Difference between revisions of "2010 AMC 12B Problems/Problem 23"
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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> | 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> | ||
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+ | ~Leonard_my_dude~ | ||
==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 18:04, 15 December 2020
Contents
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
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.