry than the roots of
. Thus, we have
, or
. Adding these gives us
, or
. Plugging this into
, we get
.
The minimum value of
is
, and the minimum value of
is
. Thus, our answer is
, or answer
.
Solution 2 (Bash)
Let and
.
Then is
, which simplifies to:
We can find by simply doing
and
to get:
The sum of the zeros of is
. From Vieta, the sum is
. Therefore,
.
The sum of the zeros of is
. From Vieta, the sum is
. Therefore,
.
Plugging in, we get:
Let's tackle the coefficients, which is the sum of the six double-products possible. Since
gives the sum of these six double products of the roots of
, we have:
Similarly with , we get:
Thus, our polynomials are and
.
The minimum value of happens at
, and is
.
The minimum value of happens at
, and is
.
The sum of these minimums is . -srisainandan6
Solution 3 (Mild Bash)
Let and
. Notice that the roots of
are
and the roots of
are
Then we get:
The two possible equations are then
and
. The solutions are
. From Vieta's we know that the total sum
so the roots are paired
and
. Then,
and
.
We can similarly get that and
, and
. Add the first two equations to get
This means
.
Once more, we can similarly obtain Therefore
.
Now we can find the minimums to be and
Summing, the answer is
~Leonard_my_dude~
Solution 4
Let ,
.
Notice how the coefficient for has to be the same for the two quadratics that are multiplied to create
, and
.
,
,
,
,
,
,
,
,
,
.
Video Solution by MOP 2024
~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.