Difference between revisions of "2010 AMC 12B Problems/Problem 23"
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<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== | + | |
+ | ==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>. | <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>. | 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>. | ||
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− | == Bash == | + | == Solution 2 (Bash) == |
Let <math>P(x) = x^2 + Bx + C</math> and <math>Q(x) = x^2 + Ex + F</math>. | Let <math>P(x) = x^2 + Bx + C</math> and <math>Q(x) = x^2 + Ex + F</math>. | ||
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The sum of these minimums is <math>2880 +297 - 54^2 - 19^2 = \boxed{-100}</math>. -srisainandan6 | The sum of these minimums is <math>2880 +297 - 54^2 - 19^2 = \boxed{-100}</math>. -srisainandan6 | ||
− | == Mild Bash == | + | == 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: | 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: | ||
Revision as of 23:22, 23 December 2020
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 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
. Let
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~
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.