Difference between revisions of "2014 AIME II Problems/Problem 5"
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==Hint== | ==Hint== | ||
<cmath>\color{red}\boxed{\boxed{\color{blue}\textbf{Use the Vieta Formula!}}}</cmath> | <cmath>\color{red}\boxed{\boxed{\color{blue}\textbf{Use the Vieta Formula!}}}</cmath> | ||
+ | |||
+ | ==Official Solution (MAA)== | ||
+ | Because the coefficient of <math>x^2</math> in both <math>p(x)</math> and <math>q(x)</math> is 0, the remaining root of <math>p(x)</math> is <math>t = -r - s</math>, and the remaining root of <math>q(x)</math> is <math>t - 1</math>. The coefficients of <math>x</math> in <math>p(x)</math> and <math>q(x)</math> are both equal to <math>a</math>, and equating the two coefficients gives <cmath>rs+st+tr = (r-4)(s-3)+(r-4)(t-1)+(s-3)(t-1)</cmath>from which <math>t = 4r - 3s + 13</math>. Furthermore, <math>b = -rst</math>, so <cmath>b + 240 = -rst + 240 = -(r + 4)(s - 3)(t - 1),</cmath>from which <math>rs-4st+ 3tr-3r+ 4s+ 12t-252 = 0</math>. Substituting <math>t = 4r-3s+ 13</math> gives <cmath>12r^2 - 24rs + 12s^2 + 84r - 84s - 96 = 0,</cmath>which is equivalent to <math>(r - s)^2 + 7(r - s) - 8 = 0</math>, and the solutions for <math>r - s</math> are 1 and <math>-8</math>. | ||
+ | |||
+ | If <math>r - s = 1</math>, then the roots of <math>p(x)</math> are <math>r</math>, <math>s = r - 1</math>, and <math>t = 4r - 3s + 13 = r + 16</math>. Because the sum of the roots is 0, <math>r = -5</math>. In this case the roots are <math>-5</math>, <math>-6</math>, and <math>11</math>, and <math>b = -rst = -330</math>. | ||
+ | |||
+ | If <math>r - s = -8</math>, then the roots of <math>p(x)</math> are <math>r</math>, <math>s = r + 8</math>, and <math>t = 4r - 3s + 13 = r - 11</math>. In this case the roots are <math>1</math>, <math>9</math>, and <math>-10</math>, and <math>b = -rst = 90</math>. | ||
+ | |||
+ | Therefore the requested sum is <math>|-330| + |90| = \boxed{420}</math>. | ||
==Solution 1== | ==Solution 1== | ||
Line 14: | Line 23: | ||
Simplifying and adding the equations gives | Simplifying and adding the equations gives | ||
− | <cmath> | + | <cmath>\begin{align}\tag{*} |
− | + | r^2 - s^2 + 4r + 3s + 49 &= 0 | |
− | + | \end{align}</cmath> | |
− | + | Now, let's deal with the <math>ax</math> terms. Plugging the roots <math>r</math>, <math>s</math>, and <math>-r-s</math> into <math>p(x)</math> yields a long polynomial, and plugging the roots <math>r+4</math>, <math>s-3</math>, and <math>-1-r-s</math> into <math>q(x)</math> yields another long polynomial. Equating the coefficients of <math>x</math> in both polynomials, we get: | |
− | Now, let's deal with the <math>ax</math> terms. Plugging the roots <math>r</math>, <math>s</math>, and <math>-r-s</math> into <math>p(x)</math> yields a long polynomial, and plugging the roots <math>r+4</math>, <math>s-3</math>, and <math>-1-r-s</math> into <math>q(x)</math> yields another long polynomial. Equating the coefficients of x in both polynomials: | ||
<cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),</cmath> | <cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),</cmath> | ||
which eventually simplifies to | which eventually simplifies to | ||
− | |||
<cmath>s = \frac{13 + 5r}{2}.</cmath> | <cmath>s = \frac{13 + 5r}{2}.</cmath> | ||
− | |||
Substitution into (*) should give <math>r = -5</math> and <math>r = 1</math>, corresponding to <math>s = -6</math> and <math>s = 9</math>, and <math>|b| = 330, 90</math>, for an answer of <math>\boxed{420}</math>. | Substitution into (*) should give <math>r = -5</math> and <math>r = 1</math>, corresponding to <math>s = -6</math> and <math>s = 9</math>, and <math>|b| = 330, 90</math>, for an answer of <math>\boxed{420}</math>. | ||
Revision as of 10:37, 26 January 2022
Contents
Problem
Real numbers and
are roots of
, and
and
are roots of
. Find the sum of all possible values of
.
Hint
Official Solution (MAA)
Because the coefficient of in both
and
is 0, the remaining root of
is
, and the remaining root of
is
. The coefficients of
in
and
are both equal to
, and equating the two coefficients gives
from which
. Furthermore,
, so
from which
. Substituting
gives
which is equivalent to
, and the solutions for
are 1 and
.
If , then the roots of
are
,
, and
. Because the sum of the roots is 0,
. In this case the roots are
,
, and
, and
.
If , then the roots of
are
,
, and
. In this case the roots are
,
, and
, and
.
Therefore the requested sum is .
Solution 1
Let ,
, and
be the roots of
(per Vieta's). Then
and similarly for
. Also,
Set up a similar equation for :
Simplifying and adding the equations gives
Now, let's deal with the
terms. Plugging the roots
,
, and
into
yields a long polynomial, and plugging the roots
,
, and
into
yields another long polynomial. Equating the coefficients of
in both polynomials, we get:
which eventually simplifies to
Substitution into (*) should give
and
, corresponding to
and
, and
, for an answer of
.
Solution 2
The roots of are
,
, and
since they sum to
by Vieta's Formula (co-efficient of
term is
).
Similarly, the roots of are
,
, and
, as they too sum to
.
Then:
and
from
and
and
from
.
From these equations, we can write that
and simplifying gives
We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get
Subtracting the first equation from the second equation gives us
.
Expanding, simplifying, substituting , and simplifying some more yields the simple quadratic
, so
. Then
.
Finally, we substitute back into to get
, or
.
The answer is .
Solution 3
By Vieta's, we know that the sum of roots of is
. Therefore,
the roots of
are
. By similar reasoning, the roots of
are
. Thus,
and
.
Since and
have the same coefficient for
, we can go ahead
and match those up to get
At this point, we can go ahead and compare the constant term in and
. Doing so is certainly valid, but we can actually do this another way. Notice that
. Therefore,
. If we plug that into
our expression, we get that
This tells us that
or
. Since
is the product of the roots, we have that the two possibilities are
and
. Adding the absolute values of these gives us
.
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.