Difference between revisions of "2014 AIME II Problems/Problem 5"
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− | ==Problem | + | ==Problem== |
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | ||
− | ==Solution== | + | ==Solution 1== |
+ | 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>-(r+s)</math>, and the remaining root of <math>q(x)</math> is <math>-(r+s+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-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 </cmath>from which <math>s = \tfrac 12 (5r+13)</math>. | ||
+ | |||
+ | ==Solution 2== | ||
Let <math>r</math>, <math>s</math>, and <math>-r-s</math> be the roots of <math>p(x)</math> (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for <math>s</math>. Also, | Let <math>r</math>, <math>s</math>, and <math>-r-s</math> be the roots of <math>p(x)</math> (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for <math>s</math>. Also, | ||
<cmath>q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0</cmath> | <cmath>q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0</cmath> | ||
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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: | ||
+ | <cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),</cmath> | ||
+ | which eventually simplifies to | ||
+ | <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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | The roots of <math>p(x)</math> are <math>r</math>, <math>s</math>, and <math>-r-s</math> since they sum to <math>0</math> by Vieta's Formula (coefficient of <math>x^2</math> term is <math>0</math>). | ||
+ | |||
+ | Similarly, the roots of <math>q(x)</math> are <math>r + 4</math>, <math>s - 3</math>, and <math>-r-s-1</math>, as they too sum to <math>0</math>. | ||
+ | |||
+ | Then: | ||
+ | |||
+ | <math>a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2</math> and <math>-b = rs(-r-s)</math> from <math>p(x)</math> and | ||
− | < | + | <math>a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2</math> and <math>-(b+240)=(r+4)(s-3)(-r-s-1)</math> from <math>q(x)</math>. |
− | + | From these equations, we can write that | |
− | < | + | <cmath>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a</cmath> |
− | + | and simplifying gives | |
+ | <cmath>2s-5r-13=0 \Rightarrow s = \frac{5r+13}{2}.</cmath> | ||
+ | |||
+ | |||
+ | |||
+ | 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 | ||
+ | <cmath>rs(r+s) = b</cmath> | ||
+ | <cmath>(r+4)(s-3)(r+s+1)=b + 240.</cmath> Subtracting the first equation from the second equation gives us <math>(r+4)(s-3)(r+s+1) - rs(r+s) = 240</math>. | ||
+ | |||
+ | Expanding, simplifying, substituting <math>s = \frac{5r+13}{2}</math>, and simplifying some more yields the simple quadratic <math>r^2 + 4r - 5 = 0</math>, so <math>r = -5, 1</math>. Then <math>s = -6, 9</math>. | ||
+ | |||
+ | Finally, we substitute back into <math>b=rs(r+s)</math> to get <math>b = (-5)(-6)(-5-6) = -330</math>, or <math>b = (1)(9)(1 + 9) = 90</math>. | ||
+ | |||
+ | The answer is <math>|-330|+|90| = \boxed{420}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | By Vieta's, we know that the sum of roots of <math>p(x)</math> is <math>0</math>. Therefore, | ||
+ | the roots of <math>p</math> are <math>r, s, -r-s</math>. By similar reasoning, the roots of <math>q(x)</math> | ||
+ | are <math>r + 4, s - 3, -r - s - 1</math>. Thus, <math>p(x) = (x - r)(x - s)(x + r + s)</math> | ||
+ | and <math>q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)</math>. | ||
− | <cmath>s = \frac{13 | + | Since <math>p(x)</math> and <math>q(x)</math> have the same coefficient for <math>x</math>, we can go ahead |
+ | and match those up to get | ||
+ | <cmath>\begin{align*} | ||
+ | rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\ | ||
+ | 0 &= -13 - 5r + 2s \\ | ||
+ | s &= \frac{5r + 13}{2} | ||
+ | \end{align*}</cmath> | ||
− | + | At this point, we can go ahead and compare the constant term in <math>p(x)</math> and | |
+ | <math>q(x)</math>. Doing so is certainly valid, but we can actually do this another way. Notice that <math>p(s) = 0</math>. Therefore, <math>q(s) = 240</math>. If we plug that into | ||
+ | our expression, we get that | ||
+ | <cmath>\begin{align*} | ||
+ | q(s) &= 3(s - r - 4)(r + 2s + 1) \\ | ||
+ | 240 &= 3(s - r - 4)(r + 2s + 1) \\ | ||
+ | 240 &= 3\left( \frac{3r + 5}{2} \right)(6r + 14) \\ | ||
+ | 80 &= (3r + 5)(3r + 7) \\ | ||
+ | 0 &= r^2 + 4r - 5 | ||
+ | \end{align*}</cmath> | ||
+ | This tells us that <math>(r, s) = (1, 9)</math> or <math>(-5, -6)</math>. Since <math>-b</math> is the product of the roots, we have that the two possibilities are <math>1 \cdot 9 \cdot (-10) = -90</math> | ||
+ | and <math>(-5)(-6)11 = 330</math>. Adding the absolute values of these gives us | ||
+ | <math>\boxed{420}</math>. | ||
− | |||
− | |||
== See also == | == See also == | ||
{{AIME box|year=2014|n=II|num-b=4|num-a=6}} | {{AIME box|year=2014|n=II|num-b=4|num-a=6}} |
Latest revision as of 22:07, 18 August 2024
Problem
Real numbers and
are roots of
, and
and
are roots of
. Find the sum of all possible values of
.
Solution 1
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
.
Solution 2
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 3
The roots of are
,
, and
since they sum to
by Vieta's Formula (coefficient 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 4
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 |
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