Difference between revisions of "2010 AMC 12B Problems/Problem 23"
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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>. | 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>. | ||
− | == | + | |
+ | == 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 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}} | {{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:42, 10 May 2020
Contents
[hide]Problem 23
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
. 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 .
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
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 | |
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