Difference between revisions of "2021 AIME II Problems/Problem 4"

Revision as of 17:01, 23 March 2021

Problem

There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$

Solution 1 (Complex Conjugate Root Theorem)

By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are a pair of complex conjugates. Let $z=m+\sqrt{n}\cdot i$ and $\overline{z}=m-\sqrt{n}\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\overline{z}.$

By Vieta's Formulas on $x^3+ax+b,$ we have $-20+z+\overline{z}=0,$ from which $\begin{align*} z+\overline{z}&=20 \ \ \ \ \ \ \ \ \ (1) \\ \left(m+\sqrt{n}\cdot i\right)+\left(m+\sqrt{n}\cdot i\right)&=20 \\ 2m&=20 \\ m&=10. \ \ \ \ \ \ \ \ (2) \end{align*}$

By Vieta's Formulas on $x^3+cx^2+d,$ we have $-21z-21\overline{z}+z\overline{z}=0,$ from which $\begin{align*} -21\left(z+\overline{z}\right)+z\overline{z}&=0 \\ -21\underbrace{\left(20\right)}_{\text{by (1)}}+\underbrace{|z|^2}_{\substack{z\overline{z}=|z|^2 \\ \text{ Identity}}}&=0 \\ |z|^2&=420 \\ m^2+n&=420 \\ {\underbrace{10}_{\text{by (2)}}}^2+n&=420 \\ n&=320. \hspace{24.5mm} (3) \end{align*}$ Finally, we get $m+n=\boxed{330}$ by $(2)$ and $(3).$

~MRENTHUSIASM

Solution 2 (Somewhat Bashy)

$(-20)^{3} + (-20)a + b = 0$ , hence $-20a + b = 8000$

Also, $(-21)^{3} + c(-21)^{2} + d = 0$ , hence $441c + d = 9261$

$m + i \sqrt{n}$ satisfies both $\Rightarrow$ we can put it in both equations and equate to 0.

In the first equation, we get $(m + i \sqrt{n})^{3} + a(m + i \sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n}) = 0$

Hence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0 \rightarrow (1)$

In the second equation, we get $(m + i \sqrt{n})^{3} + c(m + i \sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n}) = 0$

Hence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 \rightarrow (2)$

Comparing (1) and (2),

$a = 2mc$ and $am + b = m^{2}c - nc + d \rightarrow (3)$

$b = 8000 + 20a \Rightarrow b = 40mc + 8000$ ; $d = 9261 - 441c$

Substituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$

This simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261$

Hence, $c|1261 \Rightarrow c \in {1,13,97,1261}$

Consider case of $c = 1$ :

$c = 1 \Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$

$am + b = m^{2} - n + 8820$ (because c = 1) Also, $m^{2} + n + 40m = 820 \rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \rightarrow (5)$

Solving (4) and (5) simultaneously gives $m = 10, n = 320$

[AIME can not have more than one answer, so we can stop here also 😁... Not suitable for Subjective exam]

Hence, $m + n = 10 + 320 = \boxed{330}$

-Arnav Nigam

Solution 3

start off by applying vieta's and you will find that $a=m^2+n-40m$ $b=20m^2+20n$ $c=21-2m$ and $d=21m^2+21n$ . After that, we have to use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$ , respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore $(-20)^3-20(a)+b=0$ and $(-21)^3+c*(-21)^2+d=0$ and you can set these two equations equal to each other while also substituting the values of $a$ , $b$ , $c$ , and $d$ above to give you $21m^2+21n-1682m+8000=0$ , then you can rearrange the equation into $21n = -21m^2+1682m-8000$ . With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$ therefore that means $1682m-8000=0(mod 21)$ which results in $2m-20=0(mod 21)$ which finally gives us m=10 mod 21. We can test the first obvious value of $m$ which is $10$ and we see that this works as we get $m=10$ and $n=320$ . That means your answer will be $m + n = 10 + 320 = \boxed{330}$

-Jske25

@@ Line 76: / Line 76: @@
 -Arnav Nigam
+==Solution 3==
+start off by applying vieta's and you will find that <math>a=m^2+n-40m</math> <math>b=20m^2+20n</math> <math>c=21-2m</math> and <math>d=21m^2+21n</math>. After that, we have to use the fact that <math>-20</math> and <math>-21</math> are roots of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math>, respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore <math>(-20)^3-20(a)+b=0</math> and <math>(-21)^3+c*(-21)^2+d=0</math> and you can set these two equations equal to each other while also substituting the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> above to give you <math>21m^2+21n-1682m+8000=0</math>, then you can rearrange the equation into <math>21n = -21m^2+1682m-8000</math>. With this property, we know that <math>-21m^2+1682m-8000</math> is divisible by <math>21</math> therefore that means <math>1682m-8000=0(mod 21)</math> which results in <math>2m-20=0(mod 21)</math> which finally gives us m=10 mod 21. We can test the first obvious value of <math>m</math> which is <math>10</math> and we see that this works  as we get <math>m=10</math> and <math>n=320</math>. That means your answer will be <math>m + n = 10 + 320 = \boxed{330}</math>
+-Jske25
 ==See also==
 {{AIME box|year=2021|n=II|num-b=3|num-a=5}}
 {{MAA Notice}}

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